Answer :

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

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

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

Answer :

Let, $ \ f\left(x \right)=x$. Accordingly,

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

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

Answer :

Let $f\left(x \right)=99 x \quad$ Accordingly,

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

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

Answer :

(i) Let $f\left(x \right)=x^{3} - 27$

Accordingly, from the first principle,

$ \begin{aligned} f^{\prime}\left(x \right) & =\lim _ {h \rightarrow 0} \dfrac{f\left(x+h \right)-f\left(x \right)}{h} \\ \\ & =\lim _ {h \rightarrow 0} \dfrac{\left[\left(x+h \right)^{3}-27\right]-\left(x^{3}-27 \right)}{h} \\ \\ & =\lim _ {h \rightarrow 0} \dfrac{x^{3}+h^{3}+3 x^{2} h+3 x h^{2}-x^{3}}{h} \\ \\ & =\lim _ {h \rightarrow 0} \dfrac{h^{3}+3 x^{2} h+3 x h^{2}}{h} \\ \\ & =\lim _ {h \rightarrow 0}\left(h^{2}+3 x^{2}+3 x h \right) \\ \\ & =0+3 x^{2}+0=3 x^{2} \end{aligned} $

(ii) Let, $ \ f\left(x \right)=\left(x- 1 \right)\left(x - 2 \right)\quad$

Accordingly, from the first principle,

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

$\qquad\quad =\lim _ {h \rightarrow 0} \dfrac{\left(x+h-1 \right)\left(x+h-2 \right)-\left(x-1 \right)\left(x-2 \right)}{h} $

$\qquad\quad =\lim _ {h \rightarrow 0} \dfrac{\left(x^{2}+h x-2 x+h x+h^{2}-2 h-x-h+2 \right)-\left(x^{2}-2 x-x+2 \right)}{h} $

$\qquad\quad =\lim _ {h \rightarrow 0} \dfrac{\left(h x+h x+h^{2}-2 h-h \right)}{h} $

$\qquad\quad =\lim _ {h \rightarrow 0} \dfrac{2 h x+h^{2}-3 h}{h} $

$\qquad\quad =\lim _ {h \rightarrow 0}\left(2 x+h-3 \right) $

$\qquad\quad=\left(2 x+0-3 \right) $

$\qquad\quad =2 x-3 $

(iii) Let $f\left(x \right)=\dfrac{1}{x^{2}}$

Accordingly, from the first principle,

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

$\hspace{1cm} =\lim _ {h \rightarrow 0} \dfrac{\dfrac{1}{\left(x+h \right)^{2}}-\dfrac{1}{x^{2}}}{h} $

$\hspace{1cm} =\lim _ {h \rightarrow 0} \dfrac{1}{h}\left[\dfrac{x^{2}-\left(x+h \right)^{2}}{x^{2}\left(x+h \right)^{2}}\right] $

$\hspace{1cm} =\lim _ {h \rightarrow 0} \dfrac{1}{h}\left[\dfrac{x^{2}-x^{2}-h^{2}-2 h x}{x^{2}\left(x+h \right)^{2}}\right] $

$\hspace{1cm} =\lim _ {h \rightarrow 0} \dfrac{1}{h}\left[\dfrac{-h^{2}-2 h x}{x^{2}\left(x+h \right)^{2}}\right] $

$\hspace{1cm} =\lim _ {h \rightarrow 0}\left[\dfrac{-h-2 x}{x^{2}\left(x+h \right)^{2}}\right] $

$\hspace{1cm} =\dfrac{0-2 x}{x^{2}\left(x+0 \right)^{2}}=\dfrac{-2}{x^{3}} $

$\hspace{0.2cm} f\left(x \right)=\dfrac{x+1}{x-1}$

$ \text{ (iv) Let } f(x)=\dfrac{x+1}{x-1}\quad$

$ \text{ Accordingly, from the first principle, }$

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

Answer :

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

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

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

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

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

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

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

At $x=0$,

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

At $x=1$,

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

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

Answer :

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

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

$\hspace{1.3cm}=\dfrac{d}{d x}\left(x^{n} \right)+a \dfrac{d}{d x}\left(x^{n-1} \right)+a^{2} \dfrac{d}{d x}\left(x^{n-2} \right)+\ldots+a^{n-1} \dfrac{d}{d x}\left(x \right)+a^{n} \dfrac{d}{d x}\left(1 \right)$

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

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

Answer :

(i) Let $f\left(x \right)=\left(x - a \right) \left(x - b \right)$

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

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

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

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

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

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

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

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

(iii)

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

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

By quotient rule,

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

Answer :

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

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

By quotient rule,

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

Answer :

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

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

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

By Leibnitz product rule,

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

By Leibnitz product rule,

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

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

By Leibnitz product rule,

$ \begin{aligned} f^{\prime}\left(x \right) & =x^{5} \dfrac{d}{d x}\left(3-6 x^{-9} \right)+\left(3-6 x^{-9} \right) \dfrac{d}{d x}\left(x^{5} \right) \\ \\ & =x^{5} \lbrace 0-6\left(-9 \right) x^{-9-1} \rbrace +\left(3-6 x^{-9} \right)\left(5 x^{4} \right) \\ \\ & =x^{5}\left(54 x^{-10} \right)+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 $f\left(x \right)=x^{- 4}\left(3 - 4 x ^{-5} \right)$

By Leibnitz product rule,

$ \begin{aligned} f^{\prime}\left(x \right) & =x^{-4} \dfrac{d}{d x}\left(3-4 x^{-5} \right)+\left(3-4 x^{-5} \right) \dfrac{d}{d x}\left(x^{-4} \right) \\ \\ & =x^{-4} \lbrace 0-4\left(-5 \right) x^{-5-1} \rbrace +\left(3-4 x^{-5} \right)\left(-4 \right) x^{-4-1} \\ \\ & =x^{-4}\left(20 x^{-6} \right)+\left(3-4 x^{-5} \right)\left(-4 x^{-5} \right) \\ \\ & =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) $ \ \ \dfrac{2}{x+1}-\dfrac{x^2}{3x-1} $

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

By quotient rule,

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

Answer :

Let $f\left(x \right)=\cos x$.

Accordingly, from the first principle,

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

Answer :

(i) Let $f\left(x \right)=\sin x \cos x$.

Accordingly, from the first principle,

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

(ii) Let $f \left(x \right)=\sec x$.

Accordingly, from the first principle,

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

(iii) Let $f \left(x \right)=5 \sec x+4 \cos x$.

Accordingly, from the first principle,

$ \begin{aligned} f^{\prime}\left(x \right) & =\lim _ {h \rightarrow 0} \dfrac{f\left(x+h \right)-f\left(x \right)}{h} \\ \\ & =\lim _ {h \rightarrow 0} \dfrac{5 \sec \left(x+h \right)+4 \cos \left(x+h \right)-\left[5 \sec x+4 \cos x\right]}{h} \\ \\ & =5 \lim _ {h \rightarrow 0} \dfrac{\left[\sec \left(x+h \right)-\sec x\right]}{h}+4 \lim _ {h \rightarrow 0} \dfrac{\left[\cos \left(x+h \right)-\cos x\right]}{h} \\ \\ & =5 \lim _ {h \rightarrow 0} \dfrac{1}{h}\left[\dfrac{1}{\cos \left(x+h \right)}-\dfrac{1}{\cos x}\right]+4 \lim _ {h \rightarrow 0} \dfrac{1}{h}\left[\cos \left(x+h \right)-\cos x\right] \\ \\ & =5 \lim _ {h \rightarrow 0} \dfrac{1}{h}\left[\dfrac{\cos x-\cos \left(x+h \right)}{\cos x \cos \left(x+h \right)}\right]+4 \lim _ {h \rightarrow 0} \dfrac{1}{h}\left[\cos x \cos h-\sin x \sin h-\cos x\right] \\ \\ & =\dfrac{5}{\cos x} \lim _ {h \rightarrow 0} \dfrac{1}{h}\left[\dfrac{-2 \sin \left(\dfrac{x+x+h}{2} \right) \sin \left(\dfrac{x-x-h}{2} \right)}{\cos \left(x+h \right)}\right]+4 \lim _ {h \rightarrow 0} \dfrac{1}{h}\left[-\cos x\left(1-\cos h \right)-\sin x \sin h\right] \\ \\ & =\dfrac{5}{\cos x} \cdot \lim _ {h \rightarrow 0} \dfrac{1}{h}\left[\dfrac{-2 \sin \left(\dfrac{2 x+h}{2} \right) \sin \left(-\dfrac{h}{2} \right)}{\cos \left(x+h \right)}\right]+4\left[-\cos x \lim _ {h \rightarrow 0} \dfrac{\left(1-\cos h \right)}{h}-\sin x \lim _ {h \rightarrow 0} \dfrac{\sin h}{h}\right] \\ \\ & =\dfrac{5}{\cos x} \cdot \lim _ {h \rightarrow 0}\left[\dfrac{\sin \left(\dfrac{2 x+h}{2} \right) \cdot \dfrac{\sin \left(\dfrac{h}{2} \right)}{\dfrac{h}{2}}}{\cos \left(x+h \right)}\right]+4\left[\left(-\cos x \right) \cdot\left(0 \right)-\left(\sin x \right) \cdot 1\right] \\ \\ & =\dfrac{5}{\cos x} \cdot\left[\lim _ {h \rightarrow 0} \dfrac{\sin \left(\dfrac{2 x+h}{2} \right)}{\cos \left(x+h \right)} \cdot \lim _ {h \rightarrow 0} \dfrac{\sin \left(\dfrac{h}{2} \right)}{\dfrac{h}{2}}\right]-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 $f\left(x \right)=cosec \ x$.

Accordingly, from the first principle,

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

(v) Let $f\left(x \right)=3 \cot x+5 cosec \ x$.

Accordingly, from the first principle,

$ \qquad \begin{aligned} f^{\prime}\left(x \right) & =\lim _ {h \rightarrow 0} \dfrac{f\left(x+h \right)-f\left(x \right)}{h} \\ \\ & =\lim _ {h \rightarrow 0} \dfrac{3 \cot \left(x+h \right)+5 cosec \ \left(x+h \right)-3 \cot x-5 cosec \ x}{h} \\ \\ & =3 \lim _ {h \rightarrow 0} \dfrac{1}{h}\left[\cot \left(x+h \right)-\cot x\right]+5 \lim _ {h \rightarrow 0} \dfrac{1}{h}\left[cosec \ \left(x+h \right)-cosec \ x\right] \qquad …(1) \end{aligned} $

Now,

$ \begin{aligned} \lim _ {h \rightarrow 0} \dfrac{1}{h}\left[\cot \left(x+h \right)-\cot x\right] & =\lim _ {h \rightarrow 0} \dfrac{1}{h}\left[\dfrac{\cos \left(x+h \right)}{\sin \left(x+h \right)}-\dfrac{\cos x}{\sin x}\right] \\ \\ & =\lim _ {h \rightarrow 0} \dfrac{1}{h}\left[\dfrac{\cos \left(x+h \right) \sin x-\cos x \sin \left(x+h \right)}{\sin x \sin \left(x+h \right)}\right] \\ \\ & =\lim _ {h \rightarrow 0} \dfrac{1}{h}\left[\dfrac{\sin \left(x-x-h \right)}{\sin x \sin \left(x+h \right)}\right] \\ \\ & =\lim _ {h \rightarrow 0} \dfrac{1}{h}\left[\dfrac{\sin \left(-h \right)}{\sin x \sin \left(x+h \right)}\right] \\ \\ & =-\left(\lim _ {h \rightarrow 0} \dfrac{\sin h}{h} \right) \cdot\left(\lim _ {h \rightarrow 0} \dfrac{1}{\sin x \cdot \sin \left(x+h \right)} \right) \\ \\ & =-1 \cdot \dfrac{1}{\sin x \cdot \sin \left(x+0 \right)}=\dfrac{-1}{\sin ^{2} x}=-cosec \ ^{2} x \quad\quad … (2) \\ \ \lim _ {h \rightarrow 0} \dfrac{1}{h}\left[cosec \ \left(x+h \right)-cosec \ x\right] & =\lim _ {h \rightarrow 0} \dfrac{1}{h}\left[\dfrac{1}{\sin \left(x+h \right)}-\dfrac{1}{\sin x}\right] \\ \\ & =\lim _ {h \rightarrow 0} \dfrac{1}{h}\left[\dfrac{\sin x-\sin \left(x+h \right)}{\sin \left(x+h \right) \sin x}\right] \\ \\ & =\lim _ {h \rightarrow 0} \dfrac{1}{h}\left[\dfrac{2 \cos \left(\dfrac{x+x+h}{2} \right) \cdot \sin \left(\dfrac{x-x-h}{2} \right)}{\sin \left(x+h \right) \sin x}\right] \\ \\ & =\lim _ {h \rightarrow 0} \dfrac{1}{h}\left[\dfrac{2 \cos \left(\dfrac{2 x+h}{2} \right) \sin \left(-\dfrac{h}{2} \right)}{\sin \left(x+h \right) \sin x}\right] \\ \\ & \lim_ {h\rightarrow 0}\left[\dfrac{-\cos \left(\frac{2 x+h}{2} \right) \cdot \dfrac{\sin \left(\frac{h}{2} \right)}{\left(\frac{h}{2} \right)}}{\sin \left(x+h \right) \sin x}\right] \\ \\ & =\lim _ {h \rightarrow 0}\left(\dfrac{-\cos \left(\frac{2 x+h}{2} \right)}{\sin \left(x+h \right) \sin x} \right) \lim _ {\frac{h}{2} \rightarrow 0} \dfrac{\sin \left(\frac{h}{2} \right)}{\left(\frac{h}{2} \right)} \\ \\ & =\left(\dfrac{-\cos x}{\sin x \sin x} \right) \cdot 1 \\ \\ & =-cosec \ x \cot x \qquad\qquad …(3) \end{aligned} $

From $\left(1 \right), \left(2 \right),$ and $\left(3 \right),$ we obtain

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

$(vi)$ Let, $f\left(x \right)=5 \sin x$ - $6 \cos x+7$.

Accordingly, from the first principle,

$ \begin{aligned} f^{\prime}\left(x \right) & =\lim _ {h \rightarrow 0} \dfrac{f\left(x+h \right)-f\left(x \right)}{h} \\ \\ & =\lim _ {h \rightarrow 0} \dfrac{1}{h}\left[5 \sin \left(x+h \right)-6 \cos \left(x+h \right)+7-5 \sin x+6 \cos x-7\right] \\ \\ & =\lim _ {h \rightarrow 0} \dfrac{1}{h}\left[5 \lbrace \sin \left(x+h \right)-\sin x \rbrace -6 \lbrace \cos \left(x+h \right)-\cos x \rbrace \right] \\ \\ & =5 \lim _ {h \rightarrow 0} \dfrac{1}{h}\left[\sin \left(x+h \right)-\sin x\right]-6 \lim _ {h \rightarrow 0} \dfrac{1}{h}\left[\cos \left(x+h \right)-\cos x\right] \\ \\ & =5 \lim _ {h \rightarrow 0} \dfrac{1}{h}\left[2 \cos \left(\dfrac{x+h+x}{2} \right) \sin \left(\dfrac{x+h-x}{2} \right)\right]-6 \lim _ {h \rightarrow 0} \dfrac{\cos x \cos h-\sin x \sin h-\cos x}{h} \\ \\ & =5 \lim _ {h \rightarrow 0} \dfrac{1}{h}\left[2 \cos \left(\dfrac{2 x+h}{2} \right) \sin \dfrac{h}{2}\right]-6 \lim _ {h \rightarrow 0}\left[\dfrac{-\cos x\left(1-\cos h \right)-\sin x \sin h}{h}\right] \\ \\ & =5 \lim _ {h \rightarrow 0}\left(\cos \left(\dfrac{2 x+h}{2} \right) \dfrac{\sin \frac{h}{2}}{\frac{h}{2}} \right)-6 \lim _ {h \rightarrow 0}\left[\dfrac{-\cos x\left(1-\cos h \right)}{h}-\dfrac{\sin x \sin h}{h}\right] \\ \\ & =5\left[\lim _ {h \rightarrow 0} \cos \left(\dfrac{2 x+h}{2} \right)\right]\left[\lim_ {a\rightarrow 0}\dfrac{\sin \frac{h}{2}}{\frac{h}{2}}\right]-6\left[\left(-\cos x \right)\left(\lim _ {h \rightarrow 0} \dfrac{1-\cos h}{h} \right)-\sin x \lim _ {h \rightarrow 0}\left(\dfrac{\sin h}{h} \right)\right] \\ \\ & =5 \cos x \cdot 1-6\left[\left(-\cos x \right) \cdot\left(0 \right)-\sin x \cdot 1\right] \\ \\ & =5 \cos x+6 \sin x \end{aligned} $

(vii) Let $f\left(x \right)=2 \tan x$ - $7 \sec x$.

Accordingly, from the first principle,

$ \begin{aligned} f^{\prime}\left(x \right) & =\lim _ {h \rightarrow 0} \dfrac{f\left(x+h \right)-f\left(x \right)}{h} \\ \\ & =\lim _ {h \rightarrow 0} \dfrac{1}{h}\left[2 \tan \left(x+h \right)-7 \sec \left(x+h \right)-2 \tan x+7 \sec x\right] \\ \\ & =\lim _ {h \rightarrow 0} \dfrac{1}{h}\left[2 \lbrace \tan \left(x+h \right)-\tan x \rbrace -7 \lbrace \sec \left(x+h \right)-\sec x \rbrace \right] \\ \\ & =2 \lim _ {h \rightarrow 0} \dfrac{1}{h}\left[\tan \left(x+h \right)-\tan x\right]-7 \lim _ {h \rightarrow 0} \dfrac{1}{h}\left[\sec \left(x+h \right)-\sec x\right] \\ \\ & =2 \lim _ {h \rightarrow 0} \dfrac{1}{h}\left[\dfrac{\sin \left(x+h \right)}{\cos \left(x+h \right)}-\dfrac{\sin x}{\cos x}\right]-7 \lim _ {h \rightarrow 0} \dfrac{1}{h}\left[\dfrac{1}{\cos \left(x+h \right)}-\dfrac{1}{\cos x}\right] \\ \\ & =2 \lim _ {h \rightarrow 0} \dfrac{1}{h}\left[\dfrac{\sin \left(x+h \right) \cos x-\sin x \cos \left(x+h \right)}{\cos x \cos \left(x+h \right)}\right]-7 \lim _ {h \rightarrow 0} \dfrac{1}{h}\left[\dfrac{\cos x-\cos \left(x+h \right)}{\cos x \cos \left(x+h \right)}\right] \\ \\ & =2 \lim _ {h \rightarrow 0} \dfrac{1}{h}\left[\dfrac{\sin \left(x+h-x \right)}{\cos x \cos \left(x+h \right)}\right]-7 \lim _ {h \rightarrow 0} \dfrac{1}{h}\left[\dfrac{-2 \sin \left(\dfrac{x+x+h}{2} \right) \sin \left(\dfrac{x-x-h}{2} \right)}{\cos x \cos \left(x+h \right)}\right] \\ \\ & =2 \lim _ {h \rightarrow 0}\left[\left(\dfrac{\sin h}{h} \right) \dfrac{1}{\cos x \cos \left(x+h \right)}\right]-7 \lim _ {h \rightarrow 0} \dfrac{1}{h}\left[\dfrac{-2 \sin \left(\dfrac{2 x+h}{2} \right) \sin \left(-\dfrac{h}{2} \right)}{\cos x \cos \left(x+h \right)}\right] \\ \\ & =2\lim _ {h \rightarrow 0} \left(\dfrac{\sin h}{h} \right)\lim _ {h \rightarrow 0} \left(\dfrac{1}{\cos x \cos \left(x+h \right)} \right)-7\lim _ {\frac{h}{2} \rightarrow 0} \left(\dfrac{\sin \frac{h}{2}}{\frac{h}{2}} \right)\lim _ {h \rightarrow 0} \left(\dfrac{\sin \left(\frac{2 x+h}{2} \right)}{\cos x \cos \left(x+h \right)} \right) \\ \\ & =2\times 1 \times \dfrac{1}{\cos x \cos x}-7\times 1\left(\dfrac{\sin x}{\cos x \cos x} \right) \\ \\ & =2 \sec ^{2} x-7 \sec x \tan x \end{aligned} $