Answer :

Let $f(x)=x^{2} - 2$. Accordingly,

$ \begin{aligned} f^{\prime}(10) & =\lim _{h \to 0} \dfrac{f(10+h)-f(10)}{h} \\ & =\lim _{h \to 0} \dfrac{[(10+h)^{2}-2]-(10^{2}-2)}{h} \\ & =\lim _{h \to 0} \dfrac{10^{2}+2 \cdot 10 \cdot h+h^{2}-2-10^{2}+2}{h} \\ & =\lim _{h \to 0} \dfrac{20 h+h^{2}}{h} \\ & =\lim _{h \to 0}(20+h)=(20+0)=20 \end{aligned} $

Thus, the derivative of $x^{2} - 2$ at $x=10$ is 20 .

Answer :

Letf $(x)=x$. Accordingly,

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

Thus, the derivative of $x$ at $x=1$ is 1 .

Answer :

Let $f(x)=99 x$. Accordingly,

$ \begin{aligned} f^{\prime}(100) & =\lim _{h \to 0} \dfrac{f(100+h)-f(100)}{h} \\ & =\lim _{h \to 0} \dfrac{99(100+h)-99(100)}{h} \\ & =\lim _{h \to 0} \dfrac{99 \times 100+99 h-99 \times 100}{h} \\ & =\lim _{h \to 0} \dfrac{99 h}{h} \\ & =\lim _{h \to 0}(99)=99 \end{aligned} $

Thus, the derivative of $99 x$ at $x=100$ is 99 .

Answer :

(i) Let $f(x)=x^{3} - 27$. Accordingly, from the 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)^{3}-27]-(x^{3}-27)}{h} \\ & =\lim _{h \to 0} \dfrac{x^{3}+h^{3}+3 x^{2} h+3 x h^{2}-x^{3}}{h} \\ & =\lim _{h \to 0} \dfrac{h^{3}+3 x^{2} h+3 x h^{2}}{h} \\ & =\lim _{h \to 0}(h^{2}+3 x^{2}+3 x h) \\ & =0+3 x^{2}+0=3 x^{2} \end{aligned} $

(ii) Let $f(x)=(x$- 1$)(x$ - 2). Accordingly, from the 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-1)(x+h-2)-(x-1)(x-2)}{h} \\ & =\lim _{h \to 0} \dfrac{(x^{2}+h x-2 x+h x+h^{2}-2 h-x-h+2)-(x^{2}-2 x-x+2)}{h} \\ & =\lim _{h \to 0} \dfrac{(h x+h x+h^{2}-2 h-h)}{h} \\ & =\lim _{h \to 0} \dfrac{2 h x+h^{2}-3 h}{h} \\ & =\lim _{h \to 0}(2 x+h-3) \\ & =(2 x+0-3) \\ & =2 x-3 \\ \end{aligned} $

(iii) Let $f(x)=\frac{1}{x^2}$ then,

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

$=\lim _{h \to 0} \dfrac{\dfrac{1}{(x+h)^{2}}-\dfrac{1}{x^{2}}}{h} $

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

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

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

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

$=\dfrac{0-2 x}{x^{2}(x+0)^{2}}=\dfrac{-2}{x^{3}} $

(iv) Let $f(x)=\frac{x+1}{x-1}$.

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

Answer :

The given function is $f(x)=\dfrac{x^{100}}{100}+\dfrac{x^{99}}{99}+\ldots+\dfrac{x^{2}}{2}+x+1$

$\dfrac{d}{d x} f(x)=\dfrac{d}{d x}[\dfrac{x^{100}}{100}+\dfrac{x^{99}}{99}+\ldots+\dfrac{x^{2}}{2}+x+1]$

$\dfrac{d}{d x} f(x)=\dfrac{d}{d x}(\dfrac{x^{100}}{100})+\dfrac{d}{d x}(\dfrac{x^{99}}{99})+\ldots+\dfrac{d}{d x}(\dfrac{x^{2}}{2})+\dfrac{d}{d x}(x)+\dfrac{d}{d x}(1)$

On using theorem $\dfrac{d}{d x}(x^{n})=n x^{n-1}$, we obtain

$\dfrac{d}{d x} f(x)=\dfrac{100 x^{99}}{100}+\dfrac{99 x^{98}}{99}+\ldots+\dfrac{2 x}{2}+1+0$

$ =x^{99}+x^{98}+\ldots+x+1 $

$\therefore f^{\prime}(x)=x^{99}+x^{98}+\ldots+x+1$

At $x=0$,

$f^{\prime}(0)=1$

At $x=1$,

$f^{\prime}(1)=1^{99}+1^{98}+\ldots+1+1=[1+1+\ldots+1+1] _{\text{to terms }}=1 \times 100=100$

Thus, $f^{\prime}(1)=100 \times f^{\prime}(0)$

Answer :

Let $f(x)=x^{n}+a x^{n-1}+a^{2} x^{n-2}+\ldots+a^{n-1} x+a^{n}$

$\therefore f^{\prime}(x)=\dfrac{d}{d x}(x^{n}+a x^{n-1}+a^{2} x^{n-2}+\ldots+a^{n-1} x+a^{n})$

$=\dfrac{d}{d x}(x^{\prime \prime})+a \dfrac{d}{d x}(x^{n-1})+a^{2} \dfrac{d}{d x}(x^{n-2})+\ldots+a^{n-1} \dfrac{d}{d x}(x)+a^{n} \dfrac{d}{d x}(1)$

On using theorem $\dfrac{d}{d x} x^{n}=n x^{n-1}$, we obtain

$ \begin{aligned} f^{\prime}(x) & =n x^{n-1}+a(n-1) x^{n-2}+a^{2}(n-2) x^{n-3}+\ldots+a^{n-1}+a^{n}(0) \\ & =n x^{n-1}+a(n-1) x^{n-2}+a^{2}(n-2) x^{n-3}+\ldots+a^{n-1} \end{aligned} $

Answer :

(i) Let $f(x)=(x$ - a) $(x$ - $b)$

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

On using theorem $\dfrac{d}{d x}(x^{n})=n x^{n-1}$, we obtain

$f^{\prime}(x)=2 x-(a+b)+0=2 x-a-b$

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

$\Rightarrow f(x)=a^{2} x^{4}+2 a b x^{2}+b^{2}$

$\therefore f^{\prime}(x)=\dfrac{d}{d x}(a^{2} x^{4}+2 a b x^{2}+b^{2})=a^{2} \dfrac{d}{d x}(x^{4})+2 a b \dfrac{d}{d x}(x^{2})+\dfrac{d}{d x}(b^{2})$

On using theorem $\dfrac{d}{d x} x^{n}=n x^{n-1}$, we obtain

$ \begin{aligned} f^{\prime}(x) & =a^{2}(4 x^{3})+2 a b(2 x)+b^{2}(0) \\ & =4 a^{2} x^{3}+4 a b x \\ & =4 a x(a x^{2}+b) \end{aligned} $

(iii)

Let $f(x)=\dfrac{(x-a)}{(x-b)}$

$\Rightarrow f^{\prime}(x)=\dfrac{d}{d x}(\dfrac{x-a}{x-b})$

By quotient rule,

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

Answer :

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

$\Rightarrow f^{\prime}(x)=\dfrac{d}{d x}(\dfrac{x^{n}-a^{n}}{x-a})$

By quotient rule,

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

Answer :

(i) Let $f(x)=2 x-\dfrac{3}{4}$

$ \begin{aligned} f^{\prime}(x) & =\dfrac{d}{d x}(2 x-\dfrac{3}{4}) \\ & =2 \dfrac{d}{d x}(x)-\dfrac{d}{d x}(\dfrac{3}{4}) \\ & =2-0 \\ & =2 \end{aligned} $

(ii) Let $f(x)=(5 x^{3}+3 x.$ - 1$)(x$- 1$)$

By Leibnitz product rule,

$ \begin{aligned} f^{\prime}(x) & =(5 x^{3}+3 x-1) \dfrac{d}{d x}(x-1)+(x-1) \dfrac{d}{d x}(5 x^{3}+3 x-1) \\ & =(5 x^{3}+3 x-1)(1)+(x-1)(5.3 x^{2}+3-0) \\ & =(5 x^{3}+3 x-1)+(x-1)(15 x^{2}+3) \\ & =5 x^{3}+3 x-1+15 x^{3}+3 x-15 x^{2}-3 \\ & =20 x^{3}-15 x^{2}+6 x-4 \end{aligned} $

(iii) Let $f(x)=x^{-3}(5+3 x)$

By Leibnitz product rule,

$ \begin{aligned} f^{\prime}(x) & =x^{-3} \dfrac{d}{d x}(5+3 x)+(5+3 x) \dfrac{d}{d x}(x^{-3}) \\ & =x^{-3}(0+3)+(5+3 x)(-3 x^{-3-1}) \\ & =x^{-3}(3)+(5+3 x)(-3 x^{-4}) \\ & =3 x^{-3}-15 x^{-4}-9 x^{-3} \\ & =-6 x^{-3}-15 x^{-4} \\ & =-3 x^{-3}(2+\dfrac{5}{x}) \\ & =\dfrac{-3 x^{-3}}{x}(2 x+5) \\ & =\dfrac{-3}{x^{4}}(5+2 x) \end{aligned} $

(iv) Let $f(x)=x^{5}(3-6 x^{-9})$

By Leibnitz product rule,

$ \begin{aligned} f^{\prime}(x) & =x^{5} \dfrac{d}{d x}(3-6 x^{-9})+(3-6 x^{-9}) \dfrac{d}{d x}(x^{5}) \\ & =x^{5}\{0-6(-9) x^{-9-1}\}+(3-6 x^{-9})(5 x^{4}) \\ & =x^{5}(54 x^{-10})+15 x^{4}-30 x^{-5} \\ & =54 x^{-5}+15 x^{4}-30 x^{-5} \\ & =24 x^{-5}+15 x^{4} \\ & =15 x^{4}+\dfrac{24}{x^{5}} \end{aligned} $

(v) Let $f(x)=x^{-4}(3-4 x^{-5})$

By Leibnitz product rule,

$ \begin{aligned} & f^{\prime}(x)=x^{-4} \dfrac{d}{d x}(3-4 x^{-5})+(3-4 x^{-5}) \dfrac{d}{d x}(x^{-4}) \\ & =x^{-4}\{0-4(-5) x^{-5-1}\}+(3-4 x^{-5})(-4) x^{-4-1} \\ & =x^{-4}(20 x^{-6})+(3-4 x^{-5})(-4 x^{-5}) \\ & =20 x^{-10}-12 x^{-5}+16 x^{-10} \\ & =36 x^{-10}-12 x^{-5} \\ & =-\dfrac{12}{x^{5}}+\dfrac{36}{x^{10}} \end{aligned} $

(vi) Let $f(x)=\dfrac{2}{x+1}-\dfrac{x^{2}}{3 x-1}$

$f^{\prime}(x)=\dfrac{d}{d x}(\dfrac{2}{x+1})-\dfrac{d}{d x}(\dfrac{x^{2}}{3 x-1})$

By quotient rule,

$ \begin{aligned} f^{\prime}(x) & =[\dfrac{(x+1) \dfrac{d}{d x}(2)-2 \dfrac{d}{d x}(x+1)}{(x+1)^{2}}]-[\dfrac{(3 x-1) \dfrac{d}{d x}(x^{2})-x^{2} \dfrac{d}{d x}(3 x-1)}{(3 x-1)^{2}}] \\ & =[\dfrac{(x+1)(0)-2(1)}{(x+1)^{2}}]-[\dfrac{(3 x-1)(2 x)-(x^{2})(3)}{(3 x-1)^{2}}] \\ & =\dfrac{-2}{(x+1)^{2}}-[\dfrac{6 x^{2}-2 x-3 x^{2}}{(3 x-1)^{2}}] \\ & =\dfrac{-2}{(x+1)^{2}}-[\dfrac{3 x^{2}-2 x^{2}}{(3 x-1)^{2}}] \\ & =\dfrac{-2}{(x+1)^{2}}-\dfrac{x(3 x-2)}{(3 x-1)^{2}} \end{aligned} $

Answer :

Let $f(x)=\cos x$. Accordingly, from the first principle,

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

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

Answer :

(i) Let $f(x)=\sin x \cos x$. Accordingly, from the 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) \cos (x+h)-\sin x \cos x}{h} \\ & =\lim _{h \to 0} \dfrac{1}{2 h}[2 \sin (x+h) \cos (x+h)-2 \sin x \cos x] \\ & =\lim _{h \to 0} \dfrac{1}{2 h}[\sin 2(x+h)-\sin 2 x] \\ & =\lim _{h \to 0} \dfrac{1}{2 h}[2 \cos \dfrac{2 x+2 h+2 x}{2} \cdot \sin \dfrac{2 x+2 h-2 x}{2}] \\ & =\lim _{h \to 0} \dfrac{1}{h}[\cos \dfrac{4 x+2 h}{2} \sin \dfrac{2 h}{2}] \\ & =\lim _{h \to 0} \dfrac{1}{h}[\cos (2 x+h) \sin h] \\ & =\lim _{h \to 0} \cos (2 x+h) \cdot \lim _{h \to 0} \dfrac{\sin h}{h} \\ & =\cos (2 x+0) \cdot 1 \\ & =\cos 2 x \end{aligned} $

(ii) Letf $(x)=\sec x$. Accordingly, from the first principle,

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

(iii) Letf $(x)=5 \sec x+4 \cos x$. Accordingly, from the first principle,

$ \begin{aligned} & f^{\prime}(x)=\lim _{h \to 0} \dfrac{f(x+h)-f(x)}{h} \\ & =\lim _{h \to 0} \dfrac{5 \sec (x+h)+4 \cos (x+h)-[5 \sec x+4 \cos x]}{h} \\ & =5 \lim _{h \to 0} \dfrac{[\sec (x+h)-\sec x]}{h}+4 \lim _{h \to 0} \dfrac{[\cos (x+h)-\cos x]}{h} \\ & =5 \lim _{h \to 0} \dfrac{1}{h}[\dfrac{1}{\cos (x+h)}-\dfrac{1}{\cos x}]+4 \lim _{h \to 0} \dfrac{1}{h}[\cos (x+h)-\cos x] \\ & =5 \lim _{h \to 0} \dfrac{1}{h}[\dfrac{\cos x-\cos (x+h)}{\cos x \cos (x+h)}]+4 \lim _{h \to 0} \dfrac{1}{h}[\cos x \cos h-\sin x \sin h-\cos x] \\ & =\dfrac{5}{\cos x} \lim _{h \to 0} \dfrac{1}{h}[\dfrac{-2 \sin (\dfrac{x+x+h}{2}) \sin (\dfrac{x-x-h}{2})}{\cos (x+h)}]+4 \lim _{h \to 0} \dfrac{1}{h}[-\cos x(1-\cos h)-\sin x \sin h] \\ & =\dfrac{5}{\cos x} \cdot \lim _{h \to 0} \dfrac{1}{h}[\dfrac{-2 \sin (\dfrac{2 x+h}{2}) \sin (-\dfrac{h}{2})}{\cos (x+h)}]+4[-\cos x \lim _{h \to 0} \dfrac{(1-\cos h)}{h}-\sin x \lim _{h \to 0} \dfrac{\sin h}{h}] \\ & =\dfrac{5}{\cos x} \cdot \lim _{h \to 0}[\dfrac{\sin (\dfrac{2 x+h}{2}) \cdot \dfrac{\sin (\dfrac{h}{2})}{\dfrac{h}{2}}}{\cos (x+h)}]+4[(-\cos x) \cdot(0)-(\sin x) \cdot 1] \\ & =\dfrac{5}{\cos x} \cdot[\lim _{h \to 0} \dfrac{\sin (\dfrac{2 x+h}{2})}{\cos (x+h)} \cdot \lim _{h \to 0} \dfrac{\sin (\dfrac{h}{2})}{\dfrac{h}{2}}]-4 \sin x \\ & =\dfrac{5}{\cos x} \cdot \dfrac{\sin x}{\cos x} \cdot 1-4 \sin x \\ & =5 \sec x \tan x \cdot-4 \sin x \end{aligned} $

(iv) Let $f(x)=cosec x$. Accordingly, from the first principle,

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

(v) Let $f(x)=3 \cot x+5 cosec x$. Accordingly, from the first principle,

$$ \begin{align*} f^{\prime}(x) & =\lim _{h \to 0} \dfrac{f(x+h)-f(x)}{h} \\ & =\lim _{h \to 0} \dfrac{3 \cot (x+h)+5 cosec(x+h)-3 \cot x-5 cosec x}{h} \\ & =3 \lim _{h \to 0} \dfrac{1}{h}[\cot (x+h)-\cot x]+5 \lim _{h \to 0} \dfrac{1}{h}[cosec(x+h)-cosec x] \tag{1} \end{align*} $$

Now, $\lim _{h \to 0} \dfrac{1}{h}[\cot (x+h)-\cot x]$

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

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

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

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

$ f^{\prime}(x)=-3 cosec^{2} x-5 cosec x \cot x $

(vi) Let $f(x)=5 \sin x$ - $6 \cos x+7$. Accordingly, from the first principle,

$$ \begin{aligned} & f^{\prime}(x)=\lim _{h \to 0} \dfrac{f(x+h)-f(x)}{h} \\ & =\lim _{h \to 0} \dfrac{1}{h}[5 \sin (x+h)-6 \cos (x+h)+7-5 \sin x+6 \cos x-7] \\ & =\lim _{h \to 0} \dfrac{1}{h}[5\{\sin (x+h)-\sin x\}-6\{\cos (x+h)-\cos x\}] \\ & =5 \lim _{h \to 0} \dfrac{1}{h}[\sin (x+h)-\sin x]-6 \lim _{h \to 0} \dfrac{1}{h}[\cos (x+h)-\cos x] \\ & =5 \lim _{h \to 0} \dfrac{1}{h}[2 \cos (\dfrac{x+h+x}{2}) \sin (\dfrac{x+h-x}{2})]-6 \lim _{h \to 0} \dfrac{\cos x \cos h-\sin x \sin h-\cos x}{h} \\ & =5 \lim _{h \to 0} \dfrac{1}{h}[2 \cos (\dfrac{2 x+h}{2}) \sin \dfrac{h}{2}]-6 \lim _{h \to 0}[\dfrac{-\cos x(1-\cos h)-\sin x \sin h}{h}] \\ & =5 \lim _{h \to 0}(\cos (\dfrac{2 x+h}{2}) \dfrac{\sin \dfrac{h}{2}}{\dfrac{h}{2}})-6 \lim _{h \to 0}[\dfrac{-\cos x(1-\cos h)}{h}-\dfrac{\sin x \sin h}{h}] \\ & =5[\lim _{h \to 0} \cos (\dfrac{2 x+h}{2})][\dfrac{\lim _{h} \to 0}{\dfrac{\sin }{2}} \dfrac{h}{2}]-6[(-\cos x)(\lim _{h \to 0} \dfrac{1-\cos h}{h})-\sin x \lim _{h \to 0}(\dfrac{\sin h}{h})] \\ & =5 \cos x \cdot 1-6[(-\cos x) \cdot(0)-\sin x \cdot 1] \\ & =5 \cos x+6 \sin x \end{aligned} $$

(vii) Let $f(x)=2 \tan x$ - $7 \sec x$. Accordingly, from the first principle,

$$ \begin{aligned} & f^{\prime}(x)=\lim _{h \to 0} \dfrac{f(x+h)-f(x)}{h} \\ & =\lim _{h \to 0} \dfrac{1}{h}[2 \tan (x+h)-7 \sec (x+h)-2 \tan x+7 \sec x] \\ & =\lim _{h \to 0} \dfrac{1}{h}[2\{\tan (x+h)-\tan x\}-7\{\sec (x+h)-\sec x\}] \\ & =2 \lim _{h \to 0} \dfrac{1}{h}[\tan (x+h)-\tan x]-7 \lim _{h \to 0} \dfrac{1}{h}[\sec (x+h)-\sec x] \\ & =2 \lim _{h \to 0} \dfrac{1}{h}[\dfrac{\sin (x+h)}{\cos (x+h)}-\dfrac{\sin x}{\cos x}]-7 \lim _{h \to 0} \dfrac{1}{h}[\dfrac{1}{\cos (x+h)}-\dfrac{1}{\cos x}] \\ & =2 \lim _{h \to 0} \dfrac{1}{h}[\dfrac{\sin (x+h) \cos x-\sin x \cos (x+h)}{\cos x \cos (x+h)}]-7 \lim _{h \to 0} \dfrac{1}{h}[\dfrac{\cos x-\cos (x+h)}{\cos x \cos (x+h)}] \\ & =2 \lim _{h \to 0} \dfrac{1}{h}[\dfrac{\sin (x+h-x)}{\cos x \cos (x+h)}]-7 \lim _{h \to 0} \dfrac{1}{h}[\dfrac{-2 \sin (\dfrac{x+x+h}{2}) \sin (\dfrac{x-x-h}{2})}{\cos x \cos (x+h)}] \\ & =2 \lim _{h \to 0}[(\dfrac{\sin h}{h}) \dfrac{1}{\cos x \cos (x+h)}]-7 \lim _{h \to 0} \dfrac{1}{h}[\dfrac{-2 \sin (\dfrac{2 x+h}{2}) \sin (-\dfrac{h}{2})}{\cos x \cos (x+h)}] \\ & =2(\lim _{h \to 0} \dfrac{\sin h}{h})(\lim _{h \to 0} \dfrac{1}{\cos x \cos (x+h)})-7(\lim _{\dfrac{h}{2} \to 0} \dfrac{\sin \dfrac{h}{2}}{\dfrac{h}{2}})(\lim _{h \to 0} \dfrac{\sin (\dfrac{2 x+h}{2})}{\cos x \cos (x+h)}) \\ & =2.1 \cdot \dfrac{1}{\cos x \cos x}-7.1(\dfrac{\sin x}{\cos x \cos x}) \\ & =2 \sec ^{2} x-7 \sec x \tan x \end{aligned} $$