Solution

The given function is $f(x)=5 x-3$

At $x=0, f(0)=5 \times 0-3=3$

$\lim _{x \to 0} f(x)=\lim _{x \to 0}(5 x-3)=5 \times 0-3=-3$

$\therefore \lim _{x \to 0} f(x)=f(0)$

Therefore, $f$ is continuous at $x=0$

At $x=-3, f(-3)=5 \times(-3)-3=-18$

$\lim _{x \to-3} f(x)=\lim _{x \to-3}(5 x-3)=5 \times(-3)-3=-18$

$\therefore \lim _{x \to-3} f(x)=f(-3)$

Therefore, $f$ is continuous at $x=-3$

At $x=5, f(x)=f(5)=5 \times 5-3=25-3=22$

$\lim _{x \to 5} f(x)=\lim _{x \to 5}(5 x-3)=5 \times 5-3=22$

$\therefore \lim _{x \to 5} f(x)=f(5)$

Therefore, $f$ is continuous at $x=5$

Solution

The given function is $f(x)=2 x^{2}-1$

At $x=3, f(x)=f(3)=2 \times 3^{2}-1=17$

$\lim _{x \to 3} f(x)=\lim _{x \to 3}(2 x^{2}-1)=2 \times 3^{2}-1=17$

$\therefore \lim _{x \to 3} f(x)=f(3)$

Thus, $f$ is continuous at $x=3$

Solution

(a) The given function is $f(x)=x-5$

It is evident that $f$ is defined at every real number $k$ and its value at $k$ is $k-5$.

It is also observed that, $\lim _{x \to k} f(x)=\lim _{x \to k}(x-5)=k-5=f(k)$

$\therefore \lim _{x \to k} f(x)=f(k)$

Hence, $f$ is continuous at every real number and therefore, it is a continuous function.

(b) The given function is $f(x)=\frac{1}{x-5}, x \neq 5$

For any real number $k \neq 5$, we obtain

$\lim _{x \to k} f(x)=\lim _{x \to k} \frac{1}{x-5}=\frac{1}{k-5}$

Also, $f(k)=\frac{1}{k-5} \quad($ As $k \neq 5)$

$\therefore \lim _{x \to k} f(x)=f(k)$

Hence, $f$ is continuous at every point in the domain of $f$ and therefore, it is a continuous function.

(c) The given function is $f(x)=\frac{x^{2}-25}{x+5}, x \neq-5$

For any real number $c \neq-5$, we obtain

$ \begin{aligned} & \lim _{x \to c} f(x)=\lim _{x \to c} \frac{x^{2}-25}{x+5}=\lim _{x \to c} \frac{(x+5)(x-5)}{x+5}=\lim _{x \to c}(x-5)=(c-5) \\ & \text{ Also, } f(c)=\frac{(c+5)(c-5)}{c+5}=(c-5) \quad(\text{ as } c \neq-5) \\ & \therefore \lim _{x \to c} f(x)=f(c) \end{aligned} $

Hence, $f$ is continuous at every point in the domain of $f$ and therefore, it is a continuous function.

(d) The given function is $f(x)=|x-5|=\begin{cases} 5-x, \text{ if } x<5 \\ x-5, \text{ if } x \geq 5 \end{cases} .$

This function $f$ is defined at all points of the real line.

Let $c$ be a point on a real line. Then, $c<5$ or $c=5$ or $c>5$

Case I: $c<5$

Then, $f(c)=5-c$

$\lim _{x \to c} f(x)=\lim _{x \to c}(5-x)=5-c$

$\therefore \lim _{x \to c} f(x)=f(c)$

Therefore, $f$ is continuous at all real numbers less than 5 .

Case II : $c=5$

Then, $f(c)=f(5)=(5-5)=0$

$\lim _{x \to 5^{-}} f(x)=\lim _{x \to 5}(5-x)=(5-5)=0$

$\lim _{x \to 5^{+}} f(x)=\lim _{x \to 5}(x-5)=0$

$\therefore \lim _{x \to c^{-}} f(x)=\lim _{x \to c^{+}} f(x)=f(c)$

Therefore, $f$ is continuous at $x=5$

Case III: $c>5$

Then, $f(c)=f(5)=c-5$

$\lim _{x \to c} f(x)=\lim _{x \to c}(x-5)=c-5$

$\therefore \lim _{x \to c} f(x)=f(c)$

Therefore, $f$ is continuous at all real numbers greater than 5 .

Hence, $f$ is continuous at every real number and therefore, it is a continuous function.

Solution

The given function is $f(x)=x^{n}$

It is evident that $f$ is defined at all positive integers, $n$, and its value at $n$ is $n^{n}$.

Then, $\lim _{x \to n} f(n)=\lim _{x \to n}(x^{n})=n^{n}$

$\therefore \lim _{x \to n} f(x)=f(n)$

Therefore, $f$ is continuous at $n$, where $n$ is a positive integer.

Solution

The given function $f$ is $f(x)= \begin{cases}x, & \text{ if } x \leq 1 \\ 5, & \text{ if } x>1\end{cases}$

At $x=0$,

It is evident that $f$ is defined at 0 and its value at 0 is 0 .

Then, $\lim _{x \to 0} f(x)=\lim _{x \to 0} x=0$

$\therefore \lim _{x \to 0} f(x)=f(0)$

Therefore, $f$ is continuous at $x=0$

At $x=1$,

$f$ is defined at 1 and its value at 1 is 1 .

The left hand limit of $f$ at $x=1$ is,

$\lim _{x \to 1^{-}} f(x)=\lim _{x \to 1^{-}} x=1$

The right hand limit of $f$ at $x=1$ is,

$ \begin{aligned} & \lim _{x \to 1^{+}} f(x)=\lim _{x \to 1^{+}}(5)=5 \\ & \therefore \lim _{x \to 1^{-}} f(x) \neq \lim _{x \to 1^{+}} f(x) \end{aligned} $

Therefore, $f$ is not continuous at $x=1$

At $x=2$,

$f$ is defined at 2 and its value at 2 is 5 .

Then, $\lim _{x \to 2} f(x)=\lim _{x \to 2}(5)=5$

$\therefore \lim _{x \to 2} f(x)=f(2)$

Therefore, $f$ is continuous at $x=2$

Find all points of discontinuity of $f$, where $f$ is defined by

Solution

$ f(x)=\begin{cases} 2 x+3, \text{ if } x \leq 2 \\ 2 x-3, \text{ if } x>2 \end{cases} . $

It is evident that the given function $f$ is defined at all the points of the real line.

Let $c$ be a point on the real line. Then, three cases arise.

(i) $c<2$

(ii) $c>2$

(iii) $c=2$

Case (i) $c<2$

Then, $f(c)=2 c+3$

$\lim _{x \to c} f(x)=\lim _{x \to c}(2 x+3)=2 c+3$

$\therefore \lim _{x \to c} f(x)=f(c)$

Therefore, $f$ is continuous at all points $x$, such that $x<2$

Case (ii) $c>2$

Then, $f(c)=2 c-3$

$\lim _{x \to c} f(x)=\lim _{x \to c}(2 x-3)=2 c-3$

$\therefore \lim _{x \to c} f(x)=f(c)$

Therefore, $f$ is continuous at all points $x$, such that $x>2$

Case (iii) $c=2$

Then, the left hand limit of $f$ at $x=2$ is,

$\lim _{x \to 2^{-}} f(x)=\lim _{x \to 2^{-}}(2 x+3)=2 \times 2+3=7$

The right hand limit of $f$ at $x=2$ is,

$\lim _{x \to 2^{+}} f(x)=\lim _{x \to 2^{+}}(2 x-3)=2 \times 2-3=1$

It is observed that the left and right hand limit of $f$ at $x=2$ do not coincide.

Therefore, $f$ is not continuous at $x=2$

Hence, $x=2$ is the only point of discontinuity of $f$.

Solution

$ f(x)=\begin{cases} |x|+3=-x+3, \text{ if } x \leq-3 \\ -2 x, \text{ if }-3<x<3 \\ 6 x+2, \text{ if } x \geq 3 \end{cases} . $

The given function $f$ is defined at all the points of the real line.

Let $c$ be a point on the real line.

Case I:

If $c<-3$, then $f(c)=-c+3$

$\lim _{x \to c} f(x)=\lim _{x \to c}(-x+3)=-c+3$

$\therefore \lim _{x \to c} f(x)=f(c)$

Therefore, $f$ is continuous at all points $x$, such that $x<-3$

Case II:

If $c=-3$, then $f(-3)=-(-3)+3=6$

$\lim _{x \to-3^{-}} f(x)=\lim _{x \to-3^{-}}(-x+3)=-(-3)+3=6$

$\lim _{x \to-3^{+}} f(x)=\lim _{x \to-3^{+}}(-2 x)=-2 \times(-3)=6$

$\therefore \lim _{x \to-3} f(x)=f(-3)$

Therefore, $f$ is continuous at $x=-3$

Case III:

If $-3<c<3$, then $f(c)=-2 c$ and $\lim _{x \to c} f(x)=\lim _{x \to c}(-2 x)=-2 c$

$\therefore \lim _{x \to c} f(x)=f(c)$

Therefore, $f$ is continuous in $(-3,3)$.

Case IV:

If $c=3$, then the left hand limit of $f$ at $x=3$ is,

$\lim _{x \to 3^{-}} f(x)=\lim _{x \to 3^{-}}(-2 x)=-2 \times 3=-6$

The right hand limit of $f$ at $x=3$ is,

$\lim _{x \to 3^{+}} f(x)=\lim _{x \to 3^{+}}(6 x+2)=6 \times 3+2=20$

It is observed that the left and right hand limit of $f$ at $x=3$ do not coincide.

Therefore, $f$ is not continuous at $x=3$

Case V:

If $c>3$, then $f(c)=6 c+2$ and $\lim _{x \to c} f(x)=\lim _{x \to c}(6 x+2)=6 c+2$

$\therefore \lim _{x \to c} f(x)=f(c)$

Therefore, $f$ is continuous at all points $x$, such that $x>3$

Hence, $x=3$ is the only point of discontinuity of $f$.

Solution

$ f(x)=\begin{matrix} \frac{|x|}{x} \text{ if } x \neq 0 \\ 0, \text{ if } x=0 \end{matrix} . $

It is known that, $x<0 \Rightarrow|x|=-x$ and $x>0 \Rightarrow|x|=x$

Therefore, the given function can be rewritten as

$f(x)=\begin{cases} \frac{|x|}{x}=\frac{-x}{x}=-1 \text{ if } x<0 \\ 0, \text{ if } x=0 \\ \frac{|x|}{x}=\frac{x}{x}=1, \text{ if } x>0 \end{cases} .$

The given function $f$ is defined at all the points of the real line.

Let $c$ be a point on the real line.

Case I:

If $c<0$, then $f(c)=-1$

$\lim _{x \to c} f(x)=\lim _{x \to c}(-1)=-1$

$\therefore \lim _{x \to c} f(x)=f(c)$

Therefore, $f$ is continuous at all points $x<0$

Case II:

If $c=0$, then the left hand limit of $f$ at $x=0$ is,

$\lim _{x \to 0^{-}} f(x)=\lim _{x \to 0^{-}}(-1)=-1$

The right hand limit of $f$ at $x=0$ is,

$\lim _{x \to 0^{+}} f(x)=\lim _{x \to 0^{+}}(1)=1$

It is observed that the left and right hand limit of $f$ at $x=0$ do not coincide.

Therefore, $f$ is not continuous at $x=0$

Case III:

If $c>0$, then $f(c)=1$

$\lim _{x \to c} f(x)=\lim _{x \to c}(1)=1$

$\therefore \lim _{x \to c} f(x)=f(c)$

Therefore, $f$ is continuous at all points $x$, such that $x>0$

Hence, $x=0$ is the only point of discontinuity of $f$.

Solution

$ f(x)=\begin{matrix} \frac{x}{|x|}, \text{ if } x<0 \\ -1, \text{ if } x \geq 0 \end{matrix} . $

It is known that, $x<0 \Rightarrow|x|=-x$

Therefore, the given function can be rewritten as

$f(x)=\begin{cases} \frac{x}{|x|}=\frac{x}{-x}=-1, \text{ if } x<0 \\ -1, \text{ if } x \geq 0 \end{cases} .$

$\Rightarrow f(x)=-1$ for all $x \in \mathbf{R}$

Let $c$ be any real number. Then, $\lim _{x \to c} f(x)=\lim _{x \to c}(-1)=-1$

Also, $f(c)=-1=\lim _{x \to c} f(x)$

Therefore, the given function is a continuous function.

Hence, the given function has no point of discontinuity.

Solution

$ f(x)=\begin{matrix} x+1, \text{ if } x \geq 1 \\ x^{2}+1, \text{ if } x<1 \end{matrix} . $

The given function $f$ is defined at all the points of the real line.

Let $c$ be a point on the real line.

Case I:

If $c<1$, then $f(c)=c^{2}+1$ and $\lim _{x \to c} f(x)=\lim _{x \to c}(x^{2}+1)=c^{2}+1$

$\therefore \lim _{x \to c} f(x)=f(c)$

Therefore, $f$ is continuous at all points $x$, such that $x<1$

Case II:

If $c=1$, then $f(c)=f(1)=1+1=2$

The left hand limit of $f$ at $x=1$ is,

$\lim _{x \to 1^{-}} f(x)=\lim _{x \to 1^{-}}(x^{2}+1)=1^{2}+1=2$

The right hand limit of $f$ at $x=1$ is,

$\lim _{x \to 1^{+}} f(x)=\lim _{x \to 1^{+}}(x+1)=1+1=2$

$\therefore \lim _{x \to 1} f(x)=f(1)$

Therefore, $f$ is continuous at $x=1$

Case III:

If $c>1$, then $f(c)=c+1$

$\lim _{x \to c} f(x)=\lim _{x \to c}(x+1)=c+1$

$\therefore \lim _{x \to c} f(x)=f(c)$

Therefore, $f$ is continuous at all points $x$, such that $x>1$

Hence, the given function $f$ has no point of discontinuity.

Solution

$ f(x)=\begin{matrix} x^{3}-3, \text{ if } x \leq 2 \\ x^{2}+1, \text{ if } x>2 \end{matrix} . $

The given function $f$ is defined at all the points of the real line.

Let $c$ be a point on the real line.

Case I:

If $c<2$, then $f(c)=c^{3}-3$ and $\lim _{x \to c} f(x)=\lim _{x \to c}(x^{3}-3)=c^{3}-3$

$\therefore \lim _{x \to c} f(x)=f(c)$

Therefore, $f$ is continuous at all points $x$, such that $x<2$

Case II:

If $c=2$, then $f(c)=f(2)=2^{3}-3=5$

$\lim _{x \to 2^{-}} f(x)=\lim _{x \to 2^{-}}(x^{3}-3)=2^{3}-3=5$

$\lim _{x \to 2^{+}} f(x)=\lim _{x \to 2^{+}}(x^{2}+1)=2^{2}+1=5$

$\therefore \lim _{x \to 2} f(x)=f(2)$

Therefore, $f$ is continuous at $x=2$

Case III:

If $c>2$, then $f(c)=c^{2}+1$

$\lim _{x \to c} f(x)=\lim _{x \to c}(x^{2}+1)=c^{2}+1$

$\therefore \lim _{x \to c} f(x)=f(c)$

Therefore, $f$ is continuous at all points $x$, such that $x>2$

Thus, the given function $f$ is continuous at every point on the real line.

Hence, $f$ has no point of discontinuity.

Solution

$ f(x)= \begin{cases}x^{10}-1, & \text{ if } x \leq 1 \\ x^{2}, & \text{ if } x>1\end{cases} $

The given function $f$ is defined at all the points of the real line.

Let $c$ be a point on the real line.

Case I:

If $c<1$, then $f(c)=c^{10}-1$ and $\lim _{x \to c} f(x)=\lim _{x \to c}(x^{10}-1)=c^{10}-1$

$\therefore \lim _{x \to c} f(x)=f(c)$

Therefore, $f$ is continuous at all points $x$, such that $x<1$

Case II:

If $c=1$, then the left hand limit of $f$ at $x=1$ is,

$\lim _{x \to 1^{-}} f(x)=\lim _{x \to 1^{-}}(x^{10}-1)=1^{10}-1=1-1=0$

The right hand limit of $f$ at $x=1$ is,

$\lim _{x \to 1^{+}} f(x)=\lim _{x \to 1^{+}}(x^{2})=1^{2}=1$

It is observed that the left and right hand limit of $f$ at $x=1$ do not coincide.

Therefore, $f$ is not continuous at $x=1$

Case III:

If $c>1$, then $f(c)=c^{2}$

$\lim _{x \to c} f(x)=\lim _{x \to c}(x^{2})=c^{2}$

$\therefore \lim _{x \to c} f(x)=f(c)$

Therefore, $f$ is continuous at all points $x$, such that $x>1$

Thus, from the above observation, it can be concluded that $x=1$ is the only point of discontinuity of $f$.

Solution

The given function is $f(x)=\begin{cases} x+5, \text{ if } x \leq 1 \\ x-5, \text{ if } x>1 \end{cases} .$

The given function $f$ is defined at all the points of the real line.

Let $c$ be a point on the real line.

Case I:

If $c<1$, then $f(c)=c+5$ and $\lim _{x \to c} f(x)=\lim _{x \to c}(x+5)=c+5$

$\therefore \lim _{x \to c} f(x)=f(c)$

Therefore, $f$ is continuous at all points $x$, such that $x<1$

Case II:

If $c=1$, then $f(1)=1+5=6$

The left hand limit of $f$ at $x=1$ is,

$\lim _{x \to 1^{-}} f(x)=\lim _{x \to 1^{-}}(x+5)=1+5=6$

The right hand limit of $f$ at $x=1$ is,

$\lim _{x \to 1^{+}} f(x)=\lim _{x \to 1^{+}}(x-5)=1-5=-4$

It is observed that the left and right hand limit of $f$ at $x=1$ do not coincide.

Therefore, $f$ is not continuous at $x=1$

Case III:

If $c>1$, then $f(c)=c-5$ and $\lim _{x \to c} f(x)=\lim _{x \to c}(x-5)=c-5$

$\therefore \lim _{x \to c} f(x)=f(c)$

Therefore, $f$ is continuous at all points $x$, such that $x>1$

Thus, from the above observation, it can be concluded that $x=1$ is the only point of discontinuity of $f$.

Discuss the continuity of the function $f$, where $f$ is defined by

Solution

The given function is $f(x)=\begin{cases} 3, \text{ if } 0 \leq x \leq 1 \\ 4, \text{ if } 1<x<3 \\ 5, \text{ if } 3 \leq x \leq 10 \end{cases} .$

The given function is defined at all points of the interval $[0,10]$.

Let $c$ be a point in the interval $[0,10]$.

Case I:

If $0 \leq c<1$, then $f(c)=3$ and $\lim _{x \to c} f(x)=\lim _{x \to c}(3)=3$

$\therefore \lim _{x \to c} f(x)=f(c)$

Therefore, $f$ is continuous in the interval $[0,1)$.

Case II:

If $c=1$, then $f(3)=3$

The left hand limit of $f$ at $x=1$ is,

$\lim _{x \to 1^{-}} f(x)=\lim _{x \to 1^{-}}(3)=3$

The right hand limit of $f$ at $x=1$ is,

$\lim _{x \to 1^{+}} f(x)=\lim _{x \to 1^{+}}(4)=4$

It is observed that the left and right hand limits of $f$ at $x=1$ do not coincide.

Therefore, $f$ is not continuous at $x=1$

Case III:

If $1<c<3$, then $f(c)=4$ and $\lim _{x \to c} f(x)=\lim _{x \to c}(4)=4$

$\therefore \lim _{x \to c} f(x)=f(c)$

Therefore, $f$ is continuous at all points of the interval $(1,3)$.

Case IV:

If $c=3$, then $f(c)=5$

The left hand limit of $f$ at $x=3$ is,

$\lim _{x \to 3^{-}} f(x)=\lim _{x \to 3^{-}}(4)=4$

The right hand limit of $f$ at $x=3$ is,

$\lim _{x \to 3^{+}} f(x)=\lim _{x \to 3^{+}}(5)=5$

It is observed that the left and right hand limits of $f$ at $x=3$ do not coincide.

Therefore, $f$ is not continuous at $x=3$

Case V:

If $3<c \leq 10$, then $f(c)=5$ and $\lim _{x \to c} f(x)=\lim _{x \to c}(5)=5$

$\lim _{x \to c} f(x)=f(c)$

Therefore, $f$ is continuous at all points of the interval $(3,10]$.

Hence, $f$ is not continuous at $x=1$ and $x=3$

Solution

The given function is $f(x)= \begin{cases}2 x, & \text{ if } x<0 \\ 0, & \text{ if } 0 \leq x \leq 1 \\ 4 x, & \text{ if } x>1\end{cases}$

The given function is defined at all points of the real line.

Let $c$ be a point on the real line.

Case I:

If $c<0$, then $f(c)=2 c$

$\lim _{x \to c} f(x)=\lim _{x \to c}(2 x)=2 c$

$\therefore \lim _{x \to c} f(x)=f(c)$

Therefore, $f$ is continuous at all points $x$, such that $x<0$

Case II:

If $c=0$, then $f(c)=f(0)=0$

The left hand limit of $f$ at $x=0$ is,

$\lim _{x \to 0^{-}} f(x)=\lim _{x \to 0^{-}}(2 x)=2 \times 0=0$

The right hand limit of $f$ at $x=0$ is,

$\lim _{x \to 0^{+}} f(x)=\lim _{x \to 0^{+}}(0)=0$

$\therefore \lim _{x \to 0} f(x)=f(0)$

Therefore, $f$ is continuous at $x=0$

Case III:

If $0<c<1$, then $f(x)=0$ and $\lim _{x \to c} f(x)=\lim _{x \to c}(0)=0$

$\therefore \lim _{x \to c} f(x)=f(c)$

Therefore, $f$ is continuous at all points of the interval $(0,1)$.

Case IV:

If $c=1$, then $f(c)=f(1)=0$

The left hand limit of $f$ at $x=1$ is,

$\lim _{x \to 1^{-}} f(x)=\lim _{x \to 1^{-}}(0)=0$

The right hand limit of $f$ at $x=1$ is,

$\lim _{x \to 1^{+}} f(x)=\lim _{x \to 1^{+}}(4 x)=4 \times 1=4$

It is observed that the left and right hand limits of $f$ at $x=1$ do not coincide.

Therefore, $f$ is not continuous at $x=1$

Case V:

If $c<1$, then $f(c)=4 c$ and $\lim _{x \to c} f(x)=\lim _{x \to c}(4 x)=4 c$

$\therefore \lim _{x \to c} f(x)=f(c)$

Therefore, $f$ is continuous at all points $x$, such that $x>1$

Hence, $f$ is not continuous only at $x=1$

Solution

$ f(x)=\begin{matrix} -2, \text{ if } x \leq-1 \\ 2 x, \text{ if }-1<x \leq 1 \\ 2, \text{ if } x>1 \end{matrix} . $

The given function is defined at all points of the real line.

Let $c$ be a point on the real line.

Case I:

If $c<-1$, then $f(c)=-2$ and $\lim _{x \to c} f(x)=\lim _{x \to c}(-2)=-2$

$\therefore \lim _{x \to c} f(x)=f(c)$

Therefore, $f$ is continuous at all points $x$, such that $x<-1$

Case II:

If $c=-1$, then $f(c)=f(-1)=-2$

The left hand limit of $f$ at $x=-1$ is,

$\lim _{x \to-1^{-}} f(x)=\lim _{x \to-1^{-}}(-2)=-2$

The right hand limit of $f$ at $x=-1$ is,

$ \begin{aligned} & \lim _{x \to-1^{+}} f(x)=\lim _{x \to-1^{+}}(2 x)=2 \times(-1)=-2 \\ & \therefore \lim _{x \to-1} f(x)=f(-1) \end{aligned} $

Therefore, $f$ is continuous at $x=-1$

Case III:

If $-1<c<1$, then $f(c)=2 c$

$\lim _{x \to c} f(x)=\lim _{x \to c}(2 x)=2 c$

$\therefore \lim _{x \to c} f(x)=f(c)$

Therefore, $f$ is continuous at all points of the interval $(-1,1)$.

Case IV:

If $c=1$, then $f(c)=f(1)=2 \times 1=2$

The left hand limit of $f$ at $x=1$ is,

$\lim _{x \to 1^{-}} f(x)=\lim _{x \to 1^{-}}(2 x)=2 \times 1=2$

The right hand limit of $f$ at $x=1$ is,

$\lim _{x \to 1^{+}} f(x)=\lim _{x \to 1^{+}} 2=2$

$\therefore \lim _{x \to 1} f(x)=f(c)$

Therefore, $f$ is continuous at $x=2$

Case V:

If $c>1$, then $f(c)=2$ and $\lim _{x \to c} f(x)=\lim _{x \to c}(2)=2$

$\lim _{x \to c} f(x)=f(c)$

Therefore, $f$ is continuous at all points $x$, such that $x>1$

Thus, from the above observations, it can be concluded that $f$ is continuous at all points of the real line.

Solution

$ f(x)=\begin{matrix} a x+1, \text{ if } x \leq 3 \\ b x+3, \text{ if } x>3 \end{matrix} . $

If $f$ is continuous at $x=3$, then $\lim _{x \to 3^{-}} f(x)=\lim _{x \to 3^{+}} f(x)=f(3)$

Also,

$\lim _{x \to 3^{-}} f(x)=\lim _{x \to 3^{-}}(a x+1)=3 a+1$

$\lim _{x \to 3^{+}} f(x)=\lim _{x \to 3^{+}}(b x+3)=3 b+3$

$f(3)=3 a+1$

Therefore, from (1), we obtain

$3 a+1=3 b+3=3 a+1$

$\Rightarrow 3 a+1=3 b+3$

$\Rightarrow 3 a=3 b+2$

$\Rightarrow a=b+\frac{2}{3}$

Therefore, the required relationship is given by, $a=b+\frac{2}{3}$

Solution

The given function is $f(x)= \begin{cases}\lambda(x^{2}-2 x), & \text{ if } x \leq 0 \\ 4 x+1, & \text{ if } x>0\end{cases}$

If $f$ is continuous at $x=0$, then

$\lim _{x \to 0^{-}} f(x)=\lim _{x \to 0^{+}} f(x)=f(0)$

$\Rightarrow \lim _{x \to 0^{-}} \lambda(x^{2}-2 x)=\lim _{x \to 0^{+}}(4 x+1)=\lambda(0^{2}-2 \times 0)$

$\Rightarrow \lambda(0^{2}-2 \times 0)=4 \times 0+1=0$

$\Rightarrow 0=1=0$, which is not possible

Therefore, there is no value of $\lambda$ for which $f$ is continuous at $x=0$

At $x=1$,

$f(1)=4 x+1=4 \times 1+1=5$

$\lim _{x \to 1}(4 x+1)=4 \times 1+1=5$

$\therefore \lim _{x \to 1} f(x)=f(1)$

Therefore, for any values of $\lambda, f$ is continuous at $x=1$

Solution

The given function is $g(x)=x-[x]$

It is evident that $g$ is defined at all integral points.

Let $n$ be an integer.

Then,

$g(n)=n-[n]=n-n=0$

The left hand limit of $f$ at $x=n$ is,

$\lim _{x \to n^{-}} g(x)=\lim _{x \to n^{-}}(x-[x])=\lim _{x \to n^{-}}(x)-\lim _{x \to n^{-}}[x]=n-(n-1)=1$

The right hand limit of $f$ at $x=n$ is,

$\lim _{x \to n^{+}} g(x)=\lim _{x \to n^{+}}(x-[x])=\lim _{x \to n^{+}}(x)-\lim _{x \to n^{+}}[x]=n-n=0$

It is observed that the left and right hand limits of $f$ at $x=n$ do not coincide.

Therefore, $f$ is not continuous at $x=n$

Hence, $g$ is discontinuous at all integral points.

Solution

It is known that if $g$ and $h$ are two continuous functions, then

$g+h, g-h$, and $g . h$ are also continuous.

It has to proved first that $g(x)=\sin x$ and $h(x)=\cos x$ are continuous functions.

Let $g(x)=\sin x$

It is evident that $g(x)=\sin x$ is defined for every real number.

Let $c$ be a real number. Put $x=c+h$

If $x \to c$, then $h \to 0$

$ \begin{aligned} & g(c)=\sin c \\ & \begin{aligned} \lim _{x \to c} g(x) & =\lim _{x \to c} \sin x \\ & =\lim _{h \to 0} \sin (c+h) \\ & =\lim _{h \to 0}[\sin c \cos h+\cos c \sin h] \\ & =\lim _{h \to 0}(\sin c \cos h)+\lim _{h \to 0}(\cos c \sin h) \\ & =\sin c \cos 0+\cos c \sin 0 \\ & =\sin c+0 \\ & =\sin c \end{aligned} \\ & \therefore \lim _{x \to c} g(x)=g(c) \end{aligned} $

Therefore, $g$ is a continuous function.

Let $h(x)=\cos x$

It is evident that $h(x)=\cos x$ is defined for every real number.

Let $c$ be a real number. Put $x=c+h$

If $x \to c$, then $h \to 0$

$h(c)=\cos c$

$ \begin{aligned} \lim _{x \to c} h(x) & =\lim _{x \to c} \cos x \\ & =\lim _{h \to 0} \cos (c+h) \\ & =\lim _{h \to 0}[\cos c \cos h-\sin c \sin h] \\ & =\lim _{h \to 0} \cos c \cos h-\lim _{h \to 0} \sin c \sin h \\ & =\cos c \cos 0-\sin c \sin 0 \\ & =\cos c \times 1-\sin c \times 0 \\ & =\cos c \end{aligned} $

$\therefore \lim _{x \to c} h(x)=h(c)$

Therefore, $h$ is a continuous function.

Therefore, it can be concluded that

(a) $f(x)=g(x)+h(x)=\sin x+\cos x$ is a continuous function

(b) $f(x)=g(x)-h(x)=\sin x-\cos x$ is a continuous function

(c) $f(x)=g(x) \times h(x)=\sin x \times \cos x$ is a continuous function

Solution

It is known that if $g$ and $h$ are two continuous functions, then

(i) $\frac{h(x)}{g(x)}, g(x) \neq 0$ is continuous

(ii) $\frac{1}{g(x)}, g(x) \neq 0$ is continuous

(iii) $\frac{1}{h(x)}, h(x) \neq 0$ is continuous

It has to be proved first that $g(x)=\sin x$ and $h(x)=\cos x$ are continuous functions.

Let $g(x)=\sin x$

It is evident that $g(x)=\sin x$ is defined for every real number.

Let $c$ be a real number. Put $x=c+h$

If $x \to c$, then $h \to 0$

$g(c)=\sin c$

$\lim _{x \to c} g(x)=\lim _{x \to c} \sin x$

$=\lim _{h \to 0} \sin (c+h)$

$=\lim _{h \to 0}[\sin c \cos h+\cos c \sin h]$

$=\lim _{h \to 0}(\sin c \cos h)+\lim _{h \to 0}(\cos c \sin h)$

$=\sin c \cos 0+\cos c \sin 0$

$=\sin c+0$

$=\sin c$

$\therefore \lim _{x \to c} g(x)=g(c)$

Therefore, $g$ is a continuous function.

Let $h(x)=\cos x$

It is evident that $h(x)=\cos x$ is defined for every real number.

Let $c$ be a real number. Put $x=c+h$

If $x \to c$, then $h \to 0$

$h(c)=\cos c$

$ \begin{aligned} \lim _{x \to c} h(x) & =\lim _{x \to c} \cos x \\ & =\lim _{h \to 0} \cos (c+h) \\ & =\lim _{h \to 0}[\cos c \cos h-\sin c \sin h] \\ & =\lim _{h \to 0} \cos c \cos h-\lim _{h \to 0} \sin c \sin h \\ & =\cos c \cos 0-\sin c \sin 0 \\ & =\cos c \times 1-\sin c \times 0 \\ & =\cos c \end{aligned} $

$\therefore \lim _{x \to c} h(x)=h(c)$

Therefore, $h(x)=\cos x$ is continuous function.

It can be concluded that,

$cosec x=\frac{1}{\sin x}, \sin x \neq 0$ is continuous

$\Rightarrow cosec x, x \neq n \pi(n \in Z)$ is continuous

Therefore, cosecant is continuous except at $x=n p, n$ Î $\mathbf{Z}$

$\sec x=\frac{1}{\cos x}, \cos x \neq 0$ is continuous

$\Rightarrow \sec x, x \neq(2 n+1) \frac{\pi}{2}(n \in \mathbf{Z})$ is continuous

Therefore, secant is continuous except at $x=(2 n+1) \frac{\pi}{2}(n \in \mathbf{Z})$

$\cot x=\frac{\cos x}{\sin x}, \sin x \neq 0$ is continuous

$\Rightarrow \cot x, x \neq n \pi(n \in Z)$ is continuous

Therefore, cotangent is continuous except at $x=n p, n \hat{I} \mathbf{Z}$

Solution

$ f(x)=\begin{cases} \frac{\sin x}{x}, \text{ if } x<0 \\ x+1, \text{ if } x \geq 0 \end{cases} . $

It is evident that $f$ is defined at all points of the real line.

Let $c$ be a real number.

Case I:

If $c<0$, then $f(c)=\frac{\sin c}{c}$ and $\lim _{x \to c} f(x)=\lim _{x \to c}(\frac{\sin x}{x})=\frac{\sin c}{c}$

$\therefore \lim _{x \to c} f(x)=f(c)$

Therefore, $f$ is continuous at all points $x$, such that $x<0$

Case II:

If $c>0$, then $f(c)=c+1$ and $\lim _{x \to c} f(x)=\lim _{x \to c}(x+1)=c+1$

$\therefore \lim _{x \to c} f(x)=f(c)$

Therefore, $f$ is continuous at all points $x$, such that $x>0$

Case III:

If $c=0$, then $f(c)=f(0)=0+1=1$

The left hand limit of $f$ at $x=0$ is,

$\lim _{x \to 0^{-}} f(x)=\lim _{x \to 0} \frac{\sin x}{x}=1$

The right hand limit of $f$ at $x=0$ is,

$ \begin{aligned} & \lim _{x \to 0^{+}} f(x)=\lim _{x \to 0^{+}}(x+1)=1 \\ & \therefore \lim _{x \to 0^{-}} f(x)=\lim _{x \to 0^{+}} f(x)=f(0) \end{aligned} $

Therefore, $f$ is continuous at $x=0$

From the above observations, it can be concluded that $f$ is continuous at all points of the real line.

Thus, $f$ has no point of discontinuity.

Solution

$ f(x)= \begin{cases}x^{2} \sin \frac{1}{x}, & \text{ if } x \neq 0 \\ 0, & \text{ if } x=0\end{cases} $

It is evident that $f$ is defined at all points of the real line.

Let $c$ be a real number.

Case I:

If $c \neq 0$, then $f(c)=c^{2} \sin \frac{1}{c}$

$\lim _{x \to c} f(x)=\lim _{x \to c}(x^{2} \sin \frac{1}{x})=(\lim _{x \to c} x^{2})(\lim _{x \to c} \sin \frac{1}{x})=c^{2} \sin \frac{1}{c}$

$\therefore \lim _{x \to c} f(x)=f(c)$

Therefore, $f$ is continuous at all points $x \neq 0$

Case II:

If $c=0$, then $f(0)=0$ $\lim _{x \to 0^{-}} f(x)=\lim _{x \to 0^{-}}(x^{2} \sin \frac{1}{x})=\lim _{x \to 0}(x^{2} \sin \frac{1}{x})$

It is known that, $-1 \leq \sin \frac{1}{x} \leq 1, x \neq 0$

$\Rightarrow-x^{2} \leq \sin \frac{1}{x} \leq x^{2}$

$\Rightarrow \lim _{x \to 0}(-x^{2}) \leq \lim _{x \to 0}(x^{2} \sin \frac{1}{x}) \leq \lim _{x \to 0} x^{2}$

$\Rightarrow 0 \leq \lim _{x \to 0}(x^{2} \sin \frac{1}{x}) \leq 0$

$\Rightarrow \lim _{x \to 0}(x^{2} \sin \frac{1}{x})=0$

$\therefore \lim _{x \to 0^{-}} f(x)=0$

Similarly, $\lim _{x \to 0^{+}} f(x)=\lim _{x \to 0^{+}}(x^{2} \sin \frac{1}{x})=\lim _{x \to 0}(x^{2} \sin \frac{1}{x})=0$

$\therefore \lim _{x \to 0^{-}} f(x)=f(0)=\lim _{x \to 0^{+}} f(x)$

Therefore, $f$ is continuous at $x=0$

From the above observations, it can be concluded that $f$ is continuous at every point of the real line.

Thus, $f$ is a continuous function.

Solution

$ f(x)= \begin{cases}\sin x-\cos x, & \text{ if } x \neq 0 \\ -1 & \text{ if } x=0\end{cases} $

It is evident that $f$ is defined at all points of the real line.

Let $c$ be a real number.

Case I:

If $c \neq 0$, then $f(c)=\sin c-\cos c$

$\lim _{x \to c} f(x)=\lim _{x \to c}(\sin x-\cos x)=\sin c-\cos c$

$\therefore \lim _{x \to c} f(x)=f(c)$

Therefore, $f$ is continuous at all points $x$, such that $x \neq 0$

Case II:

If $c=0$, then $f(0)=-1$

$\lim _{x \to 0^{-}} f(x)=\lim _{x \to 0}(\sin x-\cos x)=\sin 0-\cos 0=0-1=-1$

$\lim _{x \to 0^{+}} f(x)=\lim _{x \to 0}(\sin x-\cos x)=\sin 0-\cos 0=0-1=-1$

$\therefore \lim _{x \to 0^{-}} f(x)=\lim _{x \to 0^{+}} f(x)=f(0)$

Therefore, $f$ is continuous at $x=0$

From the above observations, it can be concluded that $f$ is continuous at every point of the real line.

Thus, $f$ is a continuous function.

Solution

$ f(x)= \begin{cases}\frac{k \cos x}{\pi-2 x}, & \text{ if } x \neq \frac{\pi}{2} \\ 3, & \text{ if } x=\frac{\pi}{2}\end{cases} $

The given function $f$ is continuous at $x=\frac{\pi}{2}$, if $f$ is defined at $x=\frac{\pi}{2}$ and if the value of the $f$ at $x=\frac{\pi}{2}$ equals the limit of $f$ at $x=\frac{\pi}{2}$.

It is evident that $f$ is defined at $x=\frac{\pi}{2}$ and $f(\frac{\pi}{2})=3$

$\lim _{x \to \frac{\pi}{2}} f(x)=\lim _{x \to \frac{\pi}{2}} \frac{k \cos x}{\pi-2 x}$

Put $x=\frac{\pi}{2}+h$

Then, $x \to \frac{\pi}{2} \Rightarrow h \to 0$

$\begin{aligned} \therefore \lim _{x \to \frac{\pi}{2}} f(x) & =\lim _{x \to \frac{\pi}{2}} \frac{k \cos x}{\pi-2 x}=\lim _{h \to 0} \frac{k \cos (\frac{\pi}{2}+h)}{\pi-2(\frac{\pi}{2}+h)} \\ & =k \lim _{h \to 0} \frac{-\sin h}{-2 h}=\frac{k}{2} \lim _{h \to 0} \frac{\sin h}{h}=\frac{k}{2} \cdot 1=\frac{k}{2}\end{aligned}$

$\therefore \lim _{x \to \frac{\pi}{2}} f(x)=f(\frac{\pi}{2})$

$\Rightarrow \frac{k}{2}=3$

$\Rightarrow k=6$

Therefore, the required value of $k$ is 6 .

Solution

The given function is $f(x)= \begin{cases}k x^{2}, & \text{ if } x \leq 2 \\ 3, & \text{ if } x>2\end{cases}$

The given function $f$ is continuous at $x=2$, if $f$ is defined at $x=2$ and if the value of $f$ at $x=2$ equals the limit of $f$ at $x=2$

It is evident that $f$ is defined at $x=2$ and $f(2)=k(2)^{2}=4 k$

$ \begin{aligned} & \lim _{x \to 2^{-}} f(x)=\lim _{x \to 2^{+}} f(x)=f(2) \\ & \Rightarrow \lim _{x \to 2^{-}}(k x^{2})=\lim _{x \to 2^{+}}(3)=4 k \\ & \Rightarrow k \times 2^{2}=3=4 k \\ & \Rightarrow 4 k=3=4 k \\ & \Rightarrow 4 k=3 \\ & \Rightarrow k=\frac{3}{4} \end{aligned} $

Therefore, the required value of $k$ is $\frac{3}{4}$.

Solution

The given function is $f(x)=\begin{cases} k x+1, \text{ if } x \leq \pi \\ \cos x, \text{ if } x>\pi \end{cases} .$

The given function $f$ is continuous at $x=p$, if $f$ is defined at $x=p$ and if the value of $f$ at $x=p$ equals the limit of $f$ at $x=p$

It is evident that $f$ is defined at $x=p$ and $f(\pi)=k \pi+1$

$ \begin{aligned} & \lim _{x \to \pi^{-}} f(x)=\lim _{x \to \pi^{+}} f(x)=f(\pi) \\ & \Rightarrow \lim _{x \to \pi^{-}}(k x+1)=\lim _{x \to \pi^{+}} \cos x=k \pi+1 \\ & \Rightarrow k \pi+1=\cos \pi=k \pi+1 \\ & \Rightarrow k \pi+1=-1=k \pi+1 \\ & \Rightarrow k=-\frac{2}{\pi} \end{aligned} $

Therefore, the required value of $k$ is $-\frac{2}{\pi}$.

Solution

$ f(x)=\begin{cases} k x+1, \text{ if } x \leq 5 \\ 3 x-5, \text{ if } x>5 \end{cases} . $

The given function $f$ is continuous at $x=5$, if $f$ is defined at $x=5$ and if the value of $f$ at $x=5$ equals the limit of $f$ at $x=5$

It is evident that $f$ is defined at $x=5$ and $f(5)=k x+1=5 k+1$

$ \begin{aligned} & \lim _{x \to 5^{-}} f(x)=\lim _{x \to 5^{+}} f(x)=f(5) \\ & \Rightarrow \lim _{x \to 5^{-}}(k x+1)=\lim _{x \to 5^{+}}(3 x-5)=5 k+1 \\ & \Rightarrow 5 k+1=15-5=5 k+1 \\ & \Rightarrow 5 k+1=10 \\ & \Rightarrow 5 k=9 \\ & \Rightarrow k=\frac{9}{5} \end{aligned} $

Therefore, the required value of $k$ is $\frac{9}{5}$.

Solution

$ f(x)= \begin{cases}5, & \text{ if } x \leq 2 \\ a x+b, & \text{ if } 2<x<10 \\ 21, & \text{ if } x \geq 10\end{cases} $

It is evident that the given function $f$ is defined at all points of the real line.

If $f$ is a continuous function, then $f$ is continuous at all real numbers.

In particular, $f$ is continuous at $x=2$ and $x=10$

Since $f$ is continuous at $x=2$, we obtain

$ \begin{aligned} & \lim _{x \to 2^{-}} f(x)=\lim _{x \to 2^{+}} f(x)=f(2) \\ & \Rightarrow \lim _{x \to 2^{-}}(5)=\lim _{x \to 2^{+}}(a x+b)=5 \\ & \Rightarrow 5=2 a+b=5 \\ & \Rightarrow 2 a+b=5 \\ \end{aligned} $

Since $f$ is continuous at $x=10$, we obtain

$$ \begin{align*} & \lim _{x \to 10^{-}} f(x)=\lim _{x \to 10^{+}} f(x)=f(10) \\ & \Rightarrow \lim _{x \to 10^{-}}(a x+b)=\lim _{x \to 10^{+}}(21)=21 \\ & \Rightarrow 10 a+b=21=21 \\ & \Rightarrow 10 a+b=21 \tag{2} \end{align*} $$

On subtracting equation (1) from equation (2), we obtain

$8 a=16$

$\Rightarrow a=2$

By putting $a=2$ in equation (1), we obtain

$2 \times 2+b=5$

$\Rightarrow 4+b=5$ $\Rightarrow b=1$

Therefore, the values of $a$ and $b$ for which $f$ is a continuous function are 2 and 1 respectively.

Solution

The given function is $f(x)=\cos (x^{2})$

This function $f$ is defined for every real number and $f$ can be written as the composition of two functions as,

$f=g \circ h$, where $g(x)=\cos x$ and $h(x)=x^{2}$

$[\because(g \circ h)(x)=g(h(x))=g(x^{2})=\cos (x^{2})=f(x)]$

It has to be first proved that $g(x)=\cos x$ and $h(x)=x^{2}$ are continuous functions.

It is evident that $g$ is defined for every real number.

Let $c$ be a real number.

Then, $g(c)=\cos c$

Put $x=c+h$

If $x \to c$, then $h \to 0$

$\lim _{x \to c} g(x)=\lim _{x \to c} \cos x$

$=\lim _{h \to 0} \cos (c+h)$

$=\lim _{h \to 0}[\cos c \cos h-\sin c \sin h]$

$=\lim _{h \to 0} \cos c \cos h-\lim _{h \to 0} \sin c \sin h$

$=\cos c \cos 0-\sin c \sin 0$

$=\cos c \times 1-\sin c \times 0$

$=\cos c$

$\therefore \lim _{x \to c} g(x)=g(c)$

Therefore, $g(x)=\cos x$ is continuous function. $h(x)=x^{2}$

Clearly, $h$ is defined for every real number.

Let $k$ be a real number, then $h(k)=k^{2}$

$\lim _{x \to k} h(x)=\lim _{x \to k} x^{2}=k^{2}$

$\therefore \lim _{x \to k} h(x)=h(k)$

Therefore, $h$ is a continuous function.

It is known that for real valued functions $g$ and $h$, such that $(g \circ h)$ is defined at $c$, if $g$ is continuous at $c$ and if $f$ is continuous at $g(c)$, then $(f \circ g)$ is continuous at $c$.

Therefore, $f(x)=(g o h)(x)=\cos (x^{2})$ is a continuous function.

Solution

The given function is $f(x)=|\cos x|$

This function $f$ is defined for every real number and $f$ can be written as the composition of two functions as,

$f=g \circ h$, where $g(x)=|x|$ and $h(x)=\cos x$

$[\because(g \circ h)(x)=g(h(x))=g(\cos x)=|\cos x|=f(x)]$

It has to be first proved that $g(x)=|x|$ and $h(x)=\cos x$ are continuous functions.

$g(x)=|x|$ can be written as

$g(x)= \begin{cases}-x, & \text{ if } x<0 \\ x, & \text{ if } x \geq 0\end{cases}$

Clearly, $g$ is defined for all real numbers.

Let $c$ be a real number.

Case I:

If $c<0$, then $g(c)=-c$ and $\lim _{x \to c} g(x)=\lim _{x \to c}(-x)=-c$

$\therefore \lim _{x \to c} g(x)=g(c)$

Therefore, $g$ is continuous at all points $x$, such that $x<0$

Case II:

If $c>0$, then $g(c)=c$ and $\lim _{x \to c} g(x)=\lim _{x \to c} x=c$

$\therefore \lim _{x \to c} g(x)=g(c)$

Therefore, $g$ is continuous at all points $x$, such that $x>0$

Case III:

If $c=0$, then $g(c)=g(0)=0$

$ \begin{aligned} & \lim _{x \to 0^{-}} g(x)=\lim _{x \to 0^{-}}(-x)=0 \\ & \lim _{x \to 0^{+}} g(x)=\lim _{x \to 0^{+}}(x)=0 \\ & \therefore \lim _{x \to 0^{-}} g(x)=\lim _{x \to 0^{+}}(x)=g(0) \end{aligned} $

Therefore, $g$ is continuous at $x=0$

From the above three observations, it can be concluded that $g$ is continuous at all points.

$h(x)=\cos x$

It is evident that $h(x)=\cos x$ is defined for every real number.

Let $c$ be a real number. Put $x=c+h$

If $x \to c$, then $h \to 0$

$h(c)=\cos c$

$ \begin{aligned} \lim _{x \to c} h(x) & =\lim _{x \to c} \cos x \\ & =\lim _{h \to 0} \cos (c+h) \\ & =\lim _{h \to 0}[\cos c \cos h-\sin c \sin h] \\ & =\lim _{h \to 0} \cos c \cos h-\lim _{h \to 0} \sin c \sin h \\ & =\cos c \cos 0-\sin c \sin 0 \\ & =\cos c \times 1-\sin c \times 0 \\ & =\cos c \end{aligned} $

$\therefore \lim _{x \to c} h(x)=h(c)$

Therefore, $h(x)=\cos x$ is a continuous function.

It is known that for real valued functions $g$ and $h$, such that $(g \circ h)$ is defined at $c$, if $g$ is continuous at $c$ and if $f$ is continuous at $g(c)$, then $(f \circ g)$ is continuous at $c$.

Therefore, $f(x)=(g \circ h)(x)=g(h(x))=g(\cos x)=|\cos x|$ is a continuous function.

Solution

Let $f(x)=\sin |x|$

This function $f$ is defined for every real number and $f$ can be written as the composition of two functions as,

$f=g \circ h$, where $g(x)=|x|$ and $h(x)=\sin x$

$[\because(g \circ h)(x)=g(h(x))=g(\sin x)=|\sin x|=f(x)]$

It has to be proved first that $g(x)=|x|$ and $h(x)=\sin x$ are continuous functions.

$g(x)=|x|$ can be written as

$g(x)= \begin{cases}-x, & \text{ if } x<0 \\ x, & \text{ if } x \geq 0\end{cases}$

Clearly, $g$ is defined for all real numbers.

Let $c$ be a real number.

Case I:

If $c<0$, then $g(c)=-c$ and $\lim _{x \to c} g(x)=\lim _{x \to c}(-x)=-c$

$\therefore \lim _{x \to c} g(x)=g(c)$

Therefore, $g$ is continuous at all points $x$, such that $x<0$

Case II:

If $c>0$, then $g(c)=c$ and $\lim _{x \to c} g(x)=\lim _{x \to c} x=c$

$\therefore \lim _{x \to c} g(x)=g(c)$

Therefore, $g$ is continuous at all points $x$, such that $x>0$

Case III:

If $c=0$, then $g(c)=g(0)=0$

$ \begin{aligned} & \lim _{x \to 0^{-}} g(x)=\lim _{x \to 0^{-}}(-x)=0 \\ & \lim _{x \to 0^{+}} g(x)=\lim _{x \to 0^{+}}(x)=0 \\ & \therefore \lim _{x \to 0^{-}} g(x)=\lim _{x \to 0^{+}}(x)=g(0) \end{aligned} $

Therefore, $g$ is continuous at $x=0$

From the above three observations, it can be concluded that $g$ is continuous at all points.

$h(x)=\sin x$

It is evident that $h(x)=\sin x$ is defined for every real number.

Let $c$ be a real number. Put $x=c+k$

If $x \to c$, then $k \to 0$

$h(c)=\sin c$

$h(c)=\sin c$

$ \begin{aligned} \lim _{x \to c} h(x) & =\lim _{x \to c} \sin x \\ & =\lim _{k \to 0} \sin (c+k) \\ & =\lim _{k \to 0}[\sin c \cos k+\cos c \sin k] \\ & =\lim _{k \to 0}(\sin c \cos k)+\lim _{h \to 0}(\cos c \sin k) \\ & =\sin c \cos 0+\cos c \sin 0 \\ & =\sin c+0 \\ & =\sin c \end{aligned} $

$\therefore \lim _{x \to c} h(x)=g(c)$

Therefore, $h$ is a continuous function.

It is known that for real valued functions $g$ and $h$, such that $(g \circ h)$ is defined at $c$, if $g$ is continuous at $c$ and if $f$ is continuous at $g(c)$, then $(f \circ g)$ is continuous at $c$.

Therefore, $f(x)=(g \circ h)(x)=g(h(x))=g(\sin x)=|\sin x|$ is a continuous function.

Solution

The given function is $f(x)=|x|-|x+1|$

The two functions, $g$ and $h$, are defined as

$g(x)=|x|$ and $h(x)=|x+1|$

Then, $f=g-h$

The continuity of $g$ and $h$ is examined first.

$g(x)=|x|$ can be written as

$g(x)= \begin{cases}-x, & \text{ if } x<0 \\ x, & \text{ if } x \geq 0\end{cases}$

Clearly, $g$ is defined for all real numbers.

Let $c$ be a real number.

Case I:

If $c<0$, then $g(c)=-c$ and $\lim _{x \to c} g(x)=\lim _{x \to c}(-x)=-c$

$\therefore \lim _{x \to c} g(x)=g(c)$

Therefore, $g$ is continuous at all points $x$, such that $x<0$

Case II:

If $c>0$, then $g(c)=c$ and $\lim _{x \to c} g(x)=\lim _{x \to c} x=c$

$\therefore \lim _{x \to c} g(x)=g(c)$

Therefore, $g$ is continuous at all points $x$, such that $x>0$

Case III:

If $c=0$, then $g(c)=g(0)=0$

$\lim _{x \to 0^{-}} g(x)=\lim _{x \to 0^{-}}(-x)=0$

$\lim _{x \to 0^{+}} g(x)=\lim _{x \to 0^{+}}(x)=0$

$\therefore \lim _{x \to 0^{-}} g(x)=\lim _{x \to 0^{+}}(x)=g(0)$

Therefore, $g$ is continuous at $x=0$

From the above three observations, it can be concluded that $g$ is continuous at all points.

$h(x)=|x+1|$ can be written as

$h(x)= \begin{cases}-(x+1), & \text{ if, } x<-1 \\ x+1, & \text{ if } x \geq-1\end{cases}$

Clearly, $h$ is defined for every real number.

Let $c$ be a real number.

Case I:

If $c<-1$, then $h(c)=-(c+1)$ and $\lim _{x \to c} h(x)=\lim _{x \to c}[-(x+1)]=-(c+1)$

$\therefore \lim _{x \to c} h(x)=h(c)$

Therefore, $h$ is continuous at all points $x$, such that $x<-1$

Case II:

If $c>-1$, then $h(c)=c+1$ and $\lim _{x \to c} h(x)=\lim _{x \to c}(x+1)=c+1$

$\therefore \lim _{x \to c} h(x)=h(c)$

Therefore, $h$ is continuous at all points $x$, such that $x>-1$

Case III:

If $c=-1$, then $h(c)=h(-1)=-1+1=0$

$\lim _{x \to-1^{-}} h(x)=\lim _{x \to-1^{-}}[-(x+1)]=-(-1+1)=0$

$\lim _{x \to-1^{+}} h(x)=\lim _{x \to-1^{+}}(x+1)=(-1+1)=0$

$\therefore \lim _{x \to-1^{-}} h(x)=\lim _{h \to-1^{+}} h(x)=h(-1)$

Therefore, $h$ is continuous at $x=-1$

From the above three observations, it can be concluded that $h$ is continuous at all points of the real line.

$g$ and $h$ are continuous functions. Therefore, $f=g-h$ is also a continuous function.

Therefore, $f$ has no point of discontinuity.

Solution

Let $f(x)=\sin (x^{2}+5), u(x)=x^{2}+5$, and $v(t)=\sin t$

Then, $($ vou $)(x)=v(u(x))=v(x^{2}+5)=\tan (x^{2}+5)=f(x)$

Thus, $f$ is a composite of two functions.

Put $t=u(x)=x^{2}+5$

Then, we obtain

$\frac{d v}{d t}=\frac{d}{d t}(\sin t)=\cos t=\cos (x^{2}+5)$

$\frac{d t}{d x}=\frac{d}{d x}(x^{2}+5)=\frac{d}{d x}(x^{2})+\frac{d}{d x}(5)=2 x+0=2 x$

Therefore, by chain rule, $\frac{d f}{d x}=\frac{d v}{d t} \cdot \frac{d t}{d x}=\cos (x^{2}+5) \times 2 x=2 x \cos (x^{2}+5)$

Alternate method

$ \begin{aligned} \frac{d}{d x}[\sin (x^{2}+5)] & =\cos (x^{2}+5) \cdot \frac{d}{d x}(x^{2}+5) \\ & =\cos (x^{2}+5) \cdot[\frac{d}{d x}(x^{2})+\frac{d}{d x}(5)] \\ & =\cos (x^{2}+5) \cdot[2 x+0] \\ & =2 x \cos (x^{2}+5) \end{aligned} $

Solution

Let $f(x)=\cos (\sin x), u(x)=\sin x$, and $v(t)=\cos t$

Then, $($ vou $)(x)=v(u(x))=v(\sin x)=\cos (\sin x)=f(x)$

Thus, $f$ is a composite function of two functions.

Put $t=u(x)=\sin x$

$\therefore \frac{d v}{d t}=\frac{d}{d t}[\cos t]=-\sin t=-\sin (\sin x)$

$\frac{d t}{d x}=\frac{d}{d x}(\sin x)=\cos x$

By chain rule, $\frac{d f}{d x}=\frac{d v}{d t} \cdot \frac{d t}{d x}=-\sin (\sin x) \cdot \cos x=-\cos x \sin (\sin x)$

Alternate method

$\frac{d}{d x}[\cos (\sin x)]=-\sin (\sin x) \cdot \frac{d}{d x}(\sin x)=-\sin (\sin x) \cdot \cos x=-\cos x \sin (\sin x)$

Solution

Let $f(x)=\sin (a x+b), u(x)=a x+b$, and $v(t)=\sin t$

Then, $($ vou $)(x)=v(u(x))=v(a x+b)=\sin (a x+b)=f(x)$

Thus, $f$ is a composite function of two functions, $u$ and $v$.

Put $t=u(x)=a x+b$

Therefore,

$\frac{d v}{d t}=\frac{d}{d t}(\sin t)=\cos t=\cos (a x+b)$

$\frac{d t}{d x}=\frac{d}{d x}(a x+b)=\frac{d}{d x}(a x)+\frac{d}{d x}(b)=a+0=a$

Hence, by chain rule, we obtain

$\frac{d f}{d x}=\frac{d v}{d t} \cdot \frac{d t}{d x}=\cos (a x+b) \cdot a=a \cos (a x+b)$

Alternate method

$ \begin{aligned} \frac{d}{d x}[\sin (a x+b)] & =\cos (a x+b) \cdot \frac{d}{d x}(a x+b) \\ & =\cos (a x+b) \cdot[\frac{d}{d x}(a x)+\frac{d}{d x}(b)] \\ & =\cos (a x+b) \cdot(a+0) \\ & =a \cos (a x+b) \end{aligned} $

Solution

Let $f(x)=\sec (\tan \sqrt{x}), u(x)=\sqrt{x}, v(t)=\tan t$, and $w(s)=\sec s$

Then, $($ wovou $)(x)=w[v(u(x))]=w[v(\sqrt{x})]=w(\tan \sqrt{x})=\sec (\tan \sqrt{x})=f(x)$

Thus, $f$ is a composite function of three functions, $u, v$, and $w$.

Put $s=v(t)=\tan t$ and $t=u(x)=\sqrt{x}$

$ \text{ Then, } \begin{aligned} \frac{d w}{d s} & =\frac{d}{d s}(\sec s)=\sec s \tan s=\sec (\tan t) \cdot \tan (\tan t) & & {[s=\tan t] } \\ & =\sec (\tan \sqrt{x}) \cdot \tan (\tan \sqrt{x}) & & {[t=\sqrt{x}] } \end{aligned} $

$\frac{d s}{d t}=\frac{d}{d t}(\tan t)=\sec ^{2} t=\sec ^{2} \sqrt{x}$

$\frac{d t}{d x}=\frac{d}{d x}(\sqrt{x})=\frac{d}{d x}(x^{\frac{1}{2}})=\frac{1}{2} \cdot x^{\frac{1}{2}-1}=\frac{1}{2 \sqrt{x}}$

Hence, by chain rule, we obtain

$ \begin{aligned} & \frac{d t}{d x}=\frac{d w}{d s} \cdot \frac{d s}{d t} \cdot \frac{d t}{d x} \\ & =\sec (\tan \sqrt{x}) \cdot \tan (\tan \sqrt{x}) \times \sec ^{2} \sqrt{x} \times \frac{1}{2 \sqrt{x}} \\ & =\frac{1}{2 \sqrt{x}} \sec ^{2} \sqrt{x} \sec (\tan \sqrt{x}) \tan (\tan \sqrt{x}) \\ & =\frac{\sec ^{2} \sqrt{x} \sec (\tan \sqrt{x}) \tan (\tan \sqrt{x})}{2 \sqrt{x}} \end{aligned} $

Alternate method

$ \begin{aligned} \frac{d}{d x}[\sec (\tan \sqrt{x})] & =\sec (\tan \sqrt{x}) \cdot \tan (\tan \sqrt{x}) \cdot \frac{d}{d x}(\tan \sqrt{x}) \\ & =\sec (\tan \sqrt{x}) \cdot \tan (\tan \sqrt{x}) \cdot \sec ^{2}(\sqrt{x}) \cdot \frac{d}{d x}(\sqrt{x}) \\ & =\sec (\tan \sqrt{x}) \cdot \tan (\tan \sqrt{x}) \cdot \sec ^{2}(\sqrt{x}) \cdot \frac{1}{2 \sqrt{x}} \\ & =\frac{\sec (\tan \sqrt{x}) \cdot \tan (\tan \sqrt{x}) \sec ^{2}(\sqrt{x})}{2 \sqrt{x}} \end{aligned} $

Solution

The given function is $f(x)=\frac{\sin (a x+b)}{\cos (c x+d)}=\frac{g(x)}{h(x)}$, where $g(x)=\sin (a x+b)$ and

$h(x)=\cos (c x+d)$

$\therefore f^{\prime}=\frac{g^{\prime} h-g h^{\prime}}{h^{2}}$

Consider $g(x)=\sin (a x+b)$

Let $u(x)=a x+b, v(t)=\sin t$

Then, $($ vou $)(x)=v(u(x))=v(a x+b)=\sin (a x+b)=g(x)$ $\therefore g$ is a composite function of two functions, $u$ and $v$.

Put $t=u(x)=a x+b$

$\frac{d v}{d t}=\frac{d}{d t}(\sin t)=\cos t=\cos (a x+b)$

$\frac{d t}{d x}=\frac{d}{d x}(a x+b)=\frac{d}{d x}(a x)+\frac{d}{d x}(b)=a+0=a$

Therefore, by chain rule, we obtain

$g^{\prime}=\frac{d g}{d x}=\frac{d v}{d t} \cdot \frac{d t}{d x}=\cos (a x+b) \cdot a=a \cos (a x+b)$

Consider $h(x)=\cos (c x+d)$

Let $p(x)=c x+d, q(y)=\cos y$

Then, $(q o p)(x)=q(p(x))=q(c x+d)=\cos (c x+d)=h(x)$

$\therefore h$ is a composite function of two functions, $p$ and $q$.

Put $y=p(x)=c x+d$

$\frac{d q}{d y}=\frac{d}{d y}(\cos y)=-\sin y=-\sin (c x+d)$

$\frac{d y}{d x}=\frac{d}{d x}(c x+d)=\frac{d}{d x}(c x)+\frac{d}{d x}(d)=c$

Therefore, by chain rule, we obtain

$h^{\prime}=\frac{d h}{d x}=\frac{d q}{d y} \cdot \frac{d y}{d x}=-\sin (c x+d) \times c=-c \sin (c x+d)$

$ \begin{aligned} \therefore f^{\prime} & =\frac{a \cos (a x+b) \cdot \cos (c x+d)-\sin (a x+b){-c \sin (c x+d)}}{[\cos (c x+d)]^{2}} \\ & =\frac{a \cos (a x+b)}{\cos (c x+d)}+c \sin (a x+b) \cdot \frac{\sin (c x+d)}{\cos (c x+d)} \times \frac{1}{\cos (c x+d)} \\ & =a \cos (a x+b) \sec (c x+d)+c \sin (a x+b) \tan (c x+d) \sec (c x+d) \end{aligned} $

Solution

The given function is $\cos x^{3} \cdot \sin ^{2}(x^{5})$

$ \begin{aligned} & \frac{d}{d x}[\cos x^{3} \cdot \sin ^{2}(x^{5})]=\sin ^{2}(x^{5}) \times \frac{d}{d x}(\cos x^{3})+\cos x^{3} \times \frac{d}{d x}[\sin ^{2}(x^{5})] \\ & =\sin ^{2}(x^{5}) \times(-\sin x^{3}) \times \frac{d}{d x}(x^{3})+\cos x^{3} \times 2 \sin (x^{5}) \cdot \frac{d}{d x}[\sin x^{5}] \\ & =-\sin x^{3} \sin ^{2}(x^{5}) \times 3 x^{2}+2 \sin x^{5} \cos x^{3} \cdot \cos x^{5} \times \frac{d}{d x}(x^{5}) \\ & =-3 x^{2} \sin x^{3} \cdot \sin ^{2}(x^{5})+2 \sin x^{5} \cos x^{5} \cos x^{3} \cdot \times 5 x^{4} \\ & =10 x^{4} \sin x^{5} \cos x^{5} \cos x^{3}-3 x^{2} \sin x^{3} \sin ^{2}(x^{5}) \end{aligned} $

Solution

$ \begin{aligned} & \frac{d}{d x}[2 \sqrt{\cot (x^{2})}] \\ & =2 \cdot \frac{1}{2 \sqrt{\cot (x^{2})}} \times \frac{d}{d x}[\cot (x^{2})] \\ & =\sqrt{\frac{\sin (x^{2})}{\cos (x^{2})}} \times-cosec^{2}(x^{2}) \times \frac{d}{d x}(x^{2}) \\ & =-\sqrt{\frac{\sin (x^{2})}{\cos (x^{2})}} \times \frac{1}{\sin ^{2}(x^{2})} \times(2 x) \\ & =\frac{-2 x}{\sqrt{\cos x^{2}} \sqrt{\sin x^{2}} \sin x^{2}} \\ & =\frac{-2 \sqrt{2} x}{\sqrt{2 \sin x^{2} \cos x^{2}} \sin x^{2}} \\ & =\frac{-2 \sqrt{2} x}{\sin x^{2} \sqrt{\sin 2 x^{2}}} \end{aligned} $

Solution

Let $f(x)=\cos (\sqrt{x})$

Also, let $u(x)=\sqrt{x}$

And, $v(t)=\cos t$

Then, $($ vou $)(x)=v(u(x))$

$ \begin{aligned} & =v(\sqrt{x}) \\ & =\cos \sqrt{x} \\ & =f(x) \end{aligned} $

Clearly, $f$ is a composite function of two functions, $u$ and $v$, such that

$t=u(x)=\sqrt{x}$

Then, $\frac{d t}{d x}=\frac{d}{d x}(\sqrt{x})=\frac{d}{d x}(x^{\frac{1}{2}})=\frac{1}{2} x^{-\frac{1}{2}}$

$ =\frac{1}{2 \sqrt{x}} $

And, $\frac{d v}{d t}=\frac{d}{d t}(\cos t)=-\sin t$

$ =-\sin (\sqrt{x}) $

By using chain rule, we obtain

$ \begin{aligned} & \frac{d t}{d x}=\frac{d v}{d t} \cdot \frac{d t}{d x} \\ & =-\sin (\sqrt{x}) \cdot \frac{1}{2 \sqrt{x}} \\ & =-\frac{1}{2 \sqrt{x}} \sin (\sqrt{x}) \\ & =-\frac{\sin (\sqrt{x})}{2 \sqrt{x}} \end{aligned} $

Alternate method

$ \begin{aligned} \frac{d}{d x}[\cos (\sqrt{x})] & =-\sin (\sqrt{x}) \cdot \frac{d}{d x}(\sqrt{x}) \\ & =-\sin (\sqrt{x}) \times \frac{d}{d x}(x^{\frac{1}{2}}) \\ & =-\sin \sqrt{x} \times \frac{1}{2} x^{-\frac{1}{2}} \\ & =\frac{-\sin \sqrt{x}}{2 \sqrt{x}} \end{aligned} $

Solution

The given function is $f(x)=|x-1|, x \in \mathbf{R}$

It is known that a function $f$ is differentiable at a point $x=c$ in its domain if both

$\lim _{h \to 0^{-}} \frac{f(c+h)-f(c)}{h}$ and $\lim _{h \to 0^{+}} \frac{f(c+h)-f(c)}{h}$ are finite and equal.

To check the differentiability of the given function at $x=1$,

consider the left hand limit of $f$ at $x=1$

$ \begin{aligned} & \lim _{h \to 0^{-}} \frac{f(1+h)-f(1)}{h}=\lim _{h \to 0^{-}} \frac{|1+h-1|-|1-1|}{h} \\ & \begin{aligned} =\lim _{h \to 0^{-}} \frac{|h|-0}{h} & =\lim _{h \to 0^{-}} \frac{-h}{h} \quad(h<0 \Rightarrow|h|=-h) \\ & =-1 \end{aligned} \end{aligned} $

Consider the right hand limit of $f$ at $x=1$

$ \begin{aligned} & \lim _{h \to 0^{+}} \frac{f(1+h)-f(1)}{h}=\lim _{h \to 0^{+}} \frac{|1+h-1|-|1-1|}{h} \\ & \begin{aligned} \lim _{h \to 0^{+}} \frac{|h|-0}{h} & =\lim _{h \to 0^{+}} \frac{h}{h} \quad(h>0 \Rightarrow|h|=h) \\ & =1 \end{aligned} \end{aligned} $

Since the left and right hand limits of $f$ at $x=1$ are not equal, $f$ is not differentiable at $x$ $=1$

Solution

The given function $f$ is $f(x)=[x], 0<x<3$

It is known that a function $f$ is differentiable at a point $x=c$ in its domain if both

$\lim _{h \to 0^{-}} \frac{f(c+h)-f(c)}{h}$ and $\lim _{h \to 0^{+}} \frac{f(c+h)-f(c)}{h}$ are finite and equal.

To check the differentiability of the given function at $x=1$, consider the left hand limit of $f$ at $x=1$

$ \begin{aligned} & \lim _{h \to 0^{-}} \frac{f(1+h)-f(1)}{h}=\lim _{h \to 0^{-}} \frac{[1+h]-[1]}{h} \\ & =\lim _{h \to 0^{-}} \frac{0-1}{h}=\lim _{h \to 0^{-}} \frac{-1}{h}=\infty \end{aligned} $

Consider the right hand limit of $f$ at $x=1$

$\lim _{h \to 0^{+}} \frac{f(1+h)-f(1)}{h}=\lim _{h \to 0^{+}} \frac{[1+h]-[1]}{h}$

$=\lim _{h \to 0^{+}} \frac{1-1}{h}=\lim _{h \to 0^{+}} 0=0$

Since the left and right hand limits of $f$ at $x=1$ are not equal, $f$ is not differentiable at $x=1$

To check the differentiability of the given function at $x=2$, consider the left hand limit of $f$ at $x=2$

$\lim _{h \to 0^{-}} \frac{f(2+h)-f(2)}{h}=\lim _{h \to 0^{-}} \frac{[2+h]-[2]}{h}$

$=\lim _{h \to 0^{-}} \frac{1-2}{h}=\lim _{h \to 0^{-}} \frac{-1}{h}=\infty$

Consider the right hand limit of $f$ at $x=1$

$\lim _{h \to 0^{+}} \frac{f(2+h)-f(2)}{h}=\lim _{h \to 0^{+}} \frac{[2+h]-[2]}{h}$

$=\lim _{h \to 0^{+}} \frac{2-2}{h}=\lim _{h \to 0^{+}} 0=0$

Since the left and right hand limits of $f$ at $x=2$ are not equal, $f$ is not differentiable at $x$ $=2$

Solution

The given relationship is $2 x+3 y=\sin x$

Differentiating this relationship with respect to $x$, we obtain

$\frac{d}{d x}(2 x+3 y)=\frac{d}{d x}(\sin x)$

$\Rightarrow \frac{d}{d x}(2 x)+\frac{d}{d x}(3 y)=\cos x$

$\Rightarrow 2+3 \frac{d y}{d x}=\cos x$

$\Rightarrow 3 \frac{d y}{d x}=\cos x-2$

$\therefore \frac{d y}{d x}=\frac{\cos x-2}{3}$

Solution

The given relationship is $2 x+3 y=\sin y$

Differentiating this relationship with respect to $x$, we obtain

$\frac{d}{d x}(2 x)+\frac{d}{d x}(3 y)=\frac{d}{d x}(\sin y)$ $\Rightarrow 2+3 \frac{d y}{d x}=\cos y \frac{d y}{d x} \quad$ [By using chain rule]

$\Rightarrow 2=(\cos y-3) \frac{d y}{d x}$

$\therefore \frac{d y}{d x}=\frac{2}{\cos y-3}$

Solution

The given relationship is $a x+b y^{2}=\cos y$

Differentiating this relationship with respect to $x$, we obtain

$\frac{d}{d x}(a x)+\frac{d}{d x}(b y^{2})=\frac{d}{d x}(\cos y)$

$\Rightarrow a+b \frac{d}{d x}(y^{2})=\frac{d}{d x}(\cos y)$

Using chain rule, we obtain $\frac{d}{d x}(y^{2})=2 y \frac{d y}{d x}$ and $\frac{d}{d x}(\cos y)=-\sin y \frac{d y}{d x}$

From (1) and (2), we obtain

$a+b \times 2 y \frac{d y}{d x}=-\sin y \frac{d y}{d x}$

$\Rightarrow(2 b y+\sin y) \frac{d y}{d x}=-a$

$\therefore \frac{d y}{d x}=\frac{-a}{2 b y+\sin y}$

Solution

The given relationship is $x y+y^{2}=\tan x+y$

Differentiating this relationship with respect to $x$, we obtain

$\frac{d}{d x}(x y+y^{2})=\frac{d}{d x}(\tan x+y)$

$\Rightarrow \frac{d}{d x}(x y)+\frac{d}{d x}(y^{2})=\frac{d}{d x}(\tan x)+\frac{d y}{d x}$

$\Rightarrow[y \cdot \frac{d}{d x}(x)+x \cdot \frac{d y}{d x}]+2 y \frac{d y}{d x}=\sec ^{2} x+\frac{d y}{d x} \quad$ [Using product rule and chain rule]

$\Rightarrow y \cdot 1+x \cdot \frac{d y}{d x}+2 y \frac{d y}{d x}=\sec ^{2} x+\frac{d y}{d x}$

$\Rightarrow(x+2 y-1) \frac{d y}{d x}=\sec ^{2} x-y$

$\therefore \frac{d y}{d x}=\frac{\sec ^{2} x-y}{(x+2 y-1)}$

Solution

The given relationship is $x^{2}+x y+y^{2}=100$

Differentiating this relationship with respect to $x$, we obtain

$ \begin{aligned} & \frac{d}{d x}(x^{2}+x y+y^{2})=\frac{d}{d x}(100) \\ & \Rightarrow \frac{d}{d x}(x^{2})+\frac{d}{d x}(x y)+\frac{d}{d x}(y^{2})=0 \end{aligned} $

$\Rightarrow 2 x+[y \cdot \frac{d}{d x}(x)+x \cdot \frac{d y}{d x}]+2 y \frac{d y}{d x}=0 \quad$ [Using product rule and chain rule]

$\Rightarrow 2 x+y \cdot 1+x \cdot \frac{d y}{d x}+2 y \frac{d y}{d x}=0$

$\Rightarrow 2 x+y+(x+2 y) \frac{d y}{d x}=0$

$\therefore \frac{d y}{d x}=-\frac{2 x+y}{x+2 y}$

Solution

The given relationship is $x^{3}+x^{2} y+x y^{2}+y^{3}=81$

Differentiating this relationship with respect to $x$, we obtain

$ \begin{aligned} & \frac{d}{d x}(x^{3}+x^{2} y+x y^{2}+y^{3})=\frac{d}{d x}(81) \\ & \Rightarrow \frac{d}{d x}(x^{3})+\frac{d}{d x}(x^{2} y)+\frac{d}{d x}(x y^{2})+\frac{d}{d x}(y^{3})=0 \\ & \Rightarrow 3 x^{2}+[y \frac{d}{d x}(x^{2})+x^{2} \frac{d y}{d x}]+[y^{2} \frac{d}{d x}(x)+x \frac{d}{d x}(y^{2})]+3 y^{2} \frac{d y}{d x}=0 \\ & \Rightarrow 3 x^{2}+[y \cdot 2 x+x^{2} \frac{d y}{d x}]+[y^{2} \cdot 1+x \cdot 2 y \cdot \frac{d y}{d x}]+3 y^{2} \frac{d y}{d x}=0 \\ & \Rightarrow(x^{2}+2 x y+3 y^{2}) \frac{d y}{d x}+(3 x^{2}+2 x y+y^{2})=0 \\ & \therefore \frac{d y}{d x}=\frac{-(3 x^{2}+2 x y+y^{2})}{(x^{2}+2 x y+3 y^{2})} \end{aligned} $

Solution

The given relationship is $\sin ^{2} y+\cos x y=\pi$

Differentiating this relationship with respect to $x$, we obtain

$$ \begin{align*} & \frac{d}{d x}(\sin ^{2} y+\cos x y)=\frac{d}{d x}(\pi) \\ & \Rightarrow \frac{d}{d x}(\sin ^{2} y)+\frac{d}{d x}(\cos x y)=0 \tag{1} \end{align*} $$

Using chain rule, we obtain

$$ \begin{align*} \frac{d}{d x}(\sin ^{2} y) & =2 \sin y \frac{d}{d x}(\sin y)=2 \sin y \cos y \frac{d y}{d x} \tag{2}\\ \frac{d}{d x}(\cos x y) & =-\sin x y \frac{d}{d x}(x y)=-\sin x y[y \frac{d}{d x}(x)+x \frac{d y}{d x}] \\ & =-\sin x y[y \cdot 1+x \frac{d y}{d x}]=-y \sin x y-x \sin x y \frac{d y}{d x} \tag{3} \end{align*} $$

From (1), (2), and (3), we obtain

$2 \sin y \cos y \frac{d y}{d x}-y \sin x y-x \sin x y \frac{d y}{d x}=0$

$\Rightarrow(2 \sin y \cos y-x \sin x y) \frac{d y}{d x}=y \sin x y$

$\Rightarrow(\sin 2 y-x \sin x y) \frac{d y}{d x}=y \sin x y$

$\therefore \frac{d y}{d x}=\frac{y \sin x y}{\sin 2 y-x \sin x y}$

Solution

The given relationship is $\sin ^{2} x+\cos ^{2} y=1$

Differentiating this relationship with respect to $x$, we obtain $\frac{d}{d x}(\sin ^{2} x+\cos ^{2} y)=\frac{d}{d x}(1)$

$\Rightarrow \frac{d}{d x}(\sin ^{2} x)+\frac{d}{d x}(\cos ^{2} y)=0$

$\Rightarrow 2 \sin x \cdot \frac{d}{d x}(\sin x)+2 \cos y \cdot \frac{d}{d x}(\cos y)=0$

$\Rightarrow 2 \sin x \cos x+2 \cos y(-\sin y) \cdot \frac{d y}{d x}=0$

$\Rightarrow \sin 2 x-\sin 2 y \frac{d y}{d x}=0$

$\therefore \frac{d y}{d x}=\frac{\sin 2 x}{\sin 2 y}$

Solution

The given relationship is $y=\sin ^{-1}(\frac{2 x}{1+x^{2}})$

$y=\sin ^{-1}(\frac{2 x}{1+x^{2}})$

$\Rightarrow \sin y=\frac{2 x}{1+x^{2}}$

Differentiating this relationship with respect to $x$, we obtain

$\frac{d}{d x}(\sin y)=\frac{d}{d x}(\frac{2 x}{1+x^{2}})$

$\Rightarrow \cos y \frac{d y}{d x}=\frac{d}{d x}(\frac{2 x}{1+x^{2}})$

The function, $\frac{2 x}{1+x^{2}}$, is of the form of $\frac{u}{v}$.

Therefore, by quotient rule, we obtain

$$ \begin{align*} & \frac{d}{d x}(\frac{2 x}{1+x^{2}})=\frac{(1+x^{2}) \cdot \frac{d}{d x}(2 x)-2 x \cdot \frac{d}{d x}(1+x^{2})}{(1+x^{2})^{2}} \\ & =\frac{(1+x^{2}) \cdot 2-2 x \cdot[0+2 x]}{(1+x^{2})^{2}}=\frac{2+2 x^{2}-4 x^{2}}{(1+x^{2})^{2}}=\frac{2(1-x^{2})}{(1+x^{2})^{2}} \tag{2} \end{align*} $$

Also, $\sin y=\frac{2 x}{1+x^{2}}$

$$ \begin{align*} \Rightarrow \cos y & =\sqrt{1-\sin ^{2} y}=\sqrt{1-(\frac{2 x}{1+x^{2}})^{2}}=\sqrt{\frac{(1+x^{2})^{2}-4 x^{2}}{(1+x^{2})^{2}}} \\ & =\sqrt{\frac{(1-x^{2})^{2}}{(1+x^{2})^{2}}}=\frac{1-x^{2}}{1+x^{2}} \tag{3} \end{align*} $$

From (1), (2), and (3), we obtain

$ \begin{aligned} & \frac{1-x^{2}}{1+x^{2}} \times \frac{d y}{d x}=\frac{2(1-x^{2})}{(1+x^{2})^{2}} \\ & \Rightarrow \frac{d y}{d x}=\frac{2}{1+x^{2}} \end{aligned} $

Solution

The given relationship is $y=\tan ^{-1}(\frac{3 x-x^{3}}{1-3 x^{2}})$ $y=\tan ^{-1}(\frac{3 x-x^{3}}{1-3 x^{2}})$

$\Rightarrow \tan y=\frac{3 x-x^{3}}{1-3 x^{2}}$

It is known that, $\tan y=\frac{3 \tan \frac{y}{3}-\tan ^{3} \frac{y}{3}}{1-3 \tan ^{2} \frac{y}{3}}$

Comparing equations (1) and (2), we obtain

$x=\tan \frac{y}{3}$

Differentiating this relationship with respect to $x$, we obtain

$\frac{d}{d x}(x)=\frac{d}{d x}(\tan \frac{y}{3})$

$\Rightarrow 1=\sec ^{2} \frac{y}{3} \cdot \frac{d}{d x}(\frac{y}{3})$

$\Rightarrow 1=\sec ^{2} \frac{y}{3} \cdot \frac{1}{3} \cdot \frac{d y}{d x}$

$\Rightarrow \frac{d y}{d x}=\frac{3}{\sec ^{2} \frac{y}{3}}=\frac{3}{1+\tan ^{2} \frac{y}{3}}$

$\therefore \frac{d y}{d x}=\frac{3}{1+x^{2}}$

Solution

The given relationship is, $y=\cos ^{-1}(\frac{1-x^{2}}{1+x^{2}})$

$\Rightarrow \cos y=\frac{1-x^{2}}{1+x^{2}}$

$\Rightarrow \frac{1-\tan ^{2} \frac{y}{2}}{1+\tan ^{2} \frac{y}{2}}=\frac{1-x^{2}}{1+x^{2}}$

On comparing L.H.S. and R.H.S. of the above relationship, we obtain

$\tan \frac{y}{2}=x$

Differentiating this relationship with respect to $x$, we obtain

$\sec ^{2} \frac{y}{2} \cdot \frac{d}{d x}(\frac{y}{2})=\frac{d}{d x}(x)$

$\Rightarrow \sec ^{2} \frac{y}{2} \times \frac{1}{2} \frac{d y}{d x}=1$

$\Rightarrow \frac{d y}{d x}=\frac{2}{\sec ^{2} \frac{y}{2}}$

$\Rightarrow \frac{d y}{d x}=\frac{2}{1+\tan ^{2} \frac{y}{2}}$

$\therefore \frac{d y}{d x}=\frac{1}{1+x^{2}}$

Solution

The given relationship is $y=\sin ^{-1}(\frac{1-x^{2}}{1+x^{2}})$ $y=\sin ^{-1}(\frac{1-x^{2}}{1+x^{2}})$

$\Rightarrow \sin y=\frac{1-x^{2}}{1+x^{2}}$

Differentiating this relationship with respect to $x$, we obtain

$\frac{d}{d x}(\sin y)=\frac{d}{d x}(\frac{1-x^{2}}{1+x^{2}})$

Using chain rule, we obtain

$ \frac{d}{d x}(\sin y)=\cos y \cdot \frac{d y}{d x} $

$ \begin{aligned} & \cos y=\sqrt{1-\sin ^{2} y}=\sqrt{1-(\frac{1-x^{2}}{1+x^{2}})^{2}} \\ & =\sqrt{\frac{(1+x^{2})^{2}-(1-x^{2})^{2}}{(1+x^{2})^{2}}}=\sqrt{\frac{4 x^{2}}{(1+x^{2})^{2}}}=\frac{2 x}{1+x^{2}} \\ & \therefore \frac{d}{d x}(\sin y)=\frac{2 x}{1+x^{2}} \frac{d y}{d x} \end{aligned} $

$\frac{d}{d x}(\frac{1-x^{2}}{1+x^{2}})=\frac{(1+x^{2}) \cdot(1-x^{2})^{\prime}-(1-x^{2}) \cdot(1+x^{2})^{\prime}}{(1+x^{2})^{2}}$

[Using quotient rule]

$=\frac{(1+x^{2})(-2 x)-(1-x^{2}) \cdot(2 x)}{(1+x^{2})^{2}}$

$=\frac{-2 x-2 x^{3}-2 x+2 x^{3}}{(1+x^{2})^{2}}$

$=\frac{-4 x}{(1+x^{2})^{2}}$

From (1), (2), and (3), we obtain

$ \begin{aligned} & \frac{2 x}{1+x^{2}} \frac{d y}{d x}=\frac{-4 x}{(1+x^{2})^{2}} \\ & \Rightarrow \frac{d y}{d x}=\frac{-2}{1+x^{2}} \end{aligned} $

Alternate method

$ \begin{aligned} & y=\sin ^{-1}(\frac{1-x^{2}}{1+x^{2}}) \\ & \Rightarrow \sin y=\frac{1-x^{2}}{1+x^{2}} \end{aligned} $

$\Rightarrow(1+x^{2}) \sin y=1-x^{2}$

$\Rightarrow(1+\sin y) x^{2}=1-\sin y$

$\Rightarrow x^{2}=\frac{1-\sin y}{1+\sin y}$

$\Rightarrow x^{2}=\frac{(\cos \frac{y}{2}-\sin \frac{y}{2})^{2}}{(\cos \frac{y}{2}+\sin \frac{y}{2})^{2}}$

$\Rightarrow x=\frac{\cos \frac{y}{2}-\sin \frac{y}{2}}{\cos \frac{y}{2}+\sin \frac{y}{2}}$

$\Rightarrow x=\frac{1-\tan \frac{y}{2}}{1+\tan \frac{y}{2}}$

$\Rightarrow x=\tan (\frac{\pi}{4}-\frac{y}{2})$

Differentiating this relationship with respect to $x$, we obtain

$ \begin{aligned} & \frac{d}{d x}(x)=\frac{d}{d x} \cdot[\tan (\frac{\pi}{4}-\frac{y}{2})] \\ & \Rightarrow 1=\sec ^{2}(\frac{\pi}{4}-\frac{y}{2}) \cdot \frac{d}{d x}(\frac{\pi}{4}-\frac{y}{2}) \\ & \Rightarrow 1=[1+\tan ^{2}(\frac{\pi}{4}-\frac{y}{2})] \cdot(-\frac{1}{2} \frac{d y}{d x}) \\ & \Rightarrow 1=(1+x^{2})(-\frac{1}{2} \frac{d y}{d x}) \\ & \Rightarrow \frac{d y}{d x}=\frac{-2}{1+x^{2}} \end{aligned} $

Solution

The given relationship is $y=\cos ^{-1}(\frac{2 x}{1+x^{2}})$

$ \begin{aligned} & y=\cos ^{-1}(\frac{2 x}{1+x^{2}}) \\ & \Rightarrow \cos y=\frac{2 x}{1+x^{2}} \end{aligned} $

Differentiating this relationship with respect to $x$, we obtain

$ \begin{aligned} & \frac{d}{d x}(\cos y)=\frac{d}{d x} \cdot(\frac{2 x}{1+x^{2}}) \\ & \Rightarrow-\sin y \cdot \frac{d y}{d x}=\frac{(1+x^{2}) \cdot \frac{d}{d x}(2 x)-2 x \cdot \frac{d}{d x}(1+x^{2})}{(1+x^{2})^{2}} \end{aligned} $

$ \begin{aligned} & \Rightarrow-\sqrt{1-\cos ^{2} y} \frac{d y}{d x}=\frac{(1+x^{2}) \times 2-2 x \cdot 2 x}{(1+x^{2})^{2}} \\ & \Rightarrow[\sqrt{1-(\frac{2 x}{1+x^{2}})^{2}}] \frac{d y}{d x}=-[\frac{2(1-x^{2})}{(1+x^{2})^{2}}] \\ & \Rightarrow \sqrt{\frac{(1+x^{2})^{2}-4 x^{2}}{(1+x^{2})^{2}}} \frac{d y}{d x}=\frac{-2(1-x^{2})}{(1+x^{2})^{2}} \\ & \Rightarrow \sqrt{\frac{(1-x^{2})^{2}}{(1+x^{2})^{2}}} \frac{d y}{d x}=\frac{-2(1-x^{2})}{(1+x^{2})^{2}} \\ & \Rightarrow \frac{1-x^{2}}{1+x^{2}} \cdot \frac{d y}{d x}=\frac{-2(1-x^{2})}{(1+x^{2})^{2}} \\ & \Rightarrow \frac{d y}{d x}=\frac{-2}{1+x^{2}} \end{aligned} $

Solution

The given relationship is $y=\sin ^{-1}(2 x \sqrt{1-x^{2}})$

$y=\sin ^{-1}(2 x \sqrt{1-x^{2}})$

$\Rightarrow \sin y=2 x \sqrt{1-x^{2}}$

Differentiating this relationship with respect to $x$, we obtain

$ \begin{aligned} & \cos y \frac{d y}{d x}=2[x \frac{d}{d x}(\sqrt{1-x^{2}})+\sqrt{1-x^{2}} \frac{d x}{d x}] \\ & \Rightarrow \sqrt{1-\sin ^{2} y} \frac{d y}{d x}=2[\frac{x}{2} \cdot \frac{-2 x}{\sqrt{1-x^{2}}}+\sqrt{1-x^{2}}] \\ & \Rightarrow \sqrt{1-(2 x \sqrt{1-x^{2}})^{2}} \frac{d y}{d x}=2[\frac{-x^{2}+1-x^{2}}{\sqrt{1-x^{2}}}] \\ & \Rightarrow \sqrt{1-4 x^{2}(1-x^{2})} \frac{d y}{d x}=2[\frac{1-2 x^{2}}{\sqrt{1-x^{2}}}] \\ & \Rightarrow \sqrt{(1-2 x^{2})^{2}} \frac{d y}{d x}=2[\frac{1-2 x^{2}}{\sqrt{1-x^{2}}}] \\ & \Rightarrow(1-2 x^{2}) \frac{d y}{d x}=2[\frac{1-2 x^{2}}{\sqrt{1-x^{2}}}] \\ & \Rightarrow \frac{d y}{d x}=\frac{2}{\sqrt{1-x^{2}}} \end{aligned} $

Solution

The given relationship is $y=\sec ^{-1}(\frac{1}{2 x^{2}-1})$

$ y=\sec ^{-1}(\frac{1}{2 x^{2}-1}) $

$\Rightarrow \sec y=\frac{1}{2 x^{2}-1}$

$\Rightarrow \cos y=2 x^{2}-1$

$\Rightarrow 2 x^{2}=1+\cos y$

$\Rightarrow 2 x^{2}=2 \cos ^{2} \frac{y}{2}$

$\Rightarrow x=\cos \frac{y}{2}$

Differentiating this relationship with respect to $x$, we obtain

$\frac{d}{d x}(x)=\frac{d}{d x}(\cos \frac{y}{2})$

$\Rightarrow 1=-\sin \frac{y}{2} \cdot \frac{d}{d x}(\frac{y}{2})$

$\Rightarrow \frac{-1}{\sin \frac{y}{2}}=\frac{1}{2} \frac{d y}{d x}$

$\Rightarrow \frac{d y}{d x}=\frac{-2}{\sin \frac{y}{2}}=\frac{-2}{\sqrt{1-\cos ^{2} \frac{y}{2}}}$

$\Rightarrow \frac{d y}{d x}=\frac{-2}{\sqrt{1-x^{2}}}$

Solution

Let $y=\frac{e^{x}}{\sin x}$

By using the quotient rule, we obtain

$ \begin{aligned} \frac{d y}{d x} & =\frac{\sin x \frac{d}{d x}(e^{x})-e^{x} \frac{d}{d x}(\sin x)}{\sin ^{2} x} \\ & =\frac{\sin x \cdot(e^{x})-e^{x} \cdot(\cos x)}{\sin ^{2} x} \\ & =\frac{e^{x}(\sin x-\cos x)}{\sin ^{2} x}, x \neq n \pi, n \in \mathbf{Z} \end{aligned} $

Solution

Let $y=e^{\sin ^{-1} x}$

By using the chain rule, we obtain

$ \begin{aligned} & \frac{d y}{d x}=\frac{d}{d x}(e^{\sin ^{-1} x}) \\ & \begin{aligned} \Rightarrow \frac{d y}{d x} & =e^{\sin ^{-1} x} \cdot \frac{d}{d x}(\sin ^{-1} x) \\ & =e^{\sin ^{-1} x} \cdot \frac{1}{\sqrt{1-x^{2}}} \\ & =\frac{e^{\sin ^{-1} x}}{\sqrt{1-x^{2}}} \end{aligned} \\ & \therefore \frac{d y}{d x}=\frac{e^{\sin ^{-1} x}}{\sqrt{1-x^{2}}}, x \in(-1,1) \end{aligned} $

Solution

Let $y=e^{x^{3}}$

By using the chain rule, we obtain

$\frac{d y}{d x}=\frac{d}{d x}(e^{x^{3}})=e^{x^{3}} \cdot \frac{d}{d x}(x^{3})=e^{x^{3}} \cdot 3 x^{2}=3 x^{2} e^{x^{3}}$

Solution

Let $y=\sin (\tan ^{-1} e^{-x})$

By using the chain rule, we obtain

$ \begin{aligned} \frac{d y}{d x} & =\frac{d}{d x}[\sin (\tan ^{-1} e^{-x})] \\ & =\cos (\tan ^{-1} e^{-x}) \cdot \frac{d}{d x}(\tan ^{-1} e^{-x}) \\ & =\cos (\tan ^{-1} e^{-x}) \cdot \frac{1}{1+(e^{-x})^{2}} \cdot \frac{d}{d x}(e^{-x}) \\ & =\frac{\cos (\tan ^{-1} e^{-x})}{1+e^{-2 x}} \cdot e^{-x} \cdot \frac{d}{d x}(-x) \\ & =\frac{e^{-x} \cos (\tan ^{-1} e^{-x})}{1+e^{-2 x}} \times(-1) \\ & =\frac{-e^{-x} \cos (\tan ^{-1} e^{-x})}{1+e^{-2 x}} \end{aligned} $

Solution

Let $y=\log (\cos e^{x})$

By using the chain rule, we obtain

$ \begin{aligned} \frac{d y}{d x} & =\frac{d}{d x}[\log (\cos e^{x})] \\ & =\frac{1}{\cos e^{x}} \cdot \frac{d}{d x}(\cos e^{x}) \\ & =\frac{1}{\cos e^{x}} \cdot(-\sin e^{x}) \cdot \frac{d}{d x}(e^{x}) \\ & =\frac{-\sin e^{x}}{\cos e^{x}} \cdot e^{x} \\ & =-e^{x} \tan e^{x}, e^{x} \neq(2 n+1) \frac{\pi}{2}, n \in \mathbf{N} \end{aligned} $

Solution

$\frac{d}{d x}(e^{x}+e^{x^{2}}+\ldots+e^{x^{3}})$

$=\frac{d}{d x}(e^{x})+\frac{d}{d x}(e^{x^{2}})+\frac{d}{d x}(e^{x^{3}})+\frac{d}{d x}(e^{x^{4}})+\frac{d}{d x}(e^{x^{3}})$

$=e^{x}+[e^{x^{2}} \times \frac{d}{d x}(x^{2})]+[e^{x^{3}} \cdot \frac{d}{d x}(x^{3})]+[e^{x^{4}} \cdot \frac{d}{d x}(x^{4})]+[e^{x^{5}} \cdot \frac{d}{d x}(x^{5})]$

$=e^{x}+(e^{x^{2}} \times 2 x)+(e^{x^{3}} \times 3 x^{2})+(e^{x^{4}} \times 4 x^{3})+(e^{x^{3}} \times 5 x^{4})$

$=e^{x}+2 x e^{x^{2}}+3 x^{2} e^{x^{3}}+4 x^{3} e^{x^{4}}+5 x^{4} e^{x^{5}}$

Solution

Let $y=\sqrt{e^{\sqrt{x}}}$

Then, $y^{2}=e^{\sqrt{x}}$

By differentiating this relationship with respect to $x$, we obtain

$ \begin{aligned} & y^{2}=e^{\sqrt{x}} \\ & \Rightarrow 2 y \frac{d y}{d x}=e^{\sqrt{x}} \frac{d}{d x}(\sqrt{x}) \quad \text{ [By applying the chain rule] } \\ & \Rightarrow 2 y \frac{d y}{d x}=e^{\sqrt{x}} \frac{1}{2} \cdot \frac{1}{\sqrt{x}} \\ & \Rightarrow \frac{d y}{d x}=\frac{e^{\sqrt{x}}}{4 y \sqrt{x}} \\ & \Rightarrow \frac{d y}{d x}=\frac{e^{\sqrt{x}}}{4 \sqrt{e^{\sqrt{x}}} \sqrt{x}} \\ & \Rightarrow \frac{d y}{d x}=\frac{e^{\sqrt{x}}}{4 \sqrt{x e^{\sqrt{x}}}}, x>0 \end{aligned} $

Solution

Let $y=\log (\log x)$

By using the chain rule, we obtain

$ \begin{aligned} \frac{d y}{d x} & =\frac{d}{d x}[\log (\log x)] \\ & =\frac{1}{\log x} \cdot \frac{d}{d x}(\log x) \\ & =\frac{1}{\log x} \cdot \frac{1}{x} \\ = & \frac{1}{x \log x}, x>1 \end{aligned} $

Solution

Let $y=\frac{\cos x}{\log x}$

By using the quotient rule, we obtain

$ \begin{aligned} \frac{d y}{d x} & =\frac{\frac{d}{d x}(\cos x) \times \log x-\cos x \times \frac{d}{d x}(\log x)}{(\log x)^{2}} \\ & =\frac{-\sin x \log x-\cos x \times \frac{1}{x}}{(\log x)^{2}} \\ & =\frac{-[x \log x \cdot \sin x+\cos x]}{x(\log x)^{2}}, x>0 \end{aligned} $

Solution

Let $y=\cos (\log x+e^{x})$

By using the chain rule, we obtain

$ \begin{aligned} \frac{d y}{d x} & =-\sin (\log x+e^{x}) \cdot \frac{d}{d x}(\log x+e^{x}) \\ & =-\sin (\log x+e^{x}) \cdot[\frac{d}{d x}(\log x)+\frac{d}{d x}(e^{x})] \\ & =-\sin (\log x+e^{x}) \cdot(\frac{1}{x}+e^{x}) \\ & =-(\frac{1}{x}+e^{x}) \sin (\log x+e^{x}), x>0 \end{aligned} $

Solution

Let $y=\cos x \cdot \cos 2 x \cdot \cos 3 x$

Taking logarithm on both the sides, we obtain

$\log y=\log (\cos x \cdot \cos 2 x \cdot \cos 3 x)$

$\Rightarrow \log y=\log (\cos x)+\log (\cos 2 x)+\log (\cos 3 x)$

Differentiating both sides with respect to $x$, we obtain

$ \begin{aligned} & \frac{1}{y} \frac{d y}{d x}=\frac{1}{\cos x} \cdot \frac{d}{d x}(\cos x)+\frac{1}{\cos 2 x} \cdot \frac{d}{d x}(\cos 2 x)+\frac{1}{\cos 3 x} \cdot \frac{d}{d x}(\cos 3 x) \\ & \Rightarrow \frac{d y}{d x}=y[-\frac{\sin x}{\cos x}-\frac{\sin 2 x}{\cos 2 x} \cdot \frac{d}{d x}(2 x)-\frac{\sin 3 x}{\cos 3 x} \cdot \frac{d}{d x}(3 x)] \\ & \therefore \frac{d y}{d x}=-\cos x \cdot \cos 2 x \cdot \cos 3 x[\tan x+2 \tan 2 x+3 \tan 3 x] \end{aligned} $

Solution

Let $y=\sqrt{\frac{(x-1)(x-2)}{(x-3)(x-4)(x-5)}}$

Taking logarithm on both the sides, we obtain

$ \begin{aligned} & \log y=\log \sqrt{\frac{(x-1)(x-2)}{(x-3)(x-4)(x-5)}} \\ & \Rightarrow \log y=\frac{1}{2} \log [\frac{(x-1)(x-2)}{(x-3)(x-4)(x-5)}] \\ & \Rightarrow \log y=\frac{1}{2}[\log {(x-1)(x-2)}-\log {(x-3)(x-4)(x-5)}] \\ & \Rightarrow \log y=\frac{1}{2}[\log (x-1)+\log (x-2)-\log (x-3)-\log (x-4)-\log (x-5)] \end{aligned} $

Differentiating both sides with respect to $x$, we obtain

$ \begin{aligned} & \frac{1}{y} \frac{d y}{d x}=\frac{1}{2} \begin{bmatrix} \frac{1}{x-1} \cdot \frac{d}{d x}(x-1)+\frac{1}{x-2} \cdot \frac{d}{d x}(x-2)-\frac{1}{x-3} \cdot \frac{d}{d x}(x-3) \\ -\frac{1}{x-4} \cdot \frac{d}{d x}(x-4)-\frac{1}{x-5} \cdot \frac{d}{d x}(x-5) \end{bmatrix} \\ & \Rightarrow \frac{d y}{d x}=\frac{y}{2}(\frac{1}{x-1}+\frac{1}{x-2}-\frac{1}{x-3}-\frac{1}{x-4}-\frac{1}{x-5}) \\ & \therefore \frac{d y}{d x}=\frac{1}{2} \sqrt{\frac{(x-1)(x-2)}{(x-3)(x-4)(x-5)}}[\frac{1}{x-1}+\frac{1}{x-2}-\frac{1}{x-3}-\frac{1}{x-4}-\frac{1}{x-5}] \end{aligned} $

Solution

Let $y=(\log x)^{\cos x}$

Taking logarithm on both the sides, we obtain

$\log y=\cos x \cdot \log (\log x)$

Differentiating both sides with respect to $x$, we obtain

$ \begin{aligned} & \frac{1}{y} \cdot \frac{d y}{d x}=\frac{d}{d x}(\cos x) \times \log (\log x)+\cos x \times \frac{d}{d x}[\log (\log x)] \\ & \Rightarrow \frac{1}{y} \cdot \frac{d y}{d x}=-\sin x \log (\log x)+\cos x \times \frac{1}{\log x} \cdot \frac{d}{d x}(\log x) \\ & \Rightarrow \frac{d y}{d x}=y[-\sin x \log (\log x)+\frac{\cos x}{\log x} \times \frac{1}{x}] \\ & \therefore \frac{d y}{d x}=(\log x)^{\cos x}[\frac{\cos x}{x \log x}-\sin x \log (\log x)] \end{aligned} $

Solution

Let $y=x^{x}-2^{\sin x}$

Also, let $x^{x}=u$ and $2^{\sin x}=v$

$\therefore y=u-v$

$\Rightarrow \frac{d y}{d x}=\frac{d u}{d x}-\frac{d v}{d x}$

$u=x^{x}$

Taking logarithm on both the sides, we obtain

$\log u=x \log x$

Differentiating both sides with respect to $x$, we obtain

$\frac{1}{u} \frac{d u}{d x}=[\frac{d}{d x}(x) \times \log x+x \times \frac{d}{d x}(\log x)]$

$\Rightarrow \frac{d u}{d x}=u[1 \times \log x+x \times \frac{1}{x}]$

$\Rightarrow \frac{d u}{d x}=x^{x}(\log x+1)$

$\Rightarrow \frac{d u}{d x}=x^{x}(1+\log x)$

$v=2^{\sin x}$

Taking logarithm on both the sides with respect to $x$, we obtain $\log v=\sin x \cdot \log 2$

Differentiating both sides with respect to $x$, we obtain

$\frac{1}{v} \cdot \frac{d v}{d x}=\log 2 \cdot \frac{d}{d x}(\sin x)$

$\Rightarrow \frac{d v}{d x}=v \log 2 \cos x$

$\Rightarrow \frac{d v}{d x}=2^{\sin x} \cos x \log 2$

$\therefore \frac{d y}{d x}=x^{x}(1+\log x)-2^{\sin x} \cos x \log 2$

Solution

Let $y=(x+3)^{2} \cdot(x+4)^{3} \cdot(x+5)^{4}$

Taking logarithm on both the sides, we obtain

$ \begin{aligned} & \log y=\log (x+3)^{2}+\log (x+4)^{3}+\log (x+5)^{4} \\ & \Rightarrow \log y=2 \log (x+3)+3 \log (x+4)+4 \log (x+5) \end{aligned} $

Differentiating both sides with respect to $x$, we obtain

$ \begin{aligned} & \frac{1}{y} \cdot \frac{d y}{d x}=2 \cdot \frac{1}{x+3} \cdot \frac{d}{d x}(x+3)+3 \cdot \frac{1}{x+4} \cdot \frac{d}{d x}(x+4)+4 \cdot \frac{1}{x+5} \cdot \frac{d}{d x}(x+5) \\ & \Rightarrow \frac{d y}{d x}=y[\frac{2}{x+3}+\frac{3}{x+4}+\frac{4}{x+5}] \\ & \Rightarrow \frac{d y}{d x}=(x+3)^{2}(x+4)^{3}(x+5)^{4} \cdot[\frac{2}{x+3}+\frac{3}{x+4}+\frac{4}{x+5}] \\ & \Rightarrow \frac{d y}{d x}=(x+3)^{2}(x+4)^{3}(x+5)^{4} \cdot[\frac{2(x+4)(x+5)+3(x+3)(x+5)+4(x+3)(x+4)}{(x+3)(x+4)(x+5)}] \\ & \Rightarrow \frac{d y}{d x}=(x+3)(x+4)^{2}(x+5)^{3} \cdot[2(x^{2}+9 x+20)+3(x^{2}+8 x+15)+4(x^{2}+7 x+12)] \\ & \therefore \frac{d y}{d x}=(x+3)(x+4)^{2}(x+5)^{3}(9 x^{2}+70 x+133) \end{aligned} $

Solution

Let $y=(x+\frac{1}{x})^{x}+x^{(1+\frac{1}{x})}$

Also, let $u=(x+\frac{1}{x})^{x}$ and $v=x^{(1+\frac{1}{x})}$

$\therefore y=u+v$

$\Rightarrow \frac{d y}{d x}=\frac{d u}{d x}+\frac{d v}{d x}$

Then, $u=(x+\frac{1}{x})^{x}$

$\Rightarrow \log u=\log (x+\frac{1}{x})^{x}$

$\Rightarrow \log u=x \log (x+\frac{1}{x})$

Differentiating both sides with respect to $x$, we obtain $\frac{1}{u} \cdot \frac{d u}{d x}=\frac{d}{d x}(x) \times \log (x+\frac{1}{x})+x \times \frac{d}{d x}[\log (x+\frac{1}{x})]$

$\Rightarrow \frac{1}{u} \frac{d u}{d x}=1 \times \log (x+\frac{1}{x})+x \times \frac{1}{(x+\frac{1}{x})} \cdot \frac{d}{d x}(x+\frac{1}{x})$

$\Rightarrow \frac{d u}{d x}=u[\log (x+\frac{1}{x})+\frac{x}{(x+\frac{1}{x})} \times(1-\frac{1}{x^{2}})]$

$\Rightarrow \frac{d u}{d x}=(x+\frac{1}{x})^{x}[\log (x+\frac{1}{x})+\frac{(x-\frac{1}{x})}{(x+\frac{1}{x})}]$

$\Rightarrow \frac{d u}{d x}=(x+\frac{1}{x})^{x}[\log (x+\frac{1}{x})+\frac{x^{2}-1}{x^{2}+1}]$

$\Rightarrow \frac{d u}{d x}=(x+\frac{1}{x})^{x}[\frac{x^{2}-1}{x^{2}+1}+\log (x+\frac{1}{x})]$

$v=x^{(1+\frac{1}{x})}$

$\Rightarrow \log v=\log [x^{(1+\frac{1}{x})}]$

$\Rightarrow \log v=(1+\frac{1}{x}) \log x$

Differentiating both sides with respect to $x$, we obtain $\frac{1}{v} \cdot \frac{d v}{d x}=[\frac{d}{d x}(1+\frac{1}{x})] \times \log x+(1+\frac{1}{x}) \cdot \frac{d}{d x} \log x$

$\Rightarrow \frac{1}{v} \frac{d v}{d x}=(-\frac{1}{x^{2}}) \log x+(1+\frac{1}{x}) \cdot \frac{1}{x}$

$\Rightarrow \frac{1}{v} \frac{d v}{d x}=-\frac{\log x}{x^{2}}+\frac{1}{x}+\frac{1}{x^{2}}$

$\Rightarrow \frac{d v}{d x}=v[\frac{-\log x+x+1}{x^{2}}]$

$\Rightarrow \frac{d v}{d x}=x^{(1+\frac{1}{x})}(\frac{x+1-\log x}{x^{2}})$

Therefore, from (1), (2), and (3), we obtain

$\frac{d y}{d x}=(x+\frac{1}{x})^{x}[\frac{x^{2}-1}{x^{2}+1}+\log (x+\frac{1}{x})]+x^{(1+\frac{1}{x})}(\frac{x+1-\log x}{x^{2}})$

Solution

Let $y=(\log x)^{x}+x^{\log x}$

Also, let $u=(\log x)^{x}$ and $v=x^{\log x}$

$\therefore y=u+v$

$\Rightarrow \frac{d y}{d x}=\frac{d u}{d x}+\frac{d v}{d x}$

$u=(\log x)^{x}$

$\Rightarrow \log u=\log [(\log x)^{x}]$

$\Rightarrow \log u=x \log (\log x)$

Differentiating both sides with respect to $x$, we obtain $\frac{1}{u} \frac{d u}{d x}=\frac{d}{d x}(x) \times \log (\log x)+x \cdot \frac{d}{d x}[\log (\log x)]$

$\Rightarrow \frac{d u}{d x}=u[1 \times \log (\log x)+x \cdot \frac{1}{\log x} \cdot \frac{d}{d x}(\log x)]$

$\Rightarrow \frac{d u}{d x}=(\log x)^{x}[\log (\log x)+\frac{x}{\log x} \cdot \frac{1}{x}]$

$\Rightarrow \frac{d u}{d x}=(\log x)^{x}[\log (\log x)+\frac{1}{\log x}]$

$\Rightarrow \frac{d u}{d x}=(\log x)^{x}[\frac{\log (\log x) \cdot \log x+1}{\log x}]$

$\Rightarrow \frac{d u}{d x}=(\log x)^{x-1}[1+\log x \cdot \log (\log x)]$

$v=x^{\log x}$

$\Rightarrow \log v=\log (x^{\log x})$

$\Rightarrow \log v=\log x \log x=(\log x)^{2}$

Differentiating both sides with respect to $x$, we obtain

$\frac{1}{v} \cdot \frac{d v}{d x}=\frac{d}{d x}[(\log x)^{2}]$

$\Rightarrow \frac{1}{v} \cdot \frac{d v}{d x}=2(\log x) \cdot \frac{d}{d x}(\log x)$

$\Rightarrow \frac{d v}{d x}=2 v(\log x) \cdot \frac{1}{x}$

$\Rightarrow \frac{d v}{d x}=2 x^{\log x} \frac{\log x}{x}$

$\Rightarrow \frac{d v}{d x}=2 x^{\log x-1} \cdot \log x$

Therefore, from (1), (2), and (3), we obtain

$\frac{d y}{d x}=(\log x)^{x-1}[1+\log x \cdot \log (\log x)]+2 x^{\log x-1} \cdot \log x$

Solution

Let $y=(\sin x)^{x}+\sin ^{-1} \sqrt{x}$

Also, let $u=(\sin x)^{x}$ and $v=\sin ^{-1} \sqrt{x}$

$\therefore y=u+v$

$\Rightarrow \frac{d y}{d x}=\frac{d u}{d x}+\frac{d v}{d x}$

$u=(\sin x)^{x}$

$\Rightarrow \log u=\log (\sin x)^{x}$

$\Rightarrow \log u=x \log (\sin x)$

Differentiating both sides with respect to $x$, we obtain

$\Rightarrow \frac{1}{u} \frac{d u}{d x}=\frac{d}{d x}(x) \times \log (\sin x)+x \times \frac{d}{d x}[\log (\sin x)]$

$\Rightarrow \frac{d u}{d x}=u[1 \cdot \log (\sin x)+x \cdot \frac{1}{\sin x} \cdot \frac{d}{d x}(\sin x)]$

$\Rightarrow \frac{d u}{d x}=(\sin x)^{x}[\log (\sin x)+\frac{x}{\sin x} \cdot \cos x]$

$\Rightarrow \frac{d u}{d x}=(\sin x)^{x}(x \cot x+\log \sin x)$

$v=\sin ^{-1} \sqrt{x}$

Differentiating both sides with respect to $x$, we obtain

$\frac{d v}{d x}=\frac{1}{\sqrt{1-(\sqrt{x})^{2}}} \cdot \frac{d}{d x}(\sqrt{x})$

$\Rightarrow \frac{d v}{d x}=\frac{1}{\sqrt{1-x}} \cdot \frac{1}{2 \sqrt{x}}$

$\Rightarrow \frac{d v}{d x}=\frac{1}{2 \sqrt{x-x^{2}}}$

Therefore, from (1), (2), and (3), we obtain

$\frac{d y}{d x}=(\sin x)^{x}(x \cot x+\log \sin x)+\frac{1}{2 \sqrt{x-x^{2}}}$

Solution

Let $y=x^{\sin x}+(\sin x)^{\cos x}$

Also, let $u=x^{\sin x}$ and $v=(\sin x)^{\cos x}$

$\therefore y=u+v$

$\Rightarrow \frac{d y}{d x}=\frac{d u}{d x}+\frac{d v}{d x}$

$u=x^{\sin x}$

$\Rightarrow \log u=\log (x^{\sin x})$

$\Rightarrow \log u=\sin x \log x$

Differentiating both sides with respect to $x$, we obtain

$\frac{1}{u} \frac{d u}{d x}=\frac{d}{d x}(\sin x) \cdot \log x+\sin x \cdot \frac{d}{d x}(\log x)$

$\Rightarrow \frac{d u}{d x}=u[\cos x \log x+\sin x \cdot \frac{1}{x}]$

$\Rightarrow \frac{d u}{d x}=x^{\sin x}[\cos x \log x+\frac{\sin x}{x}]$

$v=(\sin x)^{\cos x}$

$\Rightarrow \log v=\log (\sin x)^{\cos x}$

$\Rightarrow \log v=\cos x \log (\sin x)$

Differentiating both sides with respect to $x$, we obtain

$\frac{1}{v} \frac{d v}{d x}=\frac{d}{d x}(\cos x) \times \log (\sin x)+\cos x \times \frac{d}{d x}[\log (\sin x)]$

$\Rightarrow \frac{d v}{d x}=v[-\sin x \cdot \log (\sin x)+\cos x \cdot \frac{1}{\sin x} \cdot \frac{d}{d x}(\sin x)]$

$\Rightarrow \frac{d v}{d x}=(\sin x)^{\cos x}[-\sin x \log \sin x+\frac{\cos x}{\sin x} \cos x]$

$\Rightarrow \frac{d v}{d x}=(\sin x)^{\cos x}[-\sin x \log \sin x+\cot x \cos x]$

$\Rightarrow \frac{d v}{d x}=(\sin x)^{\cos x}[\cot x \cos x-\sin x \log \sin x]$

From (1), (2), and (3), we obtain

$\frac{d y}{d x}=x^{\sin x}(\cos x \log x+\frac{\sin x}{x})+(\sin x)^{\cos x}[\cos x \cot x-\sin x \log \sin x]$

Solution

Let $y=x^{x \cos x}+\frac{x^{2}+1}{x^{2}-1}$

Also, let $u=x^{x \cos x}$ and $v=\frac{x^{2}+1}{x^{2}-1}$

$\therefore y=u+v$

$\Rightarrow \frac{d y}{d x}=\frac{d u}{d x}+\frac{d v}{d x}$

$u=x^{x \cos x}$

$\Rightarrow \log u=\log (x^{x \cos x})$

$\Rightarrow \log u=x \cos x \log x$

Differentiating both sides with respect to $x$, we obtain

$\frac{1}{u} \frac{d u}{d x}=\frac{d}{d x}(x) \cdot \cos x \cdot \log x+x \cdot \frac{d}{d x}(\cos x) \cdot \log x+x \cos x \cdot \frac{d}{d x}(\log x)$

$\Rightarrow \frac{d u}{d x}=u[1 \cdot \cos x \cdot \log x+x \cdot(-\sin x) \log x+x \cos x \cdot \frac{1}{x}]$

$\Rightarrow \frac{d u}{d x}=x^{x \cos x}(\cos x \log x-x \sin x \log x+\cos x)$

$\Rightarrow \frac{d u}{d x}=x^{x \cos x}[\cos x(1+\log x)-x \sin x \log x]$

$v=\frac{x^{2}+1}{x^{2}-1}$

$\Rightarrow \log v=\log (x^{2}+1)-\log (x^{2}-1)$

Differentiating both sides with respect to $x$, we obtain

$\frac{1}{v} \frac{d v}{d x}=\frac{2 x}{x^{2}+1}-\frac{2 x}{x^{2}-1}$

$\Rightarrow \frac{d v}{d x}=v[\frac{2 x(x^{2}-1)-2 x(x^{2}+1)}{(x^{2}+1)(x^{2}-1)}]$

$\Rightarrow \frac{d v}{d x}=\frac{x^{2}+1}{x^{2}-1} \times[\frac{-4 x}{(x^{2}+1)(x^{2}-1)}]$

$\Rightarrow \frac{d v}{d x}=\frac{-4 x}{(x^{2}-1)^{2}}$

From (1), (2), and (3), we obtain

$\frac{d y}{d x}=x^{x \cos x}[\cos x(1+\log x)-x \sin x \log x]-\frac{4 x}{(x^{2}-1)^{2}}$

Solution

Let $y=(x \cos x)^{x}+(x \sin x)^{\frac{1}{x}}$

Also, let $u=(x \cos x)^{x}$ and $v=(x \sin x)^{\frac{1}{x}}$

$\therefore y=u+v$

$\Rightarrow \frac{d y}{d x}=\frac{d u}{d x}+\frac{d v}{d x}$

$u=(x \cos x)^{x}$

$\Rightarrow \log u=\log (x \cos x)^{x}$

$\Rightarrow \log u=x \log (x \cos x)$

$\Rightarrow \log u=x[\log x+\log \cos x]$

$\Rightarrow \log u=x \log x+x \log \cos x$

Differentiating both sides with respect to $x$, we obtain

$\frac{1}{u} \frac{d u}{d x}=\frac{d}{d x}(x \log x)+\frac{d}{d x}(x \log \cos x)$

$\Rightarrow \frac{d u}{d x}=u[{\log x \cdot \frac{d}{d x}(x)+x \cdot \frac{d}{d x}(\log x)}+{\log \cos x \cdot \frac{d}{d x}(x)+x \cdot \frac{d}{d x}(\log \cos x)}]$

$\Rightarrow \frac{d u}{d x}=(x \cos x)^{x}[(\log x \cdot 1+x \cdot \frac{1}{x})+{\log \cos x \cdot 1+x \cdot \frac{1}{\cos x} \cdot \frac{d}{d x}(\cos x)}]$

$\Rightarrow \frac{d u}{d x}=(x \cos x)^{x}[(\log x+1)+{\log \cos x+\frac{x}{\cos x} \cdot(-\sin x)}]$

$\Rightarrow \frac{d u}{d x}=(x \cos x)^{x}[(1+\log x)+(\log \cos x-x \tan x)]$

$\Rightarrow \frac{d u}{d x}=(x \cos x)^{x}[1-x \tan x+(\log x+\log \cos x)]$

$\Rightarrow \frac{d u}{d x}=(x \cos x)^{x}[1-x \tan x+\log (x \cos x)]$

$ \begin{aligned} & v=(x \sin x)^{\frac{1}{x}} \\ & \Rightarrow \log v=\log (x \sin x)^{\frac{1}{x}} \\ & \Rightarrow \log v=\frac{1}{x} \log (x \sin x) \\ & \Rightarrow \log v=\frac{1}{x}(\log x+\log \sin x) \\ & \Rightarrow \log v=\frac{1}{x} \log x+\frac{1}{x} \log \sin x \end{aligned} $

Differentiating both sides with respect to $x$, we obtain

$$ \begin{align*} & \frac{1}{v} \frac{d v}{d x}=\frac{d}{d x}(\frac{1}{x} \log x)+\frac{d}{d x}[\frac{1}{x} \log (\sin x)] \\ & \Rightarrow \frac{1}{v} \frac{d v}{d x}=[\log x \cdot \frac{d}{d x}(\frac{1}{x})+\frac{1}{x} \cdot \frac{d}{d x}(\log x)]+[\log (\sin x) \cdot \frac{d}{d x}(\frac{1}{x})+\frac{1}{x} \cdot \frac{d}{d x}{\log (\sin x)}] \\ & \Rightarrow \frac{1}{v} \frac{d v}{d x}=[\log x \cdot(-\frac{1}{x^{2}})+\frac{1}{x}+\frac{1}{x}]+[\log (\sin x) \cdot(-\frac{1}{x^{2}})+\frac{1}{x} \cdot \frac{1}{\sin x} \cdot \frac{d}{d x}(\sin x)] \\ & \Rightarrow \frac{1}{v} \frac{d v}{d x}=\frac{1}{x^{2}}(1-\log x)+[-\frac{\log (\sin x)}{x^{2}}+\frac{1}{x \sin x} \cdot \cos x] \\ & \Rightarrow \frac{d v}{d x}=(x \sin x)^{\frac{1}{x}}[\frac{1-\log x}{x^{2}}+\frac{-\log (\sin x)+x \cot x}{x^{2}}] \\ & \Rightarrow \frac{d v}{d x}=(x \sin x)^{\frac{1}{x}}[\frac{1-\log x-\log (\sin x)+x \cot x}{x^{2}}] \\ & \Rightarrow \frac{d v}{d x}=(x \sin x)^{\frac{1}{x}}[\frac{1-\log (x \sin x)+x \cot x}{x^{2}}] \tag{3} \end{align*} $$

From (1), (2), and (3), we obtain

$\frac{d y}{d x}=(x \cos x)^{x}[1-x \tan x+\log (x \cos x)]+(x \sin x)^{\frac{1}{x}}[\frac{x \cot x+1-\log (x \sin x)}{x^{2}}]$

Solution

The given function is $x^{y}+y^{x}=1$

Let $x^{y}=u$ and $y^{x}=v$

Then, the function becomes $u+v=1$

$\therefore \frac{d u}{d x}+\frac{d v}{d x}=0$

$u=x^{y}$

$\Rightarrow \log u=\log (x^{y})$

$\Rightarrow \log u=y \log x$

Differentiating both sides with respect to $x$, we obtain

$\frac{1}{u} \frac{d u}{d x}=\log x \frac{d y}{d x}+y \cdot \frac{d}{d x}(\log x)$

$\Rightarrow \frac{d u}{d x}=u[\log x \frac{d y}{d x}+y \cdot \frac{1}{x}]$

$\Rightarrow \frac{d u}{d x}=x^{y}(\log x \frac{d y}{d x}+\frac{y}{x})$

$v=y^{x}$

$\Rightarrow \log v=\log (y^{x})$

$\Rightarrow \log v=x \log y$

Differentiating both sides with respect to $x$, we obtain

$\frac{1}{v} \cdot \frac{d v}{d x}=\log y \cdot \frac{d}{d x}(x)+x \cdot \frac{d}{d x}(\log y)$

$\Rightarrow \frac{d v}{d x}=v(\log y \cdot 1+x \cdot \frac{1}{y} \cdot \frac{d y}{d x})$

$\Rightarrow \frac{d v}{d x}=y^{x}(\log y+\frac{x}{y} \frac{d y}{d x})$

From (1), (2), and (3), we obtain

$ \begin{aligned} & x^{y}(\log x \frac{d y}{d x}+\frac{y}{x})+y^{x}(\log y+\frac{x}{y} \frac{d y}{d x})=0 \\ & \Rightarrow(x^{y} \log x+x y^{x-1}) \frac{d y}{d x}=-(y x^{y-1}+y^{x} \log y) \\ & \therefore \frac{d y}{d x}=-\frac{y x^{y-1}+y^{x} \log y}{x^{y} \log x+x y^{x-1}} \end{aligned} $

Solution

The given function is $(\cos x)^{y}=(\cos y)^{x}$

Taking logarithm on both the sides, we obtain

$y \log \cos x=x \log \cos y$

Differentiating both sides, we obtain

$\log \cos x \cdot \frac{d y}{d x}+y \cdot \frac{d}{d x}(\log \cos x)=\log \cos y \cdot \frac{d}{d x}(x)+x \cdot \frac{d}{d x}(\log \cos y)$

$\Rightarrow \log \cos x \frac{d y}{d x}+y \cdot \frac{1}{\cos x} \cdot \frac{d}{d x}(\cos x)=\log \cos y \cdot 1+x \cdot \frac{1}{\cos y} \cdot \frac{d}{d x}(\cos y)$

$\Rightarrow \log \cos x \frac{d y}{d x}+\frac{y}{\cos x} \cdot(-\sin x)=\log \cos y+\frac{x}{\cos y}(-\sin y) \cdot \frac{d y}{d x}$

$\Rightarrow \log \cos x \frac{d y}{d x}-y \tan x=\log \cos y-x \tan y \frac{d y}{d x}$

$\Rightarrow(\log \cos x+x \tan y) \frac{d y}{d x}=y \tan x+\log \cos y$

$\therefore \frac{d y}{d x}=\frac{y \tan x+\log \cos y}{x \tan y+\log \cos x}$

Solution

The given function is $y^{x}=x^{y}$

Taking logarithm on both the sides, we obtain

$x \log y=y \log x$

Differentiating both sides with respect to $x$, we obtain

$ \begin{aligned} & \log y \cdot \frac{d}{d x}(x)+x \cdot \frac{d}{d x}(\log y)=\log x \cdot \frac{d}{d x}(y)+y \cdot \frac{d}{d x}(\log x) \\ & \Rightarrow \log y \cdot 1+x \cdot \frac{1}{y} \cdot \frac{d y}{d x}=\log x \cdot \frac{d y}{d x}+y \cdot \frac{1}{x} \\ & \Rightarrow \log y+\frac{x}{y} \frac{d y}{d x}=\log x \frac{d y}{d x}+\frac{y}{x} \\ & \Rightarrow(\frac{x}{y}-\log x) \frac{d y}{d x}=\frac{y}{x}-\log y \\ & \Rightarrow(\frac{x-y \log x}{y}) \frac{d y}{d x}=\frac{y-x \log y}{x} \\ & \therefore \frac{d y}{d x}=\frac{y}{x}(\frac{y-x \log y}{x-y \log x}) \end{aligned} $

Solution

The given function is $x y=e^{(x-y)}$

Taking logarithm on both the sides, we obtain

$ \begin{aligned} & \log (x y)=\log (e^{x-y}) \\ & \Rightarrow \log x+\log y=(x-y) \log e \\ & \Rightarrow \log x+\log y=(x-y) \times 1 \\ & \Rightarrow \log x+\log y=x-y \end{aligned} $

Differentiating both sides with respect to $x$, we obtain

$\frac{d}{d x}(\log x)+\frac{d}{d x}(\log y)=\frac{d}{d x}(x)-\frac{d y}{d x}$

$\Rightarrow \frac{1}{x}+\frac{1}{y} \frac{d y}{d x}=1-\frac{d y}{d x}$

$\Rightarrow(1+\frac{1}{y}) \frac{d y}{d x}=1-\frac{1}{x}$

$\Rightarrow(\frac{y+1}{y}) \frac{d y}{d x}=\frac{x-1}{x}$

$\therefore \frac{d y}{d x}=\frac{y(x-1)}{x(y+1)}$

Solution

The given relationship is $f(x)=(1+x)(1+x^{2})(1+x^{4})(1+x^{8})$

Taking logarithm on both the sides, we obtain

$\log f(x)=\log (1+x)+\log (1+x^{2})+\log (1+x^{4})+\log (1+x^{8})$

Differentiating both sides with respect to $x$, we obtain

$ \begin{aligned} & \frac{1}{f(x)} \cdot \frac{d}{d x}[f(x)]=\frac{d}{d x} \log (1+x)+\frac{d}{d x} \log (1+x^{2})+\frac{d}{d x} \log (1+x^{4})+\frac{d}{d x} \log (1+x^{8}) \\ & \Rightarrow \frac{1}{f(x)} \cdot f^{\prime}(x)=\frac{1}{1+x} \cdot \frac{d}{d x}(1+x)+\frac{1}{1+x^{2}} \cdot \frac{d}{d x}(1+x^{2})+\frac{1}{1+x^{4}} \cdot \frac{d}{d x}(1+x^{4})+\frac{1}{1+x^{8}} \cdot \frac{d}{d x}(1+x^{8}) \\ & \Rightarrow f^{\prime}(x)=f(x)[\frac{1}{1+x}+\frac{1}{1+x^{2}} \cdot 2 x+\frac{1}{1+x^{4}} \cdot 4 x^{3}+\frac{1}{1+x^{8}} \cdot 8 x^{7}] \\ & \begin{matrix} \therefore f^{\prime}(x)=(1+x)(1+x^{2})(1+x^{4})(1+x^{8})[\frac{1}{1+x}+\frac{2 x}{1+x^{2}}+\frac{4 x^{3}}{1+x^{4}}+\frac{8 x^{7}}{1+x^{8}}] \\ \text{ Hence, } f^{\prime}(1)=(1+1)(1+1^{2})(1+1^{4})(1+1^{8})[\frac{1}{1+1}+\frac{2 \times 1}{1+1^{2}}+\frac{4 \times 1^{3}}{1+1^{4}}+\frac{8 \times 1^{7}}{1+1^{8}}] \\ =2 \times 2 \times 2 \times 2[\frac{1}{2}+\frac{2}{2}+\frac{4}{2}+\frac{8}{2}] \\ =16 \times(\frac{1+2+4+8}{2}) \\ =16 \times \frac{15}{2}=120 \end{matrix} \end{aligned} $

Solution

(i) Let $y=(x^{5}-5 x+8)(x^{3}+7 x+9)$

Let $x^{2}-5 x+8=u$ and $x^{3}+7 x+9=v$

$\therefore y=u v$

$\Rightarrow \frac{d y}{d x}=\frac{d u}{d x} \cdot v+u \cdot \frac{d v}{d x} \quad$ (By using product rule)

$\Rightarrow \frac{d y}{d x}=\frac{d}{d x}(x^{2}-5 x+8) \cdot(x^{3}+7 x+9)+(x^{2}-5 x+8) \cdot \frac{d}{d x}(x^{3}+7 x+9)$

$\Rightarrow \frac{d y}{d x}=(2 x-5)(x^{3}+7 x+9)+(x^{2}-5 x+8)(3 x^{2}+7)$

$\Rightarrow \frac{d y}{d x}=2 x(x^{3}+7 x+9)-5(x^{3}+7 x+9)+x^{2}(3 x^{2}+7)-5 x(3 x^{2}+7)+8(3 x^{2}+7)$

$\Rightarrow \frac{d y}{d x}=(2 x^{4}+14 x^{2}+18 x)-5 x^{3}-35 x-45+(3 x^{4}+7 x^{2})-15 x^{3}-35 x+24 x^{2}+56$

$\therefore \frac{d y}{d x}=5 x^{4}-20 x^{3}+45 x^{2}-52 x+11$

(ii) $y=(x^{2}-5 x+8)(x^{3}+7 x+9)$

$ \begin{aligned} & =x^{2}(x^{3}+7 x+9)-5 x(x^{3}+7 x+9)+8(x^{3}+7 x+9) \\ & =x^{5}+7 x^{3}+9 x^{2}-5 x^{4}-35 x^{2}-45 x+8 x^{3}+56 x+72 \\ & =x^{5}-5 x^{4}+15 x^{3}-26 x^{2}+11 x+72 \\ & \therefore \frac{d y}{d x}=\frac{d}{d x}(x^{5}-5 x^{4}+15 x^{3}-26 x^{2}+11 x+72) \\ & =\frac{d}{d x}(x^{5})-5 \frac{d}{d x}(x^{4})+15 \frac{d}{d x}(x^{3})-26 \frac{d}{d x}(x^{2})+11 \frac{d}{d x}(x)+\frac{d}{d x}(72) \\ & =5 x^{4}-5 \times 4 x^{3}+15 \times 3 x^{2}-26 \times 2 x+11 \times 1+0 \\ & =5 x^{4}-20 x^{3}+45 x^{2}-52 x+11 \end{aligned} $

(iii) $y=(x^{2}-5 x+8)(x^{3}+7 x+9)$

Taking logarithm on both the sides, we obtain

$\log y=\log (x^{2}-5 x+8)+\log (x^{3}+7 x+9)$

Differentiating both sides with respect to $x$, we obtain

$ \begin{aligned} & \frac{1}{y} \frac{d y}{d x}=\frac{d}{d x} \log (x^{2}-5 x+8)+\frac{d}{d x} \log (x^{3}+7 x+9) \\ & \Rightarrow \frac{1}{y} \frac{d y}{d x}=\frac{1}{x^{2}-5 x+8} \cdot \frac{d}{d x}(x^{2}-5 x+8)+\frac{1}{x^{3}+7 x+9} \cdot \frac{d}{d x}(x^{3}+7 x+9) \\ & \Rightarrow \frac{d y}{d x}=y[\frac{1}{x^{2}-5 x+8} \times(2 x-5)+\frac{1}{x^{3}+7 x+9} \times(3 x^{2}+7)] \\ & \Rightarrow \frac{d y}{d x}=(x^{2}-5 x+8)(x^{3}+7 x+9)[\frac{2 x-5}{x^{2}-5 x+8}+\frac{3 x^{2}+7}{x^{3}+7 x+9}] \\ & \Rightarrow \frac{d y}{d x}=(x^{2}-5 x+8)(x^{3}+7 x+9)[\frac{(2 x-5)(x^{3}+7 x+9)+(3 x^{2}+7)(x^{2}-5 x+8)}{(x^{2}-5 x+8)(x^{3}+7 x+9)}] \\ & \Rightarrow \frac{d y}{d x}=2 x(x^{3}+7 x+9)-5(x^{3}+7 x+9)+3 x^{2}(x^{2}-5 x+8)+7(x^{2}-5 x+8) \\ & \Rightarrow \frac{d y}{d x}=(2 x^{4}+14 x^{2}+18 x)-5 x^{3}-35 x-45+(3 x^{4}-15 x^{3}+24 x^{2})+(7 x^{2}-35 x+56) \\ & \Rightarrow \frac{d y}{d x}=5 x^{4}-20 x^{3}+45 x^{2}-52 x+11 \end{aligned} $

From the above three observations, it can be concluded that all the results of $\frac{d y}{d x}$ are same.

Solution

Let $y=u . v . w=u .(v . w)$

By applying product rule, we obtain $\frac{d y}{d x}=\frac{d u}{d x} \cdot(v \cdot w)+u \cdot \frac{d}{d x}(v \cdot w)$

$\Rightarrow \frac{d y}{d x}=\frac{d u}{d x} v \cdot w+u[\frac{d v}{d x} \cdot w+v \cdot \frac{d w}{d x}] \quad$ (Again applying product rule)

$\Rightarrow \frac{d y}{d x}=\frac{d u}{d x} \cdot v \cdot w+u \cdot \frac{d v}{d x} \cdot w+u \cdot v \cdot \frac{d w}{d x}$

By taking logarithm on both sides of the equation $y=u . v . w$, we obtain

$\log y=\log u+\log v+\log w$

Differentiating both sides with respect to $x$, we obtain

$\frac{1}{y} \cdot \frac{d y}{d x}=\frac{d}{d x}(\log u)+\frac{d}{d x}(\log v)+\frac{d}{d x}(\log w)$

$\Rightarrow \frac{1}{y} \cdot \frac{d y}{d x}=\frac{1}{u} \frac{d u}{d x}+\frac{1}{v} \frac{d v}{d x}+\frac{1}{w} \frac{d w}{d x}$

$\Rightarrow \frac{d y}{d x}=y(\frac{1}{u} \frac{d u}{d x}+\frac{1}{v} \frac{d v}{d x}+\frac{1}{w} \frac{d w}{d x})$

$\Rightarrow \frac{d y}{d x}=u . v . w \cdot(\frac{1}{u} \frac{d u}{d x}+\frac{1}{v} \frac{d v}{d x}+\frac{1}{w} \frac{d w}{d x})$

$\therefore \frac{d y}{d x}=\frac{d u}{d x} \cdot v \cdot w+u \cdot \frac{d v}{d x} \cdot w+u \cdot v \cdot \frac{d w}{d x}$

Solution

The given equations are $x=2 a t^{2}$ and $y=a t^{4}$

Then, $\frac{d x}{d t}=\frac{d}{d t}(2 a t^{2})=2 a \cdot \frac{d}{d t}(t^{2})=2 a \cdot 2 t=4 a t$

$\frac{d y}{d t}=\frac{d}{d t}(a t^{4})=a \cdot \frac{d}{d t}(t^{4})=a \cdot 4 \cdot t^{3}=4 a t^{3}$

$\therefore \frac{d y}{d x}=\frac{(\frac{d y}{d t})}{(\frac{d x}{d t})}=\frac{4 a t^{3}}{4 a t}=t^{2}$

Solution

The given equations are $x=a \cos \theta$ and $y=b \cos \theta$

Then, $\frac{d x}{d \theta}=\frac{d}{d \theta}(a \cos \theta)=a(-\sin \theta)=-a \sin \theta$

$\frac{d y}{d \theta}=\frac{d}{d \theta}(b \cos \theta)=b(-\sin \theta)=-b \sin \theta$

$\therefore \frac{d y}{d x}=\frac{(\frac{d y}{d \theta})}{(\frac{d x}{d \theta})}=\frac{-b \sin \theta}{-a \sin \theta}=\frac{b}{a}$

Solution

The given equations are $x=\sin t$ and $y=\cos 2 t$

Then, $\frac{d x}{d t}=\frac{d}{d t}(\sin t)=\cos t$

$\frac{d y}{d t}=\frac{d}{d t}(\cos 2 t)=-\sin 2 t \cdot \frac{d}{d t}(2 t)=-2 \sin 2 t$

$\therefore \frac{d y}{d x}=\frac{(\frac{d y}{d t})}{(\frac{d x}{d t})}=\frac{-2 \sin 2 t}{\cos t}=\frac{-2 \cdot 2 \sin t \cos t}{\cos t}=-4 \sin t$

Solution

The given equations are $x=4 t$ and $y=\frac{4}{t}$

$ \begin{aligned} & \frac{d x}{d t}=\frac{d}{d t}(4 t)=4 \\ & \frac{d y}{d t}=\frac{d}{d t}(\frac{4}{t})=4 \cdot \frac{d}{d t}(\frac{1}{t})=4 \cdot(\frac{-1}{t^{2}})=\frac{-4}{t^{2}} \\ & \therefore \frac{d y}{d x}=\frac{(\frac{d y}{d t})}{(\frac{d x}{d t})}=\frac{(\frac{-4}{t^{2}})}{4}=\frac{-1}{t^{2}} \end{aligned} $

Solution

The given equations are $x=\cos \theta-\cos 2 \theta$ and $y=\sin \theta-\sin 2 \theta$

Then, $\frac{d x}{d \theta}=\frac{d}{d \theta}(\cos \theta-\cos 2 \theta)=\frac{d}{d \theta}(\cos \theta)-\frac{d}{d \theta}(\cos 2 \theta)$

$ =-\sin \theta-(-2 \sin 2 \theta)=2 \sin 2 \theta-\sin \theta $

$\frac{d y}{d \theta}=\frac{d}{d \theta}(\sin \theta-\sin 2 \theta)=\frac{d}{d \theta}(\sin \theta)-\frac{d}{d \theta}(\sin 2 \theta)$

$=\cos \theta-2 \cos 2 \theta$

$\therefore \frac{d y}{d x}=\frac{(\frac{d y}{d \theta})}{(\frac{d x}{d \theta})}=\frac{\cos \theta-2 \cos 2 \theta}{2 \sin 2 \theta-\sin \theta}$

Solution

The given equations are $x=a(\theta-\sin \theta)$ and $y=a(1+\cos \theta)$

Then, $\frac{d x}{d \theta}=a[\frac{d}{d \theta}(\theta)-\frac{d}{d \theta}(\sin \theta)]=a(1-\cos \theta)$

$\frac{d y}{d \theta}=a[\frac{d}{d \theta}(1)+\frac{d}{d \theta}(\cos \theta)]=a[0+(-\sin \theta)]=-a \sin \theta$

$\therefore \frac{d y}{d x}=\frac{(\frac{d y}{d \theta})}{(\frac{d x}{d \theta})}=\frac{-a \sin \theta}{a(1-\cos \theta)}=\frac{-2 \sin \frac{\theta}{2} \cos \frac{\theta}{2}}{2 \sin ^{2} \frac{\theta}{2}}=\frac{-\cos \frac{\theta}{2}}{\sin \frac{\theta}{2}}=-\cot \frac{\theta}{2}$

Solution

The given equations are $x=\frac{\sin ^{3} t}{\sqrt{\cos 2 t}}$ and $y=\frac{\cos ^{3} t}{\sqrt{\cos 2 t}}$

$ \begin{aligned} & \text{ Then, } \frac{d x}{d t}=\frac{d}{d t}[\frac{\sin ^{3} t}{\sqrt{\cos 2 t}}] \\ & =\frac{\sqrt{\cos 2 t} \cdot \frac{d}{d t}(\sin ^{3} t)-\sin ^{3} t \cdot \frac{d}{d t} \sqrt{\cos 2 t}}{\cos 2 t} \\ & =\frac{\sqrt{\cos 2 t} \cdot 3 \sin ^{2} t \cdot \frac{d}{d t}(\sin t)-\sin ^{3} t \times \frac{1}{2 \sqrt{\cos 2 t}} \cdot \frac{d}{d t}(\cos 2 t)}{\cos 2 t} \\ & =\frac{3 \sqrt{\cos 2 t} \cdot \sin ^{2} t \cos t-\frac{\sin ^{3} t}{2 \sqrt{\cos 2 t}} \cdot(-2 \sin 2 t)}{\cos 2 t} \\ & =\frac{3 \cos 2 t \sin ^{2} t \cos t+\sin ^{3} t \sin 2 t}{\cos 2 t \sqrt{\cos 2 t}} \\ & \frac{d y}{d t}=\frac{d}{d t}[\frac{\cos ^{3} t}{\sqrt{\cos 2 t}}] \\ & =\frac{\sqrt{\cos 2 t} \cdot \frac{d}{d t}(\cos ^{3} t)-\cos ^{3} t \cdot \frac{d}{d t}(\sqrt{\cos 2 t})}{\cos 2 t} \\ & =\frac{\sqrt{\cos 2 t} \cdot 3 \cos ^{2} t \cdot \frac{d}{d t}(\cos t)-\cos ^{3} t \cdot \frac{1}{2 \sqrt{\cos 2 t}} \cdot \frac{d}{d t}(\cos 2 t)}{\cos 2 t} \\ & =\frac{3 \sqrt{\cos 2 t} \cdot \cos ^{2} t(-\sin t)-\cos ^{3} t \cdot \frac{1}{2 \sqrt{\cos 2 t}} \cdot(-2 \sin 2 t)}{\cos 2 t} \\ & =\frac{-3 \cos 2 t \cdot \cos ^{2} t \cdot \sin t+\cos ^{3} t \sin 2 t}{\cos 2 t \cdot \sqrt{\cos 2 t}} \end{aligned} $

$ \begin{matrix} \therefore \frac{d y}{d x}=\frac{(\frac{d y}{d t})}{(\frac{d x}{d t})} & =\frac{-3 \cos 2 t \cdot \cos ^{2} t \cdot \sin t+\cos ^{3} t \sin 2 t}{3 \cos 2 t \sin ^{2} t \cos t+\sin ^{3} t \sin 2 t} & \\ & =\frac{-3 \cos 2 t \cdot \cos ^{2} t \cdot \sin t+\cos ^{3} t(2 \sin t \cos t)}{3 \cos 2 t \sin ^{2} t \cos t+\sin ^{3} t(2 \sin t \cos t)} & \\ & =\frac{\sin t \cos t[-3 \cos 2 t \cdot \cos t+2 \cos ^{3} t]}{\sin t \cos t[3 \cos 2 t \sin t+2 \sin ^{3} t]} & { \begin{bmatrix} \cos 2 t=(2 \cos ^{2} t-1), \\ \cos 2 t=(1-2 \sin ^{2} t) \end{bmatrix} } \\ & =\frac{[-3(2 \cos { }^{2} t-1) \cos t+2 \cos ^{3} t]}{[3(1-2 \sin ^{2} t) \sin t+2 \sin ^{3} t]} & { \begin{bmatrix} \cos 3 t=4 \cos ^{3} t-3 \cos ^{3} t \\ \sin 3 t=3 \sin ^{3} t-4 \sin ^{3} t \end{bmatrix} } \\ & =\frac{-4 \cos t+3 \cos t}{3 \sin t-4 \sin ^{3} t} & \end{matrix} $

Solution

The given equations are $x=a(\cos t+\log \tan \frac{t}{2})$ and $y=a \sin t$

$ \begin{aligned} & \text{ Then, } \frac{d x}{d t}=a \cdot[\frac{d}{d t}(\cos t)+\frac{d}{d t}(\log \tan \frac{t}{2})] \\ & =a[-\sin t+\frac{1}{\tan \frac{t}{2}} \cdot \frac{d}{d t}(\tan \frac{t}{2})] \\ & =a[-\sin t+\cot \frac{t}{2} \cdot \sec ^{2} \frac{t}{2} \cdot \frac{d}{d t}(\frac{t}{2})] \\ & =a[-\sin t+\frac{\cos \frac{t}{2}}{\sin \frac{t}{2}} \times \frac{1}{\cos ^{2} \frac{t}{2}} \times \frac{1}{2}] \\ & =a[-\sin t+\frac{1}{2 \sin \frac{t}{2} \cos \frac{t}{2}}] \\ & =a(-\sin t+\frac{1}{\sin t}) \\ & =a(\frac{-\sin ^{2} t+1}{\sin t}) \\ & =a \frac{\cos ^{2} t}{\sin t} \\ & \frac{d y}{d t}=a \frac{d}{d t}(\sin t)=a \cos t \\ & \therefore \frac{d y}{d x}=\frac{(\frac{d y}{d t})}{(\frac{d x}{d t})}=\frac{a \cos t}{(a \frac{\cos ^{2} t}{\sin t})}=\frac{\sin t}{\cos t}=\tan t \end{aligned} $

Solution

The given equations are $x=a \sec \theta$ and $y=b \tan \theta$

Then, $\frac{d x}{d \theta}=a \cdot \frac{d}{d \theta}(\sec \theta)=a \sec \theta \tan \theta$

$\frac{d y}{d \theta}=b \cdot \frac{d}{d \theta}(\tan \theta)=b \sec ^{2} \theta$

$\therefore \frac{d y}{d x}=\frac{(\frac{d y}{d \theta})}{(\frac{d x}{d \theta})}=\frac{b \sec ^{2} \theta}{a \sec \theta \tan \theta}=\frac{b}{a} \sec \theta \cot \theta=\frac{b \cos \theta}{a \cos \theta \sin \theta}=\frac{b}{a} \times \frac{1}{\sin \theta}=\frac{b}{a} cosec \theta$

Solution

The given equations are $x=a(\cos \theta+\theta \sin \theta)$ and $y=a(\sin \theta-\theta \cos \theta)$

Then, $\frac{d x}{d \theta}=a[\frac{d}{d \theta} \cos \theta+\frac{d}{d \theta}(\theta \sin \theta)]=a[-\sin \theta+\theta \frac{d}{d \theta}(\sin \theta)+\sin \theta \frac{d}{d \theta}(\theta)]$ $=a[-\sin \theta+\theta \cos \theta+\sin \theta]=a \theta \cos \theta$

$ \begin{aligned} \frac{d y}{d \theta}=a[\frac{d}{d \theta}(\sin \theta)-\frac{d}{d \theta}(\theta \cos \theta)] & =a[\cos \theta-{\theta \frac{d}{d \theta}(\cos \theta)+\cos \theta \cdot \frac{d}{d \theta}(\theta)}] \\ & =a[\cos \theta+\theta \sin \theta-\cos \theta] \\ & =a \theta \sin \theta \end{aligned} $

$\therefore \frac{d y}{d x}=\frac{(\frac{d y}{d \theta})}{(\frac{d x}{d \theta})}=\frac{a \theta \sin \theta}{a \theta \cos \theta}=\tan \theta$

Solution

The given equations are $x=\sqrt{a^{\sin ^{-1} t}}$ and $y=\sqrt{a^{\cos ^{-1} t}}$

$x=\sqrt{a^{\sin ^{-1} t}}$ and $y=\sqrt{a^{\cos ^{-1} t}}$

$\Rightarrow x=(a^{\sin ^{-1} t})^{\frac{1}{2}}$ and $y=(a^{\cos ^{-1} t})^{\frac{1}{2}}$

$\Rightarrow x=a^{\frac{1}{2} \sin ^{-1} t}$ and $y=a^{\frac{1}{2} \cos ^{-1} t}$

Consider $x=a^{\frac{1}{2} \sin ^{-1} t}$

Taking logarithm on both the sides, we obtain

$\log x=\frac{1}{2} \sin ^{-1} t \log a$

$\therefore \frac{1}{x} \cdot \frac{d x}{d t}=\frac{1}{2} \log a \cdot \frac{d}{d t}(\sin ^{-1} t)$

$\Rightarrow \frac{d x}{d t}=\frac{x}{2} \log a \cdot \frac{1}{\sqrt{1-t^{2}}}$

$\Rightarrow \frac{d x}{d t}=\frac{x \log a}{2 \sqrt{1-t^{2}}}$

Then, consider $y=a^{\frac{1}{2} \cos ^{-1} t}$

Taking logarithm on both the sides, we obtain

$\log y=\frac{1}{2} \cos ^{-1} t \log a$

$\therefore \frac{1}{y} \cdot \frac{d y}{d x}=\frac{1}{2} \log a \cdot \frac{d}{d t}(\cos ^{-1} t)$

$\Rightarrow \frac{d y}{d t}=\frac{y \log a}{2} \cdot(\frac{-1}{\sqrt{1-t^{2}}})$

$\Rightarrow \frac{d y}{d t}=\frac{-y \log a}{2 \sqrt{1-t^{2}}}$ $\therefore \frac{d y}{d x}=\frac{(\frac{d y}{d t})}{(\frac{d x}{d t})}=\frac{(\frac{-y \log a}{2 \sqrt{1-t^{2}}})}{(\frac{x \log a}{2 \sqrt{1-t^{2}}})}=-\frac{y}{x}$.

Hence, proved.

Solution

Let $y=x^{2}+3 x+2$

Then,

$\frac{d y}{d x}=\frac{d}{d x}(x^{2})+\frac{d}{d x}(3 x)+\frac{d}{d x}(2)=2 x+3+0=2 x+3$

$\therefore \frac{d^{2} y}{d x^{2}}=\frac{d}{d x}(2 x+3)=\frac{d}{d x}(2 x)+\frac{d}{d x}(3)=2+0=2$

Solution

Let $y=x^{20}$

Then,

$\frac{d y}{d x}=\frac{d}{d x}(x^{20})=20 x^{19}$

$\therefore \frac{d^{2} y}{d x^{2}}=\frac{d}{d x}(20 x^{19})=20 \frac{d}{d x}(x^{19})=20 \cdot 19 \cdot x^{18}=380 x^{18}$

Solution

Let $y=x \cdot \cos x$

Then,

$ \begin{aligned} & \frac{d y}{d x}=\frac{d}{d x}(x \cdot \cos x)=\cos x \cdot \frac{d}{d x}(x)+x \frac{d}{d x}(\cos x)=\cos x \cdot 1+x(-\sin x)=\cos x-x \sin x \\ & \begin{aligned} \therefore \frac{d^{2} y}{d x^{2}} & =\frac{d}{d x}[\cos x-x \sin x]=\frac{d}{d x}(\cos x)-\frac{d}{d x}(x \sin x) \\ & =-\sin x-[\sin x \cdot \frac{d}{d x}(x)+x \cdot \frac{d}{d x}(\sin x)] \\ & =-\sin x-(\sin x+x \cos x) \\ & =-(x \cos x+2 \sin x) \end{aligned} \end{aligned} $

Solution

Let $y=\log x$

Then,

$\frac{d y}{d x}=\frac{d}{d x}(\log x)=\frac{1}{x}$

$\therefore \frac{d^{2} y}{d x^{2}}=\frac{d}{d x}(\frac{1}{x})=\frac{-1}{x^{2}}$

Solution

Let $y=x^{3} \log x$

Then,

$ \begin{aligned} & \frac{d y}{d x}=\frac{d}{d x}[x^{3} \log x]=\log x \cdot \frac{d}{d x}(x^{3})+x^{3} \cdot \frac{d}{d x}(\log x) \\ & =\log x \cdot 3 x^{2}+x^{3} \cdot \frac{1}{x}=\log x \cdot 3 x^{2}+x^{2} \\ & =x^{2}(1+3 \log x) \\ & \therefore \frac{d^{2} y}{d x^{2}}=\frac{d}{d x}[x^{2}(1+3 \log x)] \\ & =(1+3 \log x) \cdot \frac{d}{d x}(x^{2})+x^{2} \frac{d}{d x}(1+3 \log x) \\ & =(1+3 \log x) \cdot 2 x+x^{2} \cdot \frac{3}{x} \\ & =2 x+6 x \log x+3 x \\ & =5 x+6 x \log x \\ & =x(5+6 \log x) \end{aligned} $

Solution

Let $y=e^{x} \sin 5 x$

Then,

$ \begin{aligned} \frac{d y}{d x} & =\frac{d}{d x}(e^{x} \sin 5 x)=\sin 5 x \cdot \frac{d}{d x}(e^{x})+e^{x} \frac{d}{d x}(\sin 5 x) \\ & =\sin 5 x \cdot e^{x}+e^{x} \cdot \cos 5 x \cdot \frac{d}{d x}(5 x)=e^{x} \sin 5 x+e^{x} \cos 5 x \cdot 5 \\ & =e^{x}(\sin 5 x+5 \cos 5 x) \\ \therefore \frac{d^{2} y}{d x^{2}} & =\frac{d}{d x}[e^{x}(\sin 5 x+5 \cos 5 x)] \\ & =(\sin 5 x+5 \cos 5 x) \cdot \frac{d}{d x}(e^{x})+e^{x} \cdot \frac{d}{d x}(\sin 5 x+5 \cos 5 x) \\ & =(\sin 5 x+5 \cos 5 x) e^{x}+e^{x}[\cos 5 x \cdot \frac{d}{d x}(5 x)+5(-\sin 5 x) \cdot \frac{d}{d x}(5 x)] \\ & =e^{x}(\sin 5 x+5 \cos 5 x)+e^{x}(5 \cos 5 x-25 \sin 5 x) \\ & =e^{x}(10 \cos 5 x-24 \sin 5 x)=2 e^{x}(5 \cos 5 x-12 \sin 5 x) \end{aligned} $

Solution

Let $y=e^{6 x} \cos 3 x$

Then,

$ \begin{aligned} \frac{d y}{d x} & =\frac{d}{d x}(e^{6 x} \cdot \cos 3 x)=\cos 3 x \cdot \frac{d}{d x}(e^{6 x})+e^{6 x} \cdot \frac{d}{d x}(\cos 3 x) \\ & =\cos 3 x \cdot e^{6 x} \cdot \frac{d}{d x}(6 x)+e^{6 x} \cdot(-\sin 3 x) \cdot \frac{d}{d x}(3 x) \\ & =6 e^{6 x} \cos 3 x-3 e^{6 x} \sin 3 x \\ \therefore \frac{d^{2} y}{d x^{2}} & =\frac{d}{d x}(6 e^{6 x} \cos 3 x-3 e^{6 x} \sin 3 x)=6 \cdot \frac{d}{d x}(e^{6 x} \cos 3 x)-3 \cdot \frac{d}{d x}(e^{6 x} \sin 3 x) \\ & =6 \cdot[6 e^{6 x} \cos 3 x-3 e^{6 x} \sin 3 x]-3 \cdot[\sin 3 x \cdot \frac{d}{d x}(e^{6 x})+e^{6 x} \cdot \frac{d}{d x}(\sin 3 x)] \\ & =36 e^{6 x} \cos 3 x-18 e^{6 x} \sin 3 x-3[\sin 3 x \cdot e^{6 x} \cdot 6+e^{6 x} \cdot \cos 3 x \cdot 3] \\ & =36 e^{6 x} \cos 3 x-18 e^{6 x} \sin 3 x-18 e^{6 x} \sin 3 x-9 e^{6 x} \cos 3 x \\ & =27 e^{6 x} \cos 3 x-36 e^{6 x} \sin 3 x \\ & =9 e^{6 x}(3 \cos 3 x-4 \sin 3 x) \end{aligned} $

Solution

Let $y=\tan ^{-1} x$

Then,

$ \begin{aligned} & \frac{d y}{d x}=\frac{d}{d x}(\tan ^{-1} x)=\frac{1}{1+x^{2}} \\ & \therefore \frac{d^{2} y}{d x^{2}}=\frac{d}{d x}(\frac{1}{1+x^{2}})=\frac{d}{d x}(1+x^{2})^{-1}=(-1) \cdot(1+x^{2})^{-2} \cdot \frac{d}{d x}(1+x^{2}) \\ & \quad=\frac{-1}{(1+x^{2})^{2}} \times 2 x=\frac{-2 x}{(1+x^{2})^{2}} \end{aligned} $

Solution

Let $y=\log (\log x)$

Then,

$ \begin{aligned} & \frac{d y}{d x}=\frac{d}{d x}[\log (\log x)]=\frac{1}{\log x} \cdot \frac{d}{d x}(\log x)=\frac{1}{x \log x}=(x \log x)^{-1} \\ & \begin{aligned} \therefore \frac{d^{2} y}{d x^{2}} & =\frac{d}{d x}[(x \log x)^{-1}]=(-1) \cdot(x \log x)^{-2} \cdot \frac{d}{d x}(x \log x) \\ & =\frac{-1}{(x \log x)^{2}} \cdot[\log x \cdot \frac{d}{d x}(x)+x \cdot \frac{d}{d x}(\log x)] \\ & =\frac{-1}{(x \log x)^{2}} \cdot[\log x \cdot 1+x \cdot \frac{1}{x}]=\frac{-(1+\log x)}{(x \log x)^{2}} \end{aligned} \end{aligned} $

Solution

Let $y=\sin (\log x)$

Then,

$ \begin{aligned} & \frac{d y}{d x}=\frac{d}{d x}[\sin (\log x)]=\cos (\log x) \cdot \frac{d}{d x}(\log x)=\frac{\cos (\log x)}{x} \\ & \begin{aligned} \therefore \frac{d^{2} y}{d x^{2}} & =\frac{d}{d x}[\frac{\cos (\log x)}{x}] \\ & =\frac{x \cdot \frac{d}{d x}[\cos (\log x)]-\cos (\log x) \cdot \frac{d}{d x}(x)}{x^{2}} \\ & =\frac{x \cdot[-\sin (\log x) \cdot \frac{d}{d x}(\log x)]-\cos (\log x) \cdot 1}{x^{2}} \\ & =\frac{-x \sin (\log x) \cdot \frac{1}{x}-\cos (\log x)}{x^{2}} \\ & =\frac{-[\sin (\log x)+\cos (\log x)]}{x^{2}} \end{aligned} \end{aligned} $

Solution

It is given that, $y=5 \cos x-3 \sin x$

Then,

$ \begin{aligned} & \frac{d y}{d x}=\frac{d}{d x}(5 \cos x)-\frac{d}{d x}(3 \sin x)=5 \frac{d}{d x}(\cos x)-3 \frac{d}{d x}(\sin x) \\ & =5(-\sin x)-3 \cos x=-(5 \sin x+3 \cos x) \\ & \therefore \frac{d^{2} y}{d x^{2}}=\frac{d}{d x}[-(5 \sin x+3 \cos x)] \\ & =-[5 \cdot \frac{d}{d x}(\sin x)+3 \cdot \frac{d}{d x}(\cos x)] \\ & =-[5 \cos x+3(-\sin x)] \\ & =-[5 \cos x-3 \sin x] \\ & =-y \\ & \therefore \frac{d^{2} y}{d x^{2}}+y=0 \end{aligned} $

Hence, proved.

Solution

It is given that, $y=\cos ^{-1} x$

Then,

$ \begin{aligned} & \frac{d y}{d x}=\frac{d}{d x}(\cos ^{-1} x)=\frac{-1}{\sqrt{1-x^{2}}}=-(1-x^{2})^{\frac{-1}{2}} \\ & \frac{d^{2} y}{d x^{2}}=\frac{d}{d x}[-(1-x^{2})^{\frac{-1}{2}}] \\ & =-(-\frac{1}{2}) \cdot(1-x^{2})^{\frac{-3}{2}} \cdot \frac{d}{d x}(1-x^{2}) \\ & =\frac{1}{2 \sqrt{(1-x^{2})^{3}}} \times(-2 x) \\ & \Rightarrow \frac{d^{2} y}{d x^{2}}=\frac{-x}{\sqrt{(1-x^{2})^{3}}} \end{aligned} $

Putting $x=\cos y$ in equation (i), we obtain

$\frac{d^{2} y}{d x^{2}}=\frac{-\cos y}{\sqrt{(1-\cos ^{2} y)^{3}}}$ $\Rightarrow \frac{d^{2} y}{d x^{2}}=\frac{-\cos y}{\sqrt{(\sin ^{2} y)^{3}}}$

$ =\frac{-\cos y}{\sin ^{3} y} $

$ =\frac{-\cos y}{\sin y} \times \frac{1}{\sin ^{2} y} $

$\Rightarrow \frac{d^{2} y}{d x^{2}}=-\cot y \cdot cosec^{2} y$

Solution

It is given that, $y=3 \cos (\log x)+4 \sin (\log x)$

Then,

$ \begin{aligned} & y_1=3 \cdot \frac{d}{d x}[\cos (\log x)]+4 \cdot \frac{d}{d x}[\sin (\log x)] \\ & =3 \cdot[-\sin (\log x) \cdot \frac{d}{d x}(\log x)]+4 \cdot[\cos (\log x) \cdot \frac{d}{d x}(\log x)] \\ & \therefore y_1=\frac{-3 \sin (\log x)}{x}+\frac{4 \cos (\log x)}{x}=\frac{4 \cos (\log x)-3 \sin (\log x)}{x} \\ & \therefore y_2=\frac{d}{d x}(\frac{4 \cos (\log x)-3 \sin (\log x)}{x}) \\ & =\frac{x{4 \cos (\log x)-3 \sin (\log x)}^{\prime}-{4 \cos (\log x)-3 \sin (\log x)}(x)^{\prime}}{x^{2}} \\ & =\frac{x[4{\cos (\log x)}^{\prime}-3{\sin (\log x)}^{\prime}]-{4 \cos (\log x)-3 \sin (\log x)} \cdot 1}{x^{2}} \\ & =\frac{x[-4 \sin (\log x) \cdot(\log x)^{\prime}-3 \cos (\log x) \cdot(\log x)^{\prime}]-4 \cos (\log x)+3 \sin (\log x)}{x^{2}} \\ & =\frac{x[-4 \sin (\log x) \cdot \frac{1}{x}-3 \cos (\log x) \cdot \frac{1}{x}]-4 \cos (\log x)+3 \sin (\log x)}{x^{2}} \\ & =\frac{-4 \sin (\log x)-3 \cos (\log x)-4 \cos (\log x)+3 \sin (\log x)}{x^{2}} \\ & =\frac{-\sin (\log x)-7 \cos (\log x)}{x^{2}} \\ & \therefore x^{2} y_2+x y_1+y \\ & =x^{2}(\frac{-\sin (\log x)-7 \cos (\log x)}{x^{2}})+x(\frac{4 \cos (\log x)-3 \sin (\log x)}{x})+3 \cos (\log x)+4 \sin (\log x) \\ & =-\sin (\log x)-7 \cos (\log x)+4 \cos (\log x)-3 \sin (\log x)+3 \cos (\log x)+4 \sin (\log x) \\ & =0 \end{aligned} $

Hence, proved.

Solution

It is given that, $y=A e^{m x}+B e^{n x}$

Then,

$ \begin{aligned} & \frac{d y}{d x}=A \cdot \frac{d}{d x}(e^{m x})+B \cdot \frac{d}{d x}(e^{n x})=A \cdot e^{m x} \cdot \frac{d}{d x}(m x)+B \cdot e^{n x} \cdot \frac{d}{d x}(n x)=A m e^{n x x}+B n e^{n x} \\ & \frac{d^{2} y}{d x^{2}}=\frac{d}{d x}(A m e^{m x}+B n e^{n x})=A m \cdot \frac{d}{d x}(e^{m x})+B n \cdot \frac{d}{d x}(e^{n x}) \\ & =A m \cdot e^{m x} \cdot \frac{d}{d x}(m x)+B n \cdot e^{n x} \cdot \frac{d}{d x}(n x)=A m^{2} e^{m x}+B n^{2} e^{n x} \\ & \therefore \frac{d^{2} y}{d x^{2}}-(m+n) \frac{d y}{d x}+m n y \\ & =A m^{2} e^{m x}+B n^{2} e^{n x}-(m+n) \cdot(A m e^{m x}+B n e^{n x})+m n(A e^{m x}+B e^{n x}) \\ & =A m^{2} e^{n x x}+B n^{2} e^{n x}-A m^{2} e^{n x x}-B m n e^{n x}-A m n e^{m x x}-B n^{2} e^{n x}+A m n e^{n x x}+B m n e^{n x} \\ & =0 \end{aligned} $

Hence, proved.

Solution

It is given that, $y=500 e^{7 x}+600 e^{-7 x}$

Then,

$ \begin{aligned} \frac{d y}{d x} & =500 \cdot \frac{d}{d x}(e^{7 x})+600 \cdot \frac{d}{d x}(e^{-7 x}) \\ & =500 \cdot e^{7 x} \cdot \frac{d}{d x}(7 x)+600 \cdot e^{-7 x} \cdot \frac{d}{d x}(-7 x) \\ & =3500 e^{7 x}-4200 e^{-7 x} \\ \therefore \frac{d^{2} y}{d x^{2}} & =3500 \cdot \frac{d}{d x}(e^{7 x})-4200 \cdot \frac{d}{d x}(e^{-7 x}) \\ & =3500 \cdot e^{7 x} \cdot \frac{d}{d x}(7 x)-4200 \cdot e^{-7 x} \cdot \frac{d}{d x}(-7 x) \\ & =7 \times 3500 \cdot e^{7 x}+7 \times 4200 \cdot e^{-7 x} \\ & =49 \times 500 e^{7 x}+49 \times 600 e^{-7 x} \\ & =49(500 e^{7 x}+600 e^{-7 x}) \\ & =49 y \end{aligned} $

Hence, proved.

Solution

The given relationship is $e^{y}(x+1)=1$

$e^{y}(x+1)=1$

$\Rightarrow e^{y}=\frac{1}{x+1}$

Taking logarithm on both the sides, we obtain

$y=\log \frac{1}{(x+1)}$

Differentiating this relationship with respect to $x$, we obtain $\frac{d y}{d x}=(x+1) \frac{d}{d x}(\frac{1}{x+1})=(x+1) \cdot \frac{-1}{(x+1)^{2}}=\frac{-1}{x+1}$

$\therefore \frac{d^{2} y}{d x^{2}}=-\frac{d}{d x}(\frac{1}{x+1})=-(\frac{-1}{(x+1)^{2}})=\frac{1}{(x+1)^{2}}$

$\Rightarrow \frac{d^{2} y}{d x^{2}}=(\frac{-1}{x+1})^{2}$

$\Rightarrow \frac{d^{2} y}{d x^{2}}=(\frac{d y}{d x})^{2}$

Hence, proved.

Solution

The given relationship is $y=(\tan ^{-1} x)^{2}$

Then,

$y _1=2 \tan ^{-1} x \frac{d}{d x}(\tan ^{-1} x)$

$\Rightarrow y _1=2 \tan ^{-1} x \cdot \frac{1}{1+x^{2}}$

$\Rightarrow(1+x^{2}) y _1=2 \tan^{-1} x$

Again differentiating with respect to $x$ on both the sides, we obtain

$(1+x^{2}) y _2+2 x y _1=2(\frac{1}{1+x^{2}})$

$\Rightarrow(1+x^{2})^{2} y _2+2 x(1+x^{2}) y _1=2$

Hence, proved.

Solution

Let $y=(3 x^{2}-9 x+5)^{9}$

Using chain rule, we obtain

$ \begin{aligned} \frac{d y}{d x} & =\frac{d}{d x}(3 x^{2}-9 x+5)^{9} \\ & =9(3 x^{2}-9 x+5)^{8} \cdot \frac{d}{d x}(3 x^{2}-9 x+5) \\ & =9(3 x^{2}-9 x+5)^{8} \cdot(6 x-9) \\ & =9(3 x^{2}-9 x+5)^{8} \cdot 3(2 x-3) \\ & =27(3 x^{2}-9 x+5)^{8}(2 x-3) \end{aligned} $

Solution

$ \begin{aligned} & \text{ Let } y=\sin ^{3} x+\cos ^{6} x \\ & \therefore \frac{d y}{d x}=\frac{d}{d x}(\sin ^{3} x)+\frac{d}{d x}(\cos ^{6} x) \\ & =3 \sin ^{2} x \cdot \frac{d}{d x}(\sin x)+6 \cos ^{5} x \cdot \frac{d}{d x}(\cos x) \\ & =3 \sin ^{2} x \cdot \cos x+6 \cos ^{5} x \cdot(-\sin x) \\ & =3 \sin x \cos x(\sin x-2 \cos ^{4} x) \end{aligned} $

Solution

Let $y=(5 x)^{3 \cos 2 x}$

Taking logarithm on both the sides, we obtain

$\log y=3 \cos 2 x \log 5 x$

Differentiating both sides with respect to $x$, we obtain

$\frac{1}{y} \frac{d y}{d x}=3[\log 5 x \cdot \frac{d}{d x}(\cos 2 x)+\cos 2 x \cdot \frac{d}{d x}(\log 5 x)]$

$\Rightarrow \frac{d y}{d x}=3 y[\log 5 x(-\sin 2 x) \cdot \frac{d}{d x}(2 x)+\cos 2 x \cdot \frac{1}{5 x} \cdot \frac{d}{d x}(5 x)]$

$\Rightarrow \frac{d y}{d x}=3 y[-2 \sin 2 x \log 5 x+\frac{\cos 2 x}{x}]$

$\Rightarrow \frac{d y}{d x}=3 y[\frac{3 \cos 2 x}{x}-6 \sin 2 x \log 5 x]$

$\therefore \frac{d y}{d x}=(5 x)^{3 \cos 2 x}[\frac{3 \cos 2 x}{x}-6 \sin 2 x \log 5 x]$

Solution

Let $y=\sin ^{-1}(x \sqrt{x})$

Using chain rule, we obtain

$ \begin{aligned} \frac{d y}{d x} & =\frac{d}{d x} \sin ^{-1}(x \sqrt{x}) \\ & =\frac{1}{\sqrt{1-(x \sqrt{x})^{2}}} \times \frac{d}{d x}(x \sqrt{x}) \\ & =\frac{1}{\sqrt{1-x^{3}}} \cdot \frac{d}{d x}(x^{\frac{3}{2}}) \\ & =\frac{1}{\sqrt{1-x^{3}}} \times \frac{3}{2} \cdot x^{\frac{1}{2}} \\ & =\frac{3 \sqrt{x}}{2 \sqrt{1-x^{3}}} \\ & =\frac{3}{2} \sqrt{\frac{x}{1-x^{3}}} \end{aligned} $

Solution

Let $y=\frac{\cos ^{-1} \frac{x}{2}}{\sqrt{2 x+7}}$

By quotient rule, we obtain

$ \begin{aligned} & \frac{d y}{d x}=\frac{\sqrt{2 x+7} \frac{d}{d x}(\cos ^{-1} \frac{x}{2})-(\cos ^{-1} \frac{x}{2}) \frac{d}{d x}(\sqrt{2 x+7})}{(\sqrt{2 x+7})^{2}} \\ & =\frac{\sqrt{2 x+7}[\frac{-1}{\sqrt{1-(\frac{x}{2})^{2}}} \cdot \frac{d}{d x}(\frac{x}{2})]-(\cos ^{-1} \frac{x}{2}) \frac{1}{2 \sqrt{2 x+7}} \cdot \frac{d}{d x}(2 x+7)}{2 x+7} \\ & =\frac{\sqrt{2 x+7} \frac{-1}{\sqrt{4-x^{2}}}-(\cos ^{-1} \frac{x}{2}) \frac{2}{2 \sqrt{2 x+7}}}{2 x+7} \\ & =\frac{-\sqrt{2 x+7}}{\sqrt{4-x^{2}} \times(2 x+7)}-\frac{\cos ^{-1} \frac{x}{2}}{(\sqrt{2 x+7})(2 x+7)} \\ & =-[\frac{1}{\sqrt{4-x^{2}} \sqrt{2 x+7}}+\frac{\cos ^{-1} \frac{x}{2}}{(2 x+7)^{\frac{3}{2}}}] \end{aligned} $

Solution

Let $y=\cot ^{-1}[\frac{\sqrt{1+\sin x}+\sqrt{1-\sin x}}{\sqrt{1+\sin x}-\sqrt{1-\sin x}}]$

Then, $\frac{\sqrt{1+\sin x}+\sqrt{1-\sin x}}{\sqrt{1+\sin x}-\sqrt{1-\sin x}}$

$=\frac{(\sqrt{1+\sin x}+\sqrt{1-\sin x})^{2}}{(\sqrt{1+\sin x}-\sqrt{1-\sin x})(\sqrt{1+\sin x}+\sqrt{1-\sin x})}$

$=\frac{(1+\sin x)+(1-\sin x)+2 \sqrt{(1-\sin x)(1+\sin x)}}{(1+\sin x)-(1-\sin x)}$

$=\frac{2+2 \sqrt{1-\sin ^{2} x}}{2 \sin x}$

$=\frac{1+\cos x}{\sin x}$

$=\frac{2 \cos ^{2} \frac{x}{2}}{2 \sin \frac{x}{2} \cos \frac{x}{2}}$

$=\cot \frac{x}{2}$

Therefore, equation (1) becomes

$y=\cot ^{-1}(\cot \frac{x}{2})$

$\Rightarrow y=\frac{x}{2}$

$\therefore \frac{d y}{d x}=\frac{1}{2} \frac{d}{d x}(x)$

$\Rightarrow \frac{d y}{d x}=\frac{1}{2}$

Solution

Let $y=(\log x)^{\log x}$

Taking logarithm on both the sides, we obtain

$\log y=\log x \cdot \log (\log x)$

Differentiating both sides with respect to $x$, we obtain

$\frac{1}{y} \frac{d y}{d x}=\frac{d}{d x}[\log x \cdot \log (\log x)]$

$\Rightarrow \frac{1}{y} \frac{d y}{d x}=\log (\log x) \cdot \frac{d}{d x}(\log x)+\log x \cdot \frac{d}{d x}[\log (\log x)]$

$\Rightarrow \frac{d y}{d x}=y[\log (\log x) \cdot \frac{1}{x}+\log x \cdot \frac{1}{\log x} \cdot \frac{d}{d x}(\log x)]$

$\Rightarrow \frac{d y}{d x}=y[\frac{1}{x} \log (\log x)+\frac{1}{x}]$

$\therefore \frac{d y}{d x}=(\log x)^{\log x}[\frac{1}{x}+\frac{\log (\log x)}{x}]$

Solution

Let $y=\cos (a \cos x+b \sin x)$

By using chain rule, we obtain

$ \begin{aligned} & \frac{d y}{d x}=\frac{d}{d x} \cos (a \cos x+b \sin x) \\ & \begin{aligned} \Rightarrow \frac{d y}{d x} & =-\sin (a \cos x+b \sin x) \cdot \frac{d}{d x}(a \cos x+b \sin x) \\ & =-\sin (a \cos x+b \sin x) \cdot[a(-\sin x)+b \cos x] \\ & =(a \sin x-b \cos x) \cdot \sin (a \cos x+b \sin x) \end{aligned} \end{aligned} $

Solution

Let $y=(\sin x-\cos x)^{(\sin x-\cos x)}$

Taking logarithm on both the sides, we obtain

$\log y=\log [(\sin x-\cos x)^{(\sin x-\cos x)}]$

$\Rightarrow \log y=(\sin x-\cos x) \cdot \log (\sin x-\cos x)$

Differentiating both sides with respect to $x$, we obtain

$\frac{1}{y} \frac{d y}{d x}=\frac{d}{d x}[(\sin x-\cos x) \log (\sin x-\cos x)]$

$\Rightarrow \frac{1}{y} \frac{d y}{d x}=\log (\sin x-\cos x) \cdot \frac{d}{d x}(\sin x-\cos x)+(\sin x-\cos x) \cdot \frac{d}{d x} \log (\sin x-\cos x)$

$\Rightarrow \frac{1}{y} \frac{d y}{d x}=\log (\sin x-\cos x) \cdot(\cos x+\sin x)+(\sin x-\cos x) \cdot \frac{1}{(\sin x-\cos x)} \cdot \frac{d}{d x}(\sin x-\cos x)$

$\Rightarrow \frac{d y}{d x}=(\sin x-\cos x)^{(\sin x-\cos x)}[(\cos x+\sin x) \cdot \log (\sin x-\cos x)+(\cos x+\sin x)]$

$\therefore \frac{d y}{d x}=(\sin x-\cos x)^{(\sin x-\cos x)}(\cos x+\sin x)[1+\log (\sin x-\cos x)]$

Solution

Let $y=x^{x}+x^{a}+a^{x}+a^{a}$

Also, let $x^{x}=u, x^{a}=v, a^{x}=w$, and $a^{a}=s$

$\therefore y=u+v+w+s$

$\Rightarrow \frac{d y}{d x}=\frac{d u}{d x}+\frac{d v}{d x}+\frac{d w}{d x}+\frac{d s}{d x}$

$u=x^{x}$

$\Rightarrow \log u=\log x^{x}$

$\Rightarrow \log u=x \log x$

Differentiating both sides with respect to $x$, we obtain $\frac{1}{u} \frac{d u}{d x}=\log x \cdot \frac{d}{d x}(x)+x \cdot \frac{d}{d x}(\log x)$

$\Rightarrow \frac{d u}{d x}=u[\log x \cdot 1+x \cdot \frac{1}{x}]$

$\Rightarrow \frac{d u}{d x}=x^{x}[\log x+1]=x^{x}(1+\log x)$

$v=x^{a}$

$\therefore \frac{d v}{d x}=\frac{d}{d x}(x^{a})$

$\Rightarrow \frac{d v}{d x}=a x^{a-1}$

$w=a^{x}$

$\Rightarrow \log w=\log a^{x}$

$\Rightarrow \log w=x \log a$

Differentiating both sides with respect to $x$, we obtain

$\frac{1}{w} \cdot \frac{d w}{d x}=\log a \cdot \frac{d}{d x}(x)$

$\Rightarrow \frac{d w}{d x}=w \log a$

$\Rightarrow \frac{d w}{d x}=a^{x} \log a$

$s=a^{a}$

Since $a$ is constant, $a^{a}$ is also a constant.

$\therefore \frac{d s}{d x}=0$

From (1), (2), (3), (4), and (5), we obtain

$ \begin{aligned} \frac{d y}{d x} & =x^{x}(1+\log x)+a x^{a-1}+a^{x} \log a+0 \\ & =x^{x}(1+\log x)+a x^{a-1}+a^{x} \log a \end{aligned} $

Solution

Let $y=x^{x^{2}-3}+(x-3)^{x^{2}}$

Also, let $u=x^{x^{2}-3}$ and $v=(x-3)^{x^{2}}$

$\therefore y=u+v$

Differentiating both sides with respect to $x$, we obtain

$\frac{d y}{d x}=\frac{d u}{d x}+\frac{d v}{d x}$

$u=x^{x^{2}-3}$

$\therefore \log u=\log (x^{x^{2}-3})$

$\log u=(x^{2}-3) \log x$

Differentiating with respect to $x$, we obtain

$\frac{1}{u} \cdot \frac{d u}{d x}=\log x \cdot \frac{d}{d x}(x^{2}-3)+(x^{2}-3) \cdot \frac{d}{d x}(\log x)$

$\Rightarrow \frac{1}{u} \frac{d u}{d x}=\log x \cdot 2 x+(x^{2}-3) \cdot \frac{1}{x}$

$\Rightarrow \frac{d u}{d x}=x^{x^{2}-3} \cdot[\frac{x^{2}-3}{x}+2 x \log x]$

Also,

$v=(x-3)^{x^{2}}$

$\therefore \log v=\log (x-3)^{x^{2}}$

$\Rightarrow \log v=x^{2} \log (x-3)$

Differentiating both sides with respect to $x$, we obtain $\frac{1}{v} \cdot \frac{d v}{d x}=\log (x-3) \cdot \frac{d}{d x}(x^{2})+x^{2} \cdot \frac{d}{d x}[\log (x-3)]$

$\Rightarrow \frac{1}{v} \frac{d v}{d x}=\log (x-3) \cdot 2 x+x^{2} \cdot \frac{1}{x-3} \cdot \frac{d}{d x}(x-3)$

$\Rightarrow \frac{d v}{d x}=v[2 x \log (x-3)+\frac{x^{2}}{x-3} \cdot 1]$

$\Rightarrow \frac{d v}{d x}=(x-3)^{x^{2}}[\frac{x^{2}}{x-3}+2 x \log (x-3)]$

Substituting the expressions of $\frac{d u}{d x}$ and $\frac{d v}{d x}$ in equation (1), we obtain

$\frac{d y}{d x}=x^{x^{2}-3}[\frac{x^{2}-3}{x}+2 x \log x]+(x-3)^{x^{2}}[\frac{x^{2}}{x-3}+2 x \log (x-3)]$

Solution

It is given that, $y=12(1-\cos t), x=10(t-\sin t)$

$\therefore \frac{d x}{d t}=\frac{d}{d t}[10(t-\sin t)]=10 \cdot \frac{d}{d t}(t-\sin t)=10(1-\cos t)$

$\frac{d y}{d t}=\frac{d}{d t}[12(1-\cos t)]=12 \cdot \frac{d}{d t}(1-\cos t)=12 \cdot[0-(-\sin t)]=12 \sin t$

$\therefore \frac{d y}{d x}=\frac{(\frac{d y}{d t})}{(\frac{d x}{d t})}=\frac{12 \sin t}{10(1-\cos t)}=\frac{12 \cdot 2 \sin \frac{t}{2} \cdot \cos \frac{t}{2}}{10 \cdot 2 \sin ^{2} \frac{t}{2}}=\frac{6}{5} \cot \frac{t}{2}$

Solution

It is given that, $y=\sin ^{-1} x+\sin ^{-1} \sqrt{1-x^{2}}$

$\therefore \frac{d y}{d x}=\frac{d}{d x}[\sin ^{-1} x+\sin ^{-1} \sqrt{1-x^{2}}]$

$\Rightarrow \frac{d y}{d x}=\frac{d}{d x}(\sin ^{-1} x)+\frac{d}{d x}(\sin ^{-1} \sqrt{1-x^{2}})$

$\Rightarrow \frac{d y}{d x}=\frac{1}{\sqrt{1-x^{2}}}+\frac{1}{\sqrt{1-(\sqrt{1-x^{2}})^{2}}} \cdot \frac{d}{d x}(\sqrt{1-x^{2}})$

$\Rightarrow \frac{d y}{d x}=\frac{1}{\sqrt{1-x^{2}}}+\frac{1}{x} \cdot \frac{1}{2 \sqrt{1-x^{2}}} \cdot \frac{d}{d x}(1-x^{2})$

$\Rightarrow \frac{d y}{d x}=\frac{1}{\sqrt{1-x^{2}}}+\frac{1}{2 x \sqrt{1-x^{2}}}(-2 x)$

$\Rightarrow \frac{d y}{d x}=\frac{1}{\sqrt{1-x^{2}}}-\frac{1}{\sqrt{1-x^{2}}}$

$\therefore \frac{d y}{d x}=0$

Solution

It is given that,

$x \sqrt{1+y}+y \sqrt{1+x}=0$ $\Rightarrow x \sqrt{1+y}=-y \sqrt{1+x}$

Squaring both sides, we obtain

$x^{2}(1+y)=y^{2}(1+x)$

$\Rightarrow x^{2}+x^{2} y=y^{2}+x y^{2}$

$\Rightarrow x^{2}-y^{2}=x y^{2}-x^{2} y$

$\Rightarrow x^{2}-y^{2}=x y(y-x)$

$\Rightarrow(x+y)(x-y)=x y(y-x)$

$\therefore x+y=-x y$

$\Rightarrow(1+x) y=-x$

$\Rightarrow y=\frac{-x}{(1+x)}$

Differentiating both sides with respect to $x$, we obtain

$y=\frac{-x}{(1+x)}$

$\frac{d y}{d x}=-\frac{(1+x) \frac{d}{d x}(x)-x \frac{d}{d x}(1+x)}{(1+x)^{2}}=-\frac{(1+x)-x}{(1+x)^{2}}=-\frac{1}{(1+x)^{2}}$

Hence, proved.

Solution

It is given that, $(x-a)^{2}+(y-b)^{2}=c^{2}$

Differentiating both sides with respect to $x$, we obtain

$$ \begin{align*} & \frac{d}{d x}[(x-a)^{2}]+\frac{d}{d x}[(y-b)^{2}]=\frac{d}{d x}(c^{2}) \\ & \Rightarrow 2(x-a) \cdot \frac{d}{d x}(x-a)+2(y-b) \cdot \frac{d}{d x}(y-b)=0 \\ & \Rightarrow 2(x-a) \cdot 1+2(y-b) \cdot \frac{d y}{d x}=0 \\ & \Rightarrow \frac{d y}{d x}=\frac{-(x-a)}{y-b} \tag{1}\\ & \therefore \frac{d^{2} y}{d x^{2}}=\frac{d}{d x}[\frac{-(x-a)}{y-b}] \end{align*} $$

$ \begin{aligned} & =-[\frac{(y-b) \cdot \frac{d}{d x}(x-a)-(x-a) \cdot \frac{d}{d x}(y-b)}{(y-b)^{2}}] \\ & =-[\frac{(y-b)-(x-a) \cdot \frac{d y}{d x}}{(y-b)^{2}}] \\ & =-[\frac{(y-b)-(x-a) \cdot{\frac{-(x-a)}{y-b}}}{(y-b)^{2}}] \\ & =-[\frac{(y-b)^{2}+(x-a)^{2}}{(y-b)^{3}}] \\ & \therefore[\frac{1+(\frac{d y}{d x})^{2}}{\frac{d^{2} y}{d x^{2}}}]^{\frac{3}{2}}=\frac{[1+\frac{(x-a)^{2}}{(y-b)^{2}}]^{\frac{3}{2}}}{-[\frac{(y-b)^{2}+(x-a)^{2}}{(y-b)^{3}}]}=\frac{[\frac{(y-b)^{2}+(x-a)^{2}}{(y-b)^{2}}]^{\frac{3}{2}}}{-[\frac{(y-b)^{2}+(x-a)^{2}}{(y-b)^{3}}]} \\ & =\frac{[\frac{c^{2}}{(y-b)^{2}}]^{\frac{3}{2}}}{-\frac{c^{2}}{(y-b)^{3}}}=\frac{\frac{c^{3}}{(y-b)^{3}}}{-\frac{c^{2}}{(y-b)^{3}}} \end{aligned} $

$ =-c \text{, which is constant and is independent of } a \text{ and } b $

Hence, proved.

Solution

It is given that, $\cos y=x \cos (a+y)$

$\therefore \frac{d}{d x}[\cos y]=\frac{d}{d x}[x \cos (a+y)]$

$\Rightarrow-\sin y \frac{d y}{d x}=\cos (a+y) \cdot \frac{d}{d x}(x)+x \cdot \frac{d}{d x}[\cos (a+y)]$

$\Rightarrow-\sin y \frac{d y}{d x}=\cos (a+y)+x \cdot[-\sin (a+y)] \frac{d y}{d x}$

$\Rightarrow[x \sin (a+y)-\sin y] \frac{d y}{d x}=\cos (a+y)$

Since $\cos y=x \cos (a+y), x=\frac{\cos y}{\cos (a+y)}$

Then, equation (1) reduces to

$[\frac{\cos y}{\cos (a+y)} \cdot \sin (a+y)-\sin y] \frac{d y}{d x}=\cos (a+y)$

$\Rightarrow[\cos y \cdot \sin (a+y)-\sin y \cdot \cos (a+y)] \cdot \frac{d y}{d x}=\cos ^{2}(a+y)$

$\Rightarrow \sin (a+y-y) \frac{d y}{d x}=\cos ^{2}(a+b)$

$\Rightarrow \frac{d y}{d x}=\frac{\cos ^{2}(a+b)}{\sin a}$

Hence, proved.

Solution

It is given that, $x=a(\cos t+t \sin t)$ and $y=a(\sin t-t \cos t)$

$\therefore \frac{d x}{d t}=a \cdot \frac{d}{d t}(\cos t+t \sin t)$

$=a[-\sin t+\sin t \cdot \frac{d}{d x}(t)+t \cdot \frac{d}{d t}(\sin t)]$

$=a[-\sin t+\sin t+t \cos t]=a t \cos t$

$\frac{d y}{d t}=a \cdot \frac{d}{d t}(\sin t-t \cos t)$

$=a[\cos t-{\cos t \cdot \frac{d}{d t}(t)+t \cdot \frac{d}{d t}(\cos t)}]$

$=a[\cos t-{\cos t-t \sin t}]=a t \sin t$

$\therefore \frac{d y}{d x}=\frac{(\frac{d y}{d t})}{(\frac{d x}{d t})}=\frac{a t \sin t}{a t \cos t}=\tan t$

Then, $\frac{d^{2} y}{d x^{2}}=\frac{d}{d x}(\frac{d y}{d x})=\frac{d}{d x}(\tan t)=\sec ^{2} t \cdot \frac{d t}{d x}$

$ \begin{aligned} & =\sec ^{2} t \cdot \frac{1}{a t \cos t} \quad[\frac{d x}{d t}=a t \cos t \Rightarrow \frac{d t}{d x}=\frac{1}{a t \cos t}] \\ & =\frac{\sec ^{3} t}{a t}, 0<t<\frac{\pi}{2} \end{aligned} $

Solution

It is known that, $|x|= \begin{cases}x, & \text{ if } x \geq 0 \\ -x, & \text{ if } x<0\end{cases}$

Therefore, when $x \geq 0, f(x)=|x|^{3}=x^{3}$

In this case, $f^{\prime}(x)=3 x^{2}$ and hence, $f^{\prime \prime}(x)=6 x$

When $x<0, f(x)=|x|^{3}=(-x)^{3}=-x^{3}$

In this case, $f^{\prime}(x)=-3 x^{2}$ and hence, $f^{\prime \prime}(x)=-6 x$

Thus, for $f(x)=|x|^{3}, f^{\prime \prime}(x)$ exists for all real $x$ and is given by,

$f^{\prime \prime}(x)= \begin{cases}6 x, & \text{ if } x \geq 0 \\ -6 x, & \text{ if } x<0\end{cases}$

Solution

$\sin (A+B)=\sin A \cos B+\cos A \sin B$

Differentiating both sides with respect to $x$, we obtain

$ \begin{aligned} & \frac{d}{d x}[\sin (A+B)]=\frac{d}{d x}(\sin A \cos B)+\frac{d}{d x}(\cos A \sin B) \\ & \Rightarrow \cos (A+B) \cdot \frac{d}{d x}(A+B)=\cos B \cdot \frac{d}{d x}(\sin A)+\sin A \cdot \frac{d}{d x}(\cos B) \\ & +\sin B \cdot \frac{d}{d x}(\cos A)+\cos A \cdot \frac{d}{d x}(\sin B) \\ & \Rightarrow \cos (A+B) \cdot \frac{d}{d x}(A+B)=\cos B \cdot \cos A \frac{d A}{d x}+\sin A(-\sin B) \frac{d B}{d x} \\ & +\sin B(-\sin A) \cdot \frac{d A}{d x}+\cos A \cos B \frac{d B}{d x} \\ & \Rightarrow \cos (A+B) \cdot[\frac{d A}{d x}+\frac{d B}{d x}]=(\cos A \cos B-\sin A \sin B) \cdot[\frac{d A}{d x}+\frac{d B}{d x}] \\ & \therefore \cos (A+B)=\cos A \cos B-\sin A \sin B \end{aligned} $

Solution

$ \begin{aligned} & y= \begin{vmatrix} f(x) & g(x) & h(x) \\ l & m & n \\ a & b & c \end{vmatrix} \\ & \Rightarrow y=(m c-n b) f(x)-(l c-n a) g(x)+(l b-m a) h(x) \end{aligned} $

Then, $\frac{d y}{d x}=\frac{d}{d x}[(m c-n b) f(x)]-\frac{d}{d x}[(l c-n a) g(x)]+\frac{d}{d x}[(l b-m a) h(x)]$

$ =(m c-n b) f^{\prime}(x)-(l c-n a) g^{\prime}(x)+(l b-m a) h^{\prime}(x) $

$ = \begin{vmatrix} f^{\prime}(x) & g^{\prime}(x) & h^{\prime}(x) \\ l & m & n \\ a & b & c \end{vmatrix} $

Thus, $\frac{d y}{d x}= \begin{vmatrix} f^{\prime}(x) & g^{\prime}(x) & h^{\prime}(x) \\ l & m & n \\ a & b & c\end{vmatrix} $

Solution

It is given that, $y=e^{a \cos ^{-1} x}$

Taking logarithm on both the sides, we obtain

$\log y=a \cos ^{-1} x \log e$

$\log y=a \cos ^{-1} x$

Differentiating both sides with respect to $x$, we obtain

$\frac{1}{y} \frac{d y}{d x}=a \times \frac{-1}{\sqrt{1-x^{2}}}$

$\Rightarrow \frac{d y}{d x}=\frac{-a y}{\sqrt{1-x^{2}}}$

By squaring both the sides, we obtain

$(\frac{d y}{d x})^{2}=\frac{a^{2} y^{2}}{1-x^{2}}$

$\Rightarrow(1-x^{2})(\frac{d y}{d x})^{2}=a^{2} y^{2}$

$(1-x^{2})(\frac{d y}{d x})^{2}=a^{2} y^{2}$

Again differentiating both sides with respect to $x$, we obtain

$(\frac{d y}{d x})^{2} \frac{d}{d x}(1-x^{2})+(1-x^{2}) \times \frac{d}{d x}[(\frac{d y}{d x})^{2}]=a^{2} \frac{d}{d x}(y^{2})$

$\Rightarrow(\frac{d y}{d x})^{2}(-2 x)+(1-x^{2}) \times 2 \frac{d y}{d x} \cdot \frac{d^{2} y}{d x^{2}}=a^{2} \cdot 2 y \cdot \frac{d y}{d x}$

$\Rightarrow(\frac{d y}{d x})^{2}(-2 x)+(1-x^{2}) \times 2 \frac{d y}{d x} \cdot \frac{d^{2} y}{d x^{2}}=a^{2} \cdot 2 y \cdot \frac{d y}{d x}$

$\Rightarrow-x \frac{d y}{d x}+(1-x^{2}) \frac{d^{2} y}{d x^{2}}=a^{2} \cdot y \quad[\frac{d y}{d x} \neq 0]$

$\Rightarrow(1-x^{2}) \frac{d^{2} y}{d x^{2}}-x \frac{d y}{d x}-a^{2} y=0$

Hence, proved.