Answer :

(i) Let $f(x)=- x$. Accordingly, $f(x+h)=-(x+h)$

By first principle,

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

(ii) Let

$ f(x)=(-x)^{-1}=\dfrac{1}{-x}=\dfrac{-1}{x} \text{.Accordingly, } f(x+h)=\dfrac{-1}{(x+h)} $

By first principle,

$ \begin{aligned} f^{\prime}(x) & =\lim _{h \to 0} \dfrac{f(x+h)-f(x)}{h} \\ & =\lim _{h \to 0} \dfrac{1}{h}[\dfrac{-1}{x+h}-(\dfrac{-1}{x})] \\ & =\lim _{h \to 0} \dfrac{1}{h}[\dfrac{-1}{x+h}+\dfrac{1}{x}] \\ & =\lim _{h \to 0} \dfrac{1}{h}[\dfrac{-x+(x+h)}{x(x+h)}] \\ & =\lim _{h \to 0} \dfrac{1}{h}[\dfrac{-x+x+h}{x(x+h)}] \\ & =\lim _{h \to 0} \dfrac{1}{h}[\dfrac{h}{x(x+h)}] \\ & =\lim _{h \to 0} \dfrac{1}{x(x+h)} \\ & =\dfrac{1}{x \cdot x}=\dfrac{1}{x^{2}} \end{aligned} $

(iii) Let $f(x)=\sin (x+1)$. Accordingly, $f(x+h)=\sin (x+h+1)$

By first principle,

$ \begin{aligned} f^{\prime}(x) & =\lim _{h \to 0} \dfrac{f(x+h)-f(x)}{h} \\ & =\lim _{h \to 0} \dfrac{1}{h}[\sin (x+h+1)-\sin (x+1)] \\ & =\lim _{h \to 0} \dfrac{1}{h}[2 \cos (\dfrac{x+h+1+x+1}{2}) \sin (\dfrac{x+h+1-x-1}{2})] \\ & =\lim _{h \to 0} \dfrac{1}{h}[2 \cos (\dfrac{2 x+h+2}{2}) \sin (\dfrac{h}{2})] \\ & =\lim _{h \to 0}[\cos (\dfrac{2 x+h+2}{2}) \cdot \dfrac{\sin (\dfrac{h}{2})}{(\dfrac{h}{2})}] \\ &=\lim _{h \rightarrow 0} \cos (\frac{2 x+h+2}{2}) \cdot \lim _{\frac{h}{2}\rightarrow 0} \frac{ (sin{\frac{h}{2}})}{(\frac{h}{2})} as h \rightarrow 0 \Rightarrow \frac{h}{2} \rightarrow 0\\ & =\cos (\dfrac{2 x+0+2}{2}) \cdot 1 \\ & =\cos (x+1) \\ \end{aligned} $

(iv) Let $ f(x)=\cos \left(x-\frac{\pi}{8}\right) \cdot \text { Accordingly }, f(x+h)=\cos \left(x+h-\frac{\pi}{8}\right) $

By first principle,

$ \begin{aligned} f^{\prime}(x) & =\lim _{h \to 0} \dfrac{f(x+h)-f(x)}{h} \\ & =\lim _{h \to 0} \dfrac{1}{h}[\cos (x+h-\dfrac{\pi}{8})-\cos (x-\dfrac{\pi}{8})] \end{aligned} $

$ \begin{aligned} & =\lim _{h \to 0} \dfrac{1}{h}[-2 \sin \dfrac{(x+h-\dfrac{\pi}{8}+x-\dfrac{\pi}{8})}{2} \sin (\dfrac{x+h-\dfrac{\pi}{8}-x+\dfrac{\pi}{8}}{2})] \\ & =\lim _{h \to 0} \dfrac{1}{h}[-2 \sin (\dfrac{2 x+h-\dfrac{\pi}{4}}{2}) \sin \dfrac{h}{2}] \\ & =\lim _{h \to 0}[-\sin (\dfrac{2 x+h-\dfrac{\pi}{4}}{2}) \dfrac{\sin (\dfrac{h}{2})}{(\dfrac{h}{2})}] \end{aligned} $

$ \begin{aligned} =\lim _{h \to 0}[-\sin (\dfrac{2 x+h-\dfrac{\pi}{4}}{2})] .\lim _{\dfrac{h}{2} \to 0}\dfrac{\sin (\dfrac{h}{2})}{(\dfrac{h}{2})} \hspace{30px} [\text{As } h \to 0 \Rightarrow \dfrac{h}{2} \to 0] \end{aligned} $

$ \begin{aligned} & =-\sin (\dfrac{2 x+0-\dfrac{\pi}{4}}{2}) .1 \\ & =-\sin (x-\dfrac{\pi}{8}) \end{aligned} $

Answer :

Let $f(x)=x+a$. Accordingly, $f(x+h)=x+h+a$

By first principle,

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

Answer :

Let $f(x)=(p x+q)(\dfrac{r}{x}+s)$

By Leibnitz product rule,

$ \begin{aligned} f^{\prime}(x) & =(p x+q)(\dfrac{r}{x}+s)^{\prime}+(\dfrac{r}{x}+s)(p x+q)^{\prime} \\ & =(p x+q)(r x^{-1}+s)^{\prime}+(\dfrac{r}{x}+s)(p) \\ & =(p x+q)(-r x^{-2})+(\dfrac{r}{x}+s) p \\ & =(p x+q)(\dfrac{-r}{x^{2}})+(\dfrac{r}{x}+s) p \\ & =\dfrac{-p r}{x}-\dfrac{q r}{x^{2}}+\dfrac{p r}{x}+p s \\ & =p s-\dfrac{q r}{x^{2}} \end{aligned} $

Answer :

Let $f(x)=(a x+b)(c x+d)^{2}$

By Leibnitz product rule,

$ \begin{aligned} f^{\prime}(x) & =(a x+b) \dfrac{d}{d x}(c x+d)^{2}+(c x+d)^{2} \dfrac{d}{d x}(a x+b) \\ & =(a x+b) \dfrac{d}{d x}(c^{2} x^{2}+2 c d x+d^{2})+(c x+d)^{2} \dfrac{d}{d x}(a x+b) \\ & =(a x+b)[\dfrac{d}{d x}(c^{2} x^{2})+\dfrac{d}{d x}(2 c d x)+\dfrac{d}{d x} d^{2}]+(c x+d)^{2}[\dfrac{d}{d x} a x+\dfrac{d}{d x} b] \\ & =(a x+b)(2 c^{2} x+2 c d)+(c x+d^{2}) a \\ & =2 c(a x+b)(c x+d)+a(c x+d)^{2} \end{aligned} $

Answer :

Let $f(x)=\dfrac{a x+b}{c x+d}$

By quotient rule,

$ \begin{aligned} f^{\prime}(x) & =\dfrac{(c x+d) \dfrac{d}{d x}(a x+b)-(a x+b) \dfrac{d}{d x}(c x+d)}{(c x+d)^{2}} \\ & =\dfrac{(c x+d)(a)-(a x+b)(c)}{(c x+d)^{2}} \\ & =\dfrac{a c x+a d-a c x-b c}{(c x+d)^{2}} \\ & =\dfrac{a d-b c}{(c x+d)^{2}} \end{aligned} $

Answer :

Let $f(x)=\dfrac{1+\dfrac{1}{x}}{1-\dfrac{1}{x}}=\dfrac{\dfrac{x+1}{x}}{\dfrac{x-1}{x}}=\dfrac{x+1}{x-1}$, where $x \neq 0$

By quotient rule,

$ \begin{aligned} f^{\prime}(x) & =\dfrac{(x-1) \dfrac{d}{d x}(x+1)-(x+1) \dfrac{d}{d x}(x-1)}{(x-1)^{2}}, x \neq 0,1 \\ & =\dfrac{(x-1)(1)-(x+1)(1)}{(x-1)^{2}}, x \neq 0,1 \\ & =\dfrac{x-1-x-1}{(x-1)^{2}}, x \neq 0,1 \\ & =\dfrac{-2}{(x-1)^{2}}, x \neq 0,1 \end{aligned} $

Answer :

Let $f(x)=\dfrac{1}{a x^{2}+b x+c}$

By quotient rule,

$ \begin{aligned} f^{\prime}(x) & =\dfrac{(a x^{2}+b x+c) \dfrac{d}{d x}(1)-\dfrac{d}{d x}(a x^{2}+b x+c)}{(a x^{2}+b x+c)^{2}} \\ & =\dfrac{(a x^{2}+b x+c)(0)-(2 a x+b)}{(a x^{2}+b x+c)^{2}} \\ & =\dfrac{-(2 a x+b)}{(a x^{2}+b x+c)^{2}} \end{aligned} $

Answer :

Let $f(x)=\dfrac{a x+b}{p x^{2}+q x+r}$

By quotient rule,

$ \begin{aligned} f^{\prime}(x) & =\dfrac{(p x^{2}+q x+r) \dfrac{d}{d x}(a x+b)-(a x+b) \dfrac{d}{d x}(p x^{2}+q x+r)}{(p x^{2}+q x+r)^{2}} \\ & =\dfrac{(p x^{2}+q x+r)(a)-(a x+b)(2 p x+q)}{(p x^{2}+q x+r)^{2}} \\ & =\dfrac{a p x^{2}+a q x+a r-2 a p x^{2}-a q x-2 b p x-b q}{(p x^{2}+q x+r)^{2}} \\ & =\dfrac{-a p x^{2}-2 b p x+a r-b q}{(p x^{2}+q x+r)^{2}} \end{aligned} $

Answer :

Let $f(x)=\dfrac{p x^{2}+q x+r}{a x+b}$

By quotient rule,

$ \begin{aligned} f^{\prime}(x) & =\dfrac{(a x+b) \dfrac{d}{d x}(p x^{2}+q x+r)-(p x^{2}+q x+r) \dfrac{d}{d x}(a x+b)}{(a x+b)^{2}} \\ & =\dfrac{(a x+b)(2 p x+q)-(p x^{2}+q x+r)(a)}{(a x+b)^{2}} \\ & =\dfrac{2 a p x^{2}+a q x+2 b p x+b q-a p x^{2}-a q x-a r}{(a x+b)^{2}} \\ & =\dfrac{a p x^{2}+2 b p x+b q-a r}{(a x+b)^{2}} \end{aligned} $

Answer :

$ \begin{aligned} & \text{ Let } f(x)=\dfrac{a}{x^{4}}-\dfrac{b}{x^{2}}+\cos x \\ & f^{\prime}(x)=\dfrac{d}{d x}(\dfrac{a}{x^{4}})-\dfrac{d}{d x}(\dfrac{b}{x^{2}})+\dfrac{d}{d x}(\cos x) \\ & =a \dfrac{d}{d x}(x^{-4})-b \dfrac{d}{d x}(x^{-2})+\dfrac{d}{d x}(\cos x) \\ & =a(-4 x^{-5})-b(-2 x^{-3})+(-\sin x) \quad[\dfrac{d}{d x}(x^{n})=n x^{n-1} \text{ and } \dfrac{d}{d x}(\cos x)=-\sin x] \\ & =\dfrac{-4 a}{x^{5}}+\dfrac{2 b}{x^{3}}-\sin x \end{aligned} $

Answer :

$ \begin{aligned} & \text{ Let } f(x)=4 \sqrt{x}-2 \\ & \begin{aligned} f^{\prime}(x) & =\dfrac{d}{d x}(4 \sqrt{x}-2)=\dfrac{d}{d x}(4 \sqrt{x})-\dfrac{d}{d x}(2) \\ & =4 \dfrac{d}{d x}(x^{\dfrac{1}{2}})-0=4(\dfrac{1}{2} x^{\dfrac{1}{2}-1}) \\ & =(2 x^{-\dfrac{1}{2}})=\dfrac{2}{\sqrt{x}} \end{aligned} \end{aligned} $

Answer :

Let $f(x)=(a x+b)^{n}$. Accordingly, $f(x+h)=\{a(x+h)+b\}^{n}=(a x+a h+b)^{n}$ By first principle,

$ \begin{aligned} f^{\prime}(x) & =\lim _{h \to 0} \dfrac{f(x+h)-f(x)}{h} \\ & =\lim _{h \to 0} \dfrac{(a x+a h+b)^{n}-(a x+b)^{n}}{h} \\ & =\lim _{h \to 0} \dfrac{(a x+b)^{n}(1+\dfrac{a h}{a x+b})^{n}-(a x+b)^{n}}{h} \\ & =(a x+b)^{n} \lim _{h \to 0} \dfrac{(1+\dfrac{a h}{a x+b})^{n}-1}{h} \\ & =(a x+b)^{n} \lim _{h \to 0} \dfrac{1}{n}[\{1+n(\dfrac{a h}{a x+b})+\dfrac{n(n-1)}{\lfloor 2}(\dfrac{a h}{a x+b})^{2}+\ldots\}-1] \\ & =(a x+b)^{n} \lim _{h \to 0} \dfrac{1}{h}[\dfrac{n}{a h})^{n}]+\dfrac{n(n-1) a^{2} h^{2}}{\lfloor(a x+b)^{2}.}+\ldots(\text{ Terms containing higher degrees of } h)] \\ & =(a x+b)^{n} \lim _{h \to 0}[\dfrac{n a}{(a x+b)}+\dfrac{n(n-1) a^{2} h}{\lfloor(a x+b)^{2}.}+\ldots] \\ & =(a x+b)^{n}[\dfrac{n a}{(a x+b)}+0] \\ & =n a (a x+b)^{n-1} \end{aligned} $

Answer :

Let $f(x)=(a x+b)^{n}(c x+d)^{m}$

By Leibnitz product rule,

$$ \begin{equation*} f^{\prime}(x)=(a x+b)^{n} \dfrac{d}{d x}(c x+d)^{m}+(c x+d)^{m} \dfrac{d}{d x}(a x+b)^{n} \tag{1} \end{equation*} $$

Now, let $f_1(x)=(c x+d)^{m}$

$f_1(x+h)=(c x+c h+d)^{m}$

$f_1^{\prime}(x)=\lim _{h \to 0} \dfrac{f_1(x+h)-f_1^{\prime}(x)}{h}$

$=\lim _{h \to 0} \dfrac{(c x+c h+d)^{m}-(c x+d)^{m}}{h}$

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

$=(c x+d)^{m} \lim _{h \to 0} \dfrac{1}{h}[(1+\dfrac{m c h}{(c x+d)}+\dfrac{m(m-1)}{2} \dfrac{(c^{2} h^{2})}{(c x+d)^{2}}+\ldots)-1]$

$=(c x+d)^{m} \lim _{h \to 0} \dfrac{1}{h}[\dfrac{m c h}{(c x+d)}+\dfrac{m(m-1) c^{2} h^{2}}{2(c x+d)^{2}}+\ldots(.$ Terms containing higher degrees of $.h)]$

$=(c x+d)^{m} \lim _{h \to 0}[\dfrac{m c}{(c x+d)}+\dfrac{m(m-1) c^{2} h}{2(c x+d)^{2}}+\ldots]$

$=(c x+d)^{m}[\dfrac{m c}{c x+d}+0]$

$=\dfrac{m c(c x+d)^{m}}{(c x+d)}$

$=m c(c x+d)^{m-1}$

$\dfrac{d}{d x}(c x+d)^{It}=m c(c x+d)^{m-1}$

Similarly, $\dfrac{d}{d x}(a x+b)^{n}=n a(a x+b)^{n-1}$

Therefore, from (1), (2), and (3), we obtain

$ \begin{aligned} f^{\prime}(x) & =(a x+b)^{n}\{m c(c x+d)^{m-1}\}+(c x+d)^{m}\{n a(a x+b)^{n-1}\} \\ & =(a x+b)^{n-1}(c x+d)^{m-1}[m c(a x+b)+n a(c x+d)] \end{aligned} $

Answer :

Let $f(x)=\sin (x+a)$

$f(x+h)=\sin (x+h+a)$

By first principle,

$ \begin{aligned} & f^{\prime}(x)=\lim _{h \to 0} \dfrac{f(x+h)-f(x)}{h} \\ & =\lim _{h \to 0} \dfrac{\sin (x+h+a)-\sin (x+a)}{h} \\ & =\lim _{h \to 0} \dfrac{1}{h}[2 \cos (\dfrac{x+h+a+x+a}{2}) \sin (\dfrac{x+h+a-x-a}{2})] \\ & =\lim _{h \to \infty} \dfrac{1}{h}[2 \cos (\dfrac{2 x+2 a+h}{2}) \sin (\dfrac{h}{2})] \\ & =\lim _{h \to 0}[\cos (\dfrac{2 x+2 a+h}{2})\{\dfrac{\sin (\dfrac{h}{2})}{(\dfrac{h}{2})}\}] \\ & =\lim _{h \to 0} \cos (\dfrac{2 x+2 a+h}{2}) \lim _{\dfrac{h}{2} \to 0}\{\dfrac{\sin (\dfrac{h}{2})}{(\dfrac{h}{2})}\} \quad[\text{ As } h \to 0 \Rightarrow \dfrac{h}{2} \to 0] \\ & =\cos (\dfrac{2 x+2 a}{2}) \times 1 \quad[\lim _{x \to 0} \dfrac{\sin x}{x}=1] \\ & =\cos (x+a) \end{aligned} $

Answer :

Let $f(x)=cosec x \cot x$

By Leibnitz product rule,

$f^{\prime}(x)=cosec x(\cot x)^{\prime}+\cot x(cosec x)^{\prime}$

Let $f_1(x)=\cot x$. Accordingly, $f_1(x+h)=\cot (x+h)$

By first principle,

$$ \begin{align*} f_1^{\prime}(x) & =\lim _{h \to 0} \dfrac{f_1(x+h)-f_1(x)}{h} \\ & =\lim _{h \to 0} \dfrac{\cot (x+h)-\cot x}{h} \\ & =\lim _{h \to 0} \dfrac{1}{h}(\dfrac{\cos (x+h)}{\sin (x+h)}-\dfrac{\cos x}{\sin x}) \\ & =\lim _{h \to 0} \dfrac{1}{h}[\dfrac{\sin x \cos (x+h)-\cos x \sin (x+h)}{\sin x \sin (x+h)}] \\ & =\lim _{h \to 0} \dfrac{1}{h}[\dfrac{\sin (x-x-h)}{\sin x \sin (x+h)}] \\ & =\dfrac{1}{\sin _{x} x} \cdot \lim _{h \to 0} \dfrac{1}{h}[\dfrac{\sin (-h)}{\sin (x+h)}] \\ & =\dfrac{-1}{\sin ^{x} x} \cdot(\lim _{h \to 0} \dfrac{\sin h}{h})(\lim _{h \to 0} \dfrac{1}{\sin (x+h)}) \\ & =\dfrac{-1}{\sin ^{x}} \cdot 1 \cdot(\dfrac{1}{\sin (x+0)}) \\ & =\dfrac{-1}{\sin ^{2} x} \\ & =-cosec^{2} x \tag{2} \end{align*} $$

$\therefore(\cot x)^{\prime}=-cosec^{2} x$

Now, let $f_2(x)=cosec x$. Accordingly, $f_2(x+h)=cosec(x+h)$

By first principle,

$ \begin{aligned} f_2^{\prime}(x) & =\lim _{h \to 0} \dfrac{f_2(x+h)-f_2(x)}{h} \\ & =\lim _{h \to 0} \dfrac{1}{h}[cosec(x+h)-cosec x] \end{aligned} $

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

$$ \begin{align*} & =\dfrac{1}{\sin x} \cdot \lim _{h \to 0} \dfrac{1}{h}[\dfrac{2 \cos (\dfrac{x+x+h}{2}) \sin (\dfrac{x-x-h}{2})}{\sin (x+h)}] \\ & =\dfrac{1}{\sin x} \cdot \lim _{h \to 0} \dfrac{1}{h}[\dfrac{.2 \cos (\dfrac{2 x+h}{2}) \sin (\dfrac{-h}{2})]}{\sin (x+h)}] \\ & =\dfrac{1}{\sin x} \cdot \lim _{h \to 0}[\dfrac{-\sin (\dfrac{h}{2})}{(\dfrac{h}{2})} \cdot \dfrac{\cos (\dfrac{2 x+h}{2})}{\sin (x+h)}] \\ & =\dfrac{-1}{\sin x} \cdot \lim _{h \to 0} \dfrac{\sin (\dfrac{h}{2})}{(\dfrac{h}{2})} \cdot \lim _{h \to 0} \dfrac{\cos (\dfrac{2 x+h}{2})}{\sin (x+h)} \\ & =\dfrac{-1}{\sin x} \cdot 1 \cdot \dfrac{\cos (\dfrac{2 x+0}{2})}{\sin (x+0)} \\ & =\dfrac{-1}{\sin x} \cdot \dfrac{\cos x}{\sin x} \\ & =-\cos ec x \cdot \cot x \tag{3} \end{align*} $$

$\therefore(cosec x)^{\prime}=-cosec x \cdot \cot x$

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

$ \begin{aligned} f^{\prime}(x) & =cosec x(-cosec^{2} x)+\cot x(-cosec x \cot x) \\ & =-cosec^{3} x-\cot ^{2} x cosec x \end{aligned} $

Answer :

Let $f(x)=\dfrac{\cos x}{1+\sin x}$

By quotient rule,

$ \begin{aligned} f^{\prime}(x) & =\dfrac{(1+\sin x) \dfrac{d}{d x}(\cos x)-(\cos x) \dfrac{d}{d x}(1+\sin x)}{(1+\sin x)^{2}} \\ & =\dfrac{(1+\sin x)(-\sin x)-(\cos x)(\cos x)}{(1+\sin x)^{2}} \\ & =\dfrac{-\sin x-\sin ^{2} x-\cos ^{2} x}{(1+\sin x)^{2}} \\ & =\dfrac{-\sin x-(\sin ^{2} x+\cos ^{2} x)}{(1+\sin x)^{2}} \\ & =\dfrac{-\sin x-1}{(1+\sin x)^{2}} \\ & =\dfrac{-(1+\sin x)}{(1+\sin x)^{2}} \\ & =\dfrac{-1}{(1+\sin x)} \end{aligned} $

Answer :

Let $f(x)=\dfrac{\sin x+\cos x}{\sin x-\cos x}$

By quotient rule,

$ \begin{aligned} f^{\prime}(x) & =\dfrac{(\sin x-\cos x) \dfrac{d}{d x}(\sin x+\cos x)-(\sin x+\cos x) \dfrac{d}{d x}(\sin x-\cos x)}{(\sin x-\cos x)^{2}} \\ & =\dfrac{(\sin x-\cos x)(\cos x-\sin x)-(\sin x+\cos x)(\cos x+\sin x)}{(\sin x-\cos x)^{2}} \\ & =\dfrac{-(\sin x-\cos x)^{2}-(\sin x+\cos x)^{2}}{(\sin x-\cos x)^{2}} \\ & =\dfrac{-[\sin ^{2} x+\cos ^{2} x-2 \sin x \cos x+\sin ^{2} x+\cos ^{2} x+2 \sin x \cos x]}{(\sin x-\cos x)^{2}} \\ & =\dfrac{-[1+1]}{(\sin x-\cos x)^{2}} \\ & =\dfrac{-2}{(\sin x-\cos x)^{2}} \end{aligned} $

Answer :

Let $f(x)=\dfrac{\sec x-1}{\sec x+1}$

$f(x)=\dfrac{\dfrac{1}{\cos x}-1}{\dfrac{1}{\cos x}+1}=\dfrac{1-\cos x}{1+\cos x}$

By quotient rule,

$ \begin{aligned} & f^{\prime}(x)=\dfrac{(1+\cos x) \dfrac{d}{d x}(1-\cos x)-(1-\cos x) \dfrac{d}{d x}(1+\cos x)}{(1+\cos x)^{2}} \\ & =\dfrac{(1+\cos x)(\sin x)-(1-\cos x)(-\sin x)}{(1+\cos x)^{2}} \\ & =\dfrac{\sin x+\cos x \sin x+\sin x-\sin x \cos x}{(1+\cos x)^{2}} \\ & =\dfrac{2 \sin x}{(1+\cos x)^{2}} \\ & =\dfrac{2 \sin x}{(1+\dfrac{1}{\sec x})^{2}}=\dfrac{2 \sin x}{\dfrac{(\sec x+1)^{2}}{\sec ^{2} x}} \\ & =\dfrac{2 \sin x \sec ^{2} x}{(\sec x+1)^{2}} \\ & =\dfrac{\dfrac{2 \sin x}{\cos x} \sec x}{(\sec x+1)^{2}} \\ & =\dfrac{2 \sec x \tan x}{(\sec x+1)^{2}} \end{aligned} $

Answer :

Let $y=\sin ^{n} x$.

Accordingly, for $n=1, y=\sin x$.

$\therefore \dfrac{d y}{d x}=\cos x$, i.e., $\dfrac{d}{d x} \sin x=\cos x$

For $n=2, y=\sin ^{2} x$.

$$ \begin{align*} \therefore \dfrac{d y}{d x} & =\dfrac{d}{d x}(\sin x \sin x) \\ & =(\sin x)^{\prime} \sin x+\sin x(\sin x)^{\prime} \\ & =\cos x \sin x+\sin x \cos x \\ & =2 \sin x \cos x \tag{1} \end{align*} $$

For $n=3, y=\sin ^{3} x$.

$ \therefore \dfrac{d y}{d x}=\dfrac{d}{d x}(\sin x \sin ^{2} x) $

$$ \begin{matrix} =(\sin x)^{\prime} \sin ^{2} x+\sin x(\sin ^{2} x)^{\prime} & \text{ [By Leibnitz product rule] } \\ =\cos x \sin ^{2} x+\sin x(2 \sin x \cos x) & \text{ [Using (1)] } \\ =\cos x \sin ^{2} x+2 \sin ^{2} x \cos x & \\ =3 \sin ^{2} x \cos x & \end{matrix} $$

We assert that $\dfrac{d}{d x}(\sin ^{n} x)=n \sin ^{(n-1)} x \cos x$

Let our assertion be true for $n=k$.

i.e., $\dfrac{d}{d x}(\sin ^{k} x)=k \sin ^{(k-1)} x \cos x$

Consider

$ \begin{aligned} \dfrac{d}{d x}(\sin ^{k+1} x) & =\dfrac{d}{d x}(\sin x \sin ^{k} x) \\ & =(\sin x)^{\prime} \sin ^{k} x+\sin x(\sin ^{k}. \\ & =\cos x \sin ^{k} x+\sin x(k \sin ^{(k-1}. \\ & =\cos x \sin ^{k} x+k \sin ^{k} x \cos x \\ & =(k+1) \sin ^{k} x \cos x \end{aligned} $

$ \begin{matrix} =(\sin x)^{\prime} \sin ^{k} x+\sin x(\sin ^{k} x)^{\prime} & \text{ [By Leibnitz product rule] } \\ =\cos x \sin ^{k} x+\sin x(k \sin ^{(k-1)} x \cos x) & {[\text{ Using (2)] }} \end{matrix} $

Thus, our assertion is true for $n=k+1$.

Hence, by mathematical induction, $\dfrac{d}{d x}(\sin ^{n} x)=n \sin ^{(n-1)} x \cos x$

Answer :

Let $f(x)=\dfrac{a+b \sin x}{c+d \cos x}$

By quotient rule,

$ \begin{aligned} f^{\prime}(x) & =\dfrac{(c+d \cos x) \dfrac{d}{d x}(a+b \sin x)-(a+b \sin x) \dfrac{d}{d x}(c+d \cos x)}{(c+d \cos x)^{2}} \\ & =\dfrac{(c+d \cos x)(b \cos x)-(a+b \sin x)(-d \sin x)}{(c+d \cos x)^{2}} \\ & =\dfrac{c b \cos x+b d \cos ^{2} x+a d \sin x+b d \sin ^{2} x}{(c+d \cos x)^{2}} \\ & =\dfrac{b c \cos x+a d \sin x+b d(\cos ^{2} x+\sin ^{2} x)}{(c+d \cos x)^{2}} \\ & =\dfrac{b c \cos x+a d \sin x+b d}{(c+d \cos x)^{2}} \end{aligned} $

Answer :

Let $f(x)=\dfrac{\sin (x+a)}{\cos x}$

By quotient rule,

$$ \begin{align*} f^{\prime}(x) & =\dfrac{\cos x \dfrac{d}{d x}[\sin (x+a)]-\sin (x+a) \dfrac{d}{d x} \cos x}{\cos ^{2} x} \\ f^{\prime}(x) & =\dfrac{\cos x \dfrac{d}{d x}[\sin (x+a)]-\sin (x+a)(-\sin x)}{\cos ^{2} x} \tag{i} \end{align*} $$

Let $g(x)=\sin (x+a)$. Accordingly, $g(x+h)=\sin (x+h+a)$

By first principle,

$$ \begin{align*} g^{\prime}(x) & =\lim _{h \to 0} \dfrac{g(x+h)-g(x)}{h} \\ & =\lim _{h \to 0} \dfrac{1}{h}[\sin (x+h+a)-\sin (x+a)] \\ & =\lim _{h \to 0} \dfrac{1}{h}[2 \cos (\dfrac{x+h+a+x+a}{2}) \sin (\dfrac{x+h+a-x-a}{2})] \\ & =\lim _{h \to 0} \dfrac{1}{h}[2 \cos (\dfrac{2 x+2 a+h}{2}) \sin (\dfrac{h}{2})] \\ & =\lim _{h \to 0}{\operatorname { c o s } ( \dfrac { 2 x + 2 a + h } { 2 } ) [\dfrac{\sin (\dfrac{h}{2})}.(\dfrac{h}{2})]} \quad[\text{ As } h \to 0 \Rightarrow \dfrac{h}{2} \to 0].. \\ & =\lim _{h \to 0} \cos (\dfrac{2 x+2 a+h}{2}) \lim _{\dfrac{h}{2} \to 0}\{\dfrac{\sin (\dfrac{h}{2})}{(\dfrac{h}{2})}\} \quad[\lim _{h \to 0} \dfrac{\sin h}{h}=1] \\ & =(\cos \dfrac{2 x+2 a}{2}) \times 1 \\ & =\cos (x+a) \tag{ii} \end{align*} \text{ (ii) } \quad $$

From (i) and (ii), we obtain

$ \begin{aligned} f^{\prime}(x) & =\dfrac{\cos x \cdot \cos (x+a)+\sin x \sin (x+a)}{\cos ^{2} x} \\ & =\dfrac{\cos (x+a-x)}{\cos ^{2} x} \\ & =\dfrac{\cos a}{\cos ^{2} x} \end{aligned} $

Answer :

Let $f(x)=x^{4}(5 \sin x-3 \cos x)$

By product rule,

$ \begin{aligned} f^{\prime}(x) & =x^{4} \dfrac{d}{d x}(5 \sin x-3 \cos x)+(5 \sin x-3 \cos x) \dfrac{d}{d x}(x^{4}) \\ & =x^{4}[5 \dfrac{d}{d x}(\sin x)-3 \dfrac{d}{d x}(\cos x)]+(5 \sin x-3 \cos x) \dfrac{d}{d x}(x^{4}) \\ & =x^{4}[5 \cos x-3(-\sin x)]+(5 \sin x-3 \cos x)(4 x^{3}) \\ & =x^{3}[5 x \cos x+3 x \sin x+20 \sin x-12 \cos x] \end{aligned} $

Answer:

Let $f(x)=(x^{2}+1) \cos x$

By product rule,

$ \begin{aligned} f^{\prime}(x) & =(x^{2}+1) \dfrac{d}{d x}(\cos x)+\cos x \dfrac{d}{d x}(x^{2}+1) \\ & =(x^{2}+1)(-\sin x)+\cos x(2 x) \\ & =-x^{2} \sin x-\sin x+2 x \cos x \end{aligned} $

Answer :

Let $f(x)=(a x^{2}+\sin x)(p+q \cos x)$

By product rule,

$ \begin{aligned} f^{\prime}(x) & =(a x^{2}+\sin x) \dfrac{d}{d x}(p+q \cos x)+(p+q \cos x) \dfrac{d}{d x}(a x^{2}+\sin x) \\ & =(a x^{2}+\sin x)(-q \sin x)+(p+q \cos x)(2 a x+\cos x) \\ & =-q \sin x(a x^{2}+\sin x)+(p+q \cos x)(2 a x+\cos x) \end{aligned} $

Answer :

Let $f(x)=(x+\cos x)(x-\tan x)$

By product rule,

$ \begin{align*} & f^{\prime}(x) =(x+\cos x) \dfrac{d}{d x}(x-\tan x)+(x-\tan x) \dfrac{d}{d x}(x+\cos x) \\ & =(x+\cos x)[\dfrac{d}{d x}(x)-\dfrac{d}{d x}(\tan x)]+(x-\tan x)(1-\sin x) \\ & =(x+\cos x)[1-\dfrac{d}{d x} \tan x]+(x-\tan x)(1-\sin x)\tag{i}\\ \end{align*} $

$ \begin{align*} & \text{ Let } g(x)=\tan x \text{. Accordingly, } g(x+h)=\tan (x+h) \end{align*} $

By first principle,

$$ \begin{align*} g^{\prime}(x) & =\lim _{h \to 0} \dfrac{g(x+h)-g(x)}{h} \\ & =\lim _{h \to 0}(\dfrac{\tan (x+h)-\tan x}{h}) \\ & =\lim _{h \to 0} \dfrac{1}{h}[\dfrac{\sin (x+h)}{\cos (x+h)}-\dfrac{\sin x}{\cos x}] \\ & =\lim _{h \to 0} \dfrac{1}{h}[\dfrac{\sin (x+h) \cos x-\sin x \cos (x+h)}{\cos (x+h) \cos x}] \\ & =\dfrac{1}{\cos x} \cdot \lim _{h \to 0} \dfrac{1}{h}[\dfrac{\sin (x+h-x)}{\cos (x+h)}] \\ & =\dfrac{1}{\cos x} \cdot \lim _{h \to 0} \dfrac{1}{h}[\dfrac{\sin h}{\cos (x+h)}] \\ & =\dfrac{1}{\cos ^{x}} \cdot(\lim _{h \to 0} \dfrac{\sin h}{h}) \cdot(\lim _{h \to 0} \dfrac{1}{\cos (x+h)}) \\ & =\dfrac{1}{\cos ^{2}} \cdot 1 \cdot \dfrac{1}{\cos (x+0)} \\ & =\dfrac{1}{\cos ^{2} x} \\ & =\sec ^{2} x \tag{ii} \end{align*} $$

Therefore, from (i) and (ii), we obtain

$$ \begin{aligned} f^{\prime}(x) & =(x+\cos x)(1-\sec ^{2} x)+(x-\tan x)(1-\sin x) \\ & =(x+\cos x)(-\tan ^{2} x)+(x-\tan x)(1-\sin x) \\ & =-\tan ^{2} x(x+\cos x)+(x-\tan x)(1-\sin x) \end{aligned} $$

Answer :

Let $f(x)=\dfrac{4 x+5 \sin x}{3 x+7 \cos x}$

By quotient rule,

$ \begin{aligned} f^{\prime}(x) & =\dfrac{(3 x+7 \cos x) \dfrac{d}{d x}(4 x+5 \sin x)-(4 x+5 \sin x) \dfrac{d}{d x}(3 x+7 \cos x)}{(3 x+7 \cos x)^{2}} \\ & =\dfrac{(3 x+7 \cos x)[4 \dfrac{d}{d x}(x)+5 \dfrac{d}{d x}(\sin x)]-(4 x+5 \sin x)[3 \dfrac{d}{d x} x+7 \dfrac{d}{d x} \cos x]}{(3 x+7 \cos x)^{2}} \\ & =\dfrac{(3 x+7 \cos x)(4+5 \cos x)-(4 x+5 \sin x)(3-7 \sin x)}{(3 x+7 \cos x)^{2}} \\ & =\dfrac{12 x+15 x \cos x+28 \cos x+35 \cos ^{2} x-12 x+28 x \sin x-15 \sin x+35 \sin ^{2} x}{(3 x+7 \cos x)^{2}} \\ & =\dfrac{15 x \cos x+28 \cos x+28 x \sin x-15 \sin x+35(\cos ^{2} x+\sin ^{2} x)}{(3 x+7 \cos x)^{2}} \\ & =\dfrac{35+15 x \cos x+28 \cos x+28 x \sin x-15 \sin x}{(3 x+7 \cos x)^{2}} \end{aligned} $

Answer :

Let $f(x)=\dfrac{x^{2} \cos (\dfrac{\pi}{4})}{\sin x}$

By quotient rule,

$ \begin{aligned} f^{\prime}(x) & =\cos \dfrac{\pi}{4} \cdot[\dfrac{\sin x \dfrac{d}{d x}(x^{2})-x^{2} \dfrac{d}{d x}(\sin x)}{\sin ^{2} x}] \\ & =\cos \dfrac{\pi}{4} \cdot[\dfrac{\sin x \cdot 2 x-x^{2} \cos x}{\sin ^{2} x}] \\ & =\dfrac{x \cos \dfrac{\pi}{4}[2 \sin x-x \cos x]}{\sin ^{2} x} \end{aligned} $

Answer :

Let $f(x)=\dfrac{x}{1+\tan x}$

$f^{\prime}(x)=\dfrac{(1+\tan x) \dfrac{d}{d x}(x)-x \dfrac{d}{d x}(1+\tan x)}{(1+\tan x)^{2}}$

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

Let $g(x)=1+\tan x$. Accordingly, $g(x+h)=1+\tan (x+h)$.

By first principle,

$$ \begin{align*} g^{\prime}(x) & =\lim _{h \to 0} \dfrac{g(x+h)-g(x)}{h} \\ & =\lim _{h \to 0}[\dfrac{1+\tan (x+h)-1-\tan x}{h}] \\ & =\lim _{h \to 0} \dfrac{1}{h}[\dfrac{\sin (x+h)}{\cos (x+h)}-\dfrac{\sin x}{\cos x}] \\ & =\lim _{h \to 0} \dfrac{1}{h}[\dfrac{\sin (x+h) \cos x-\sin x \cos (x+h)}{\cos (x+h) \cos x}] \\ & =\lim _{h \to 0} \dfrac{1}{h}[\dfrac{\sin (x+h-x)}{\cos (x+h) \cos x}] \\ & =\lim _{h \to 0} \dfrac{1}{h}[\dfrac{\sin h}{\cos (x+h) \cos x}] \\ & =(\lim _{h \to 0} \dfrac{\sin h}{h}) \cdot(\lim _{h \to 0} \dfrac{1}{\cos (x+h) \cos x}) \\ & =1 \times \dfrac{1}{\cos ^{2} x}=\sec ^{2} x \\ \Rightarrow \dfrac{d}{d x} & (1+\tan x)=\sec ^{2} x \tag{ii} \end{align*} $$

From (i) and (ii), we obtain

$ f^{\prime}(x)=\dfrac{1+\tan x-x \sec ^{2} x}{(1+\tan x)^{2}} $

Answer :

Let $f(x)=(x+\sec x)(x-\tan x)$

By product rule,

$$ \begin{align*} f^{\prime}(x) & =(x+\sec x) \dfrac{d}{d x}(x-\tan x)+(x-\tan x) \dfrac{d}{d x}(x+\sec x) \\ & =(x+\sec x)[\dfrac{d}{d x}(x)-\dfrac{d}{d x} \tan x]+(x-\tan x)[\dfrac{d}{d x}(x)+\dfrac{d}{d x} \sec x] \\ & =(x+\sec x)[1-\dfrac{d}{d x} \tan x]+(x-\tan x)[1+\dfrac{d}{d x} \sec x] \tag{i} \end{align*} $$

Let $f_1(x)=\tan x, f_2(x)=\sec x$

Accordingly, $f_1(x+h)=\tan (x+h)$ and $f_2(x+h)=\sec (x+h)$

$$ \begin{align*} f_1^{\prime}(x) & =\lim _{h \to 0}(\dfrac{f_1(x+h)-f_1(x)}{h}) \\ & =\lim _{h \to 0}(\dfrac{\tan (x+h)-\tan x}{h}) \\ & =\lim _{h \to 0}[\dfrac{\tan (x+h)-\tan x}{h}] \\ & =\lim _{h \to 0} \dfrac{1}{h}[\dfrac{\sin (x+h)}{\cos (x+h)}-\dfrac{\sin x}{\cos x}] \\ & =\lim _{h \to 0} \dfrac{1}{h}[\dfrac{\sin (x+h) \cos x-\sin x \cos (x+h)}{\cos (x+h) \cos x}] \\ & =\lim _{h \to 0} \dfrac{1}{h}[\dfrac{\sin (x+h-x)}{\cos (x+h) \cos x}] \\ & =\lim _{h \to 0} \dfrac{1}{h}[\dfrac{\sin h}{\cos (x+h) \cos x}] \\ & =(\lim _{h \to 0} \dfrac{\sin h}{h}) \cdot(\lim _{h \to 0} \dfrac{1}{\cos (x+h) \cos x}) \\ & =1 \times \dfrac{1}{\cos ^{2} x}=\sec ^{2} x \\ \Rightarrow \dfrac{d}{d x} & \tan x=\sec ^{2} x \tag{ii} \end{align*} $$

$$ \begin{aligned} & f_2^{\prime}(x)=\lim _{h \to 0}(\dfrac{f_2(x+h)-f_2(x)}{h}) \\ & =\lim _{h \to 0}(\dfrac{\sec (x+h)-\sec x}{h}) \\ & =\lim _{h \to 0} \dfrac{1}{h}[\dfrac{1}{\cos (x+h)}-\dfrac{1}{\cos x}] \\ & =\lim _{h \to 0} \dfrac{1}{h}[\dfrac{\cos x-\cos (x+h)}{\cos (x+h) \cos x}] \\ & =\dfrac{1}{\cos x} \cdot \lim _{h \to 0} \dfrac{1}{h}[\dfrac{-2 \sin (\dfrac{x+x+h}{2}) \cdot \sin (\dfrac{x-x-h}{2})}{\cos (x+h)}] \\ & =\dfrac{1}{\cos x} \cdot \lim _{h \to 0} \dfrac{1}{h}[\dfrac{-2 \sin (\dfrac{2 x+h}{2}) \cdot \sin (\dfrac{-h}{2})}{\cos (x+h)}] \\ & =\dfrac{1}{\cos x} \cdot \lim _{h \to 0}[\dfrac{\sin (\dfrac{2 x+h}{2})\{\dfrac{\sin (\dfrac{h}{2})}{\dfrac{h}{2}}\}}{\cos (x+h)}] \\ & =\sec x \dfrac{\{\lim _{h \to 0} \sin (\dfrac{2 x+h}{2})\}\{\lim _{\dfrac{h}{2} \to 0} \dfrac{\sin (\dfrac{h}{2})}{\dfrac{h}{2}}\}}{\lim _{h \to 0} \cos (x+h)} \\ & =\sec x \cdot \dfrac{\sin x \cdot 1}{\cos x} \\ & \Rightarrow \dfrac{d}{d x} \sec x=\sec x \tan x \end{aligned} $$

From (i), (ii), and (iii), we obtain

$f^{\prime}(x)=(x+\sec x)(1-\sec ^{2} x)+(x-\tan x)(1+\sec x \tan x)$

Answer :

Let $f(x)=\dfrac{x}{\sin ^{n} x}$

By quotient rule,

$f^{\prime}(x)=\dfrac{\sin ^{n} x \dfrac{d}{d x} x-x \dfrac{d}{d x} \sin ^{n} x}{\sin ^{2 n} x}$

It can be easily shown that $\dfrac{d}{d x} \sin ^{n} x=n \sin ^{n-1} x \cos x$

Therefore,

$$ \begin{aligned} f^{\prime}(x) & =\dfrac{\sin ^{n} x \dfrac{d}{d x} x-x \dfrac{d}{d x} \sin ^{n} x}{\sin ^{2 n} x} \\ & =\dfrac{\sin ^{n} x \cdot 1-x(n \sin ^{n-1} x \cos x)}{\sin ^{2 n} x} \\ & =\dfrac{\sin ^{n-1} x(\sin x-n x \cos x)}{\sin ^{2 n} x} \\ & =\dfrac{\sin x-n x \cos x}{\sin ^{n+1} x} \end{aligned} $$