Answer :

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

By first principle,

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

$\mathrm{(ii) \ Let,} \quad f \left(x\right)=\left(-x\right)^{-1}=\dfrac{1}{-x}=\dfrac{-1}{x} $

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

By first principle,

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

$\mathrm{(iii) ~ Let,} \quad f\left(x\right)=\sin \left(x+1\right)$

Accordingly, $f\left(x+h\right)=\sin \left(x+h+1\right)$

By first principle,

$ \begin{aligned} f^{\prime}\left(x\right) & =\lim _ {h \to 0} \dfrac{f\left(x+h\right)-f\left(x\right)}{h} \\ \\ & =\lim _ {h \to 0} \dfrac{1}{h}\left[\sin \left(x+h+1\right)-\sin \left(x+1\right)\right] \\ \\ & =\lim _ {h \to 0} \dfrac{1}{h}\left[2 \cos \left(\dfrac{x+h+1+x+1}{2}\right) \sin \left(\dfrac{x+h+1-x-1}{2}\right)\right] \\ \\ & =\lim _ {h \to 0} \dfrac{1}{h}\left[2 \cos \left(\dfrac{2 x+h+2}{2}\right) \sin \left(\dfrac{h}{2}\right)\right] \\ \\ & =\lim _ {h \to 0}\left[\cos \left(\dfrac{2 x+h+2}{2}\right) \cdot \dfrac{\sin \left(\frac{h}{2}\right)}{\left(\frac{h}{2}\right)}\right] \\ \\ & =\lim _ {h \to 0} \cos \left(\dfrac{2 x+h+2}{2}\right) \times \lim _ {\frac{h}{2} \to 0} \left[\dfrac{\sin\frac{h}{2}}{\frac{h}{2}}\right] \\ \\ & =\cos \left(\dfrac{2 x+0+2}{2}\right) \times 1 \\ \\ & =\cos \left(x+1\right) \\ \\ f\left(x\right) & =\cos \left(x-\dfrac{\pi}{8}\right) \quad \text{ Accordingly, } \end{aligned} $

By first principle,

$ \begin{aligned} f^{\prime}\left(x\right) & =\lim _ {h \to 0} \dfrac{f\left(x+h\right)-f\left(x\right)}{h} \\ \\ & \ =\lim _ {h \to 0} \dfrac{1}{h}\left[\cos \left(x+h-\dfrac{\pi}{8}\right)-\cos \left(x-\dfrac{\pi}{8}\right)\right] \\ \\ & =\lim _ {h \to 0} \dfrac{1}{h}\left[-2 \sin \dfrac{\left(x+h-\dfrac{\pi}{8}+x-\dfrac{\pi}{8}\right)}{2} \sin \left(\dfrac{x+h-\dfrac{\pi}{8}-x+\dfrac{\pi}{8}}{2}\right)\right] \\ \\ & =\lim _ {h \to 0} \dfrac{1}{h}\left[-2 \sin \left(\dfrac{2 x+h-\dfrac{\pi}{4}}{2}\right) \sin \dfrac{h}{2}\right] \\ \\ & =\lim _ {h \to 0}\left[-\sin \left(\dfrac{2 x+h-\dfrac{\pi}{4}}{2}\right) \dfrac{\sin \left(\frac{h}{2}\right)}{\left(\frac{h}{2}\right)}\right] \\ \\ & =\lim _ {h \to 0}\left[-\sin \left(\dfrac{2 x+h-\frac{\pi}{4}}{2}\right)\right] \times\lim _ {\frac{h}{2} \to 0}\left(\dfrac{\sin \left(\frac{h}{2}\right)}{\left(\frac{h}{2}\right)}\right) \hspace{30px} \left[\because \ \ \text{As } h \to 0 \Rightarrow \dfrac{h}{2} \to 0\right] \end{aligned} $

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

Answer :

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

By first principle,

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

Answer :

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

By Leibnitz product rule,

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

By Leibnitz product rule,

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

Answer :

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

By quotient rule,

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

Answer :

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

By quotient rule,

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

Answer :

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

By quotient rule,

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

Answer :

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

By quotient rule,

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

Answer :

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

By quotient rule,

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

Answer :

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

Answer :

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

Answer :

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

Accordingly, $f\left(x+h\right)=\left \lbrace a\left(x+h\right)+b\right \rbrace ^{n}=\left(a x+a h+b\right)^{n}$ By first principle,

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

Answer :

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

By Leibnitz product rule,

$ \begin{aligned} f^{\prime}\left(x\right)=\left(a x+b\right)^{n} \dfrac{d}{d x}\left(c x+d\right)^{m}+\left(c x+d\right)^{m} \dfrac{d}{d x}\left(a x+b\right)^{n} \qquad \ldots{(1)} \end{aligned} $

$\text{Let,} \ \ f_1\left(x\right)=\left(c x+d\right)^{m}$

$\text{Now, } \ \ f_1\left(x+h\right)=\left(c x+c h+d\right)^{m}$

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

$\hspace{1cm}=\lim _ {h \to 0} \dfrac{\left(c x+c h+d\right)^{m}-\left(c x+d\right)^{m}}{h}$

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

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

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

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

$\hspace{1cm}=\left(c x+d\right)^{m}\left[\dfrac{m c}{c x+d}+0\right]$

$\hspace{1cm}=\dfrac{m c\left(c x+d\right)^{m}}{\left(c x+d\right)}$

$\hspace{1cm}=m c\left(c x+d\right)^{m-1}$

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

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

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

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

Answer :

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

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

By first principle,

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

Answer :

$\text{Let,}\quad f\left(x\right)=cosec \ x \cot x$

By Leibnitz product rule,

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

$\text{Let,}\quad f_1\left(x\right)=\cot x$. Accordingly, $f_1\left(x+h\right)=\cot \left(x+h\right)$

By first principle,

$ \begin{aligned} f_1^{\prime}\left(x\right) & =\lim _ {h \to 0} \dfrac{f_1\left(x+h\right)-f_1\left(x\right)}{h} \\ \\ & =\lim _ {h \to 0} \dfrac{\cot \left(x+h\right)-\cot x}{h} \\ \\ & =\lim _ {h \to 0} \dfrac{1}{h}\left(\dfrac{\cos \left(x+h\right)}{\sin \left(x+h\right)}-\dfrac{\cos x}{\sin x}\right) \\ \\ & =\lim _ {h \to 0} \dfrac{1}{h}\left[\dfrac{\sin x \cos \left(x+h\right)-\cos x \sin \left(x+h\right)}{\sin x \sin \left(x+h\right)}\right] \\ \\ & =\lim _ {h \to 0} \dfrac{1}{h}\left[\dfrac{\sin \left(x-x-h\right)}{\sin x \sin \left(x+h\right)}\right] \\ \\ & =\dfrac{1}{\sin x} \cdot \lim _ {h \to 0} \dfrac{1}{h}\left[\dfrac{\sin \left(-h\right)}{\sin \left(x+h\right)}\right] \\ \\ & =\dfrac{-1}{\sin x} \cdot\left(\lim _ {h \to 0} \dfrac{\sin h}{h}\right)\left(\lim _ {h \to 0} \dfrac{1}{\sin \left(x+h\right)}\right) \\ \\ & =\dfrac{-1}{\sin {x}} \cdot 1 \cdot\left(\dfrac{1}{\sin \left(x+0\right)}\right) \\ \\ & =\dfrac{-1}{\sin {2} x} \\ \\ & =-cosec \ ^{2} x \qquad \ldots{(2)} \end{aligned} $

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

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

By first principle,

$ \begin{aligned} f_2^{\prime}\left(x\right) & =\lim _ {h \to 0} \dfrac{f_2\left(x+h\right)-f_2\left(x\right)}{h} \\ \\ & =\lim _ {h \to 0} \dfrac{1}{h}\left[cosec \ \left(x+h\right)-cosec \ x\right] \\ \\ & =\lim _ {h \to 0} \dfrac{1}{h}\left[\dfrac{1}{\sin \left(x+h\right)}-\dfrac{1}{\sin x}\right] \\ \\ & =\lim _ {h \to 0} \dfrac{1}{h}\left[\dfrac{\sin x-\sin \left(x+h\right)}{\sin x \sin \left(x+h\right)}\right] \\ \\ & =\dfrac{1}{\sin x} \cdot \lim _ {h \to 0} \dfrac{1}{h}\left[\dfrac{2 \cos \left(\dfrac{x+x+h}{2}\right) \sin \left(\dfrac{x-x-h}{2}\right)}{\sin \left(x+h\right)}\right] \\ \\ & =\dfrac{1}{\sin x} \cdot \lim _ {h \to 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)}\right] \\ \\ & =\dfrac{1}{\sin x} \cdot \lim _ {h \to 0}\left[\dfrac{-\sin \left(\dfrac{h}{2}\right)}{\left(\dfrac{h}{2}\right)} \cdot \dfrac{\cos \left(\dfrac{2 x+h}{2}\right)}{\sin \left(x+h\right)}\right] \\ \\ & =\dfrac{-1}{\sin x} \cdot \lim _ {h \to 0} \dfrac{\sin \left(\dfrac{h}{2}\right)}{\left(\dfrac{h}{2}\right)} \cdot \lim _ {h \to 0} \dfrac{\cos \left(\dfrac{2 x+h}{2}\right)}{\sin \left(x+h\right)} \\ \\ & =\dfrac{-1}{\sin x} \cdot 1 \cdot \dfrac{\cos \left(\dfrac{2 x+0}{2}\right)}{\sin \left(x+0\right)} \\ \\ & =\dfrac{-1}{\sin x} \cdot \dfrac{\cos x}{\sin x} \\ \\ & =-\cos ec x \cdot \cot x \qquad \ldots{(3)} \end{aligned} $

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

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

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

Answer :

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

By quotient rule,

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

Answer :

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

By quotient rule,

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

Answer :

$\text{Let,}$

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

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

By quotient rule,

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

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

$ \begin{aligned} \therefore \ \ \dfrac{d y}{d x} & =\dfrac{d}{d x}\left(\sin x \sin ^{2} x\right) \\ \\ & =\left(\sin x\right)^{\prime} \sin ^{2} x+\sin x\left(\sin ^{2} x\right)^{\prime} \left[\text{ By Leibnitz product rule }\right] \\ \\ & =\cos x \sin ^{2} x+\sin x\left(2 \sin x \cos x\right) \qquad \left[\text{ Using (1) }\right] \\ \\ & =\cos x \sin ^{2} x+2 \sin ^{2} x \cos x & \\ \\ & =3 \sin ^{2} x \cos x & \end{aligned} $

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

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

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

Consider

$ \begin{aligned} \dfrac{d}{d x}\left(\sin ^{k+1} x\right) & =\dfrac{d}{d x}\left(\sin x \sin ^{k} x\right) \\ \\ & =\left(\sin x\right)^{\prime} \sin ^{k} x+\sin x\left(\sin ^{k}\right)’ \\ \\ & =\cos x \sin ^{k} x+\sin x\left(k \sin ^{(k-1)}\right) \\ \\ & =\cos x \sin ^{k} x+k \sin ^{k} x \cos x \\ \\ & =\left(k+1\right) \sin ^{k} x \cos x \\ \\ & =\left(\sin x\right)^{\prime} \sin ^{k} x+\sin x\left(\sin ^{k} x\right)^{\prime} \qquad \left[\text{ By Leibnitz product rule }\right] \\ \\ & =\cos x \sin ^{k} x+\sin x\left(k \sin ^{(k-1)} x \cos x\right) \qquad {\left[\text{ Using (2) }\right]} \end{aligned} $

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

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

Answer :

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

By quotient rule,

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

Answer :

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

By quotient rule,

$ \begin{aligned} f^{\prime}\left(x\right) & =\dfrac{\cos x \dfrac{d}{d x}\left[\sin \left(x+a\right)\right]-\sin \left(x+a\right) \dfrac{d}{d x} \cos x}{\cos ^{2} x} \\ \\ f^{\prime}\left(x\right) & =\dfrac{\cos x \dfrac{d}{d x}\left[\sin \left(x+a\right)\right]-\sin \left(x+a\right)\left(-\sin x\right)}{\cos ^{2} x} \qquad …{(i)} \end{aligned} $

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

By first principle,

$ \begin{aligned} g^{\prime}\left(x\right) & =\lim _ {h \to 0} \dfrac{g\left(x+h\right)-g\left(x\right)}{h} \\ \\ & =\lim _ {h \to 0} \dfrac{1}{h}\left[\sin \left(x+h+a\right)-\sin \left(x+a\right)\right] \\ \\ & =\lim _ {h \to 0} \dfrac{1}{h}\left[2 \cos \left(\dfrac{x+h+a+x+a}{2}\right) \sin \left(\dfrac{x+h+a-x-a}{2}\right)\right] \\ \\ & =\lim _ {h \to 0} \dfrac{1}{h}\left[2 \cos \left(\dfrac{2 x+2 a+h}{2}\right) \sin \left(\dfrac{h}{2}\right)\right] \\ \\ & =\lim _ {h \to 0}{ \cos \left( \dfrac { 2 x + 2 a + h } { 2 } \right) \left[\dfrac{\sin \left(\dfrac{h}{2}\right)}{\left(\dfrac{h}{2}\right)}\right]} \quad\left[\text{ As } h \to 0 \Rightarrow \dfrac{h}{2} \to 0\right] \\ \\ & =\lim _ {h \to 0} \cos \left(\dfrac{2 x+2 a+h}{2}\right) \lim _ {\frac{h}{2} \to 0}\left \lbrace \dfrac{\sin \left(\dfrac{h}{2}\right)}{\left(\dfrac{h}{2}\right)}\right \rbrace \quad\left[\lim _ {h \to 0} \dfrac{\sin h}{h}=1\right] \\ \\ & =\left(\cos \dfrac{2 x+2 a}{2}\right) \times 1 \\ \\ & =\cos \left(x+a\right) \qquad …{(ii)} \end{aligned} $

From (i) and (ii), we obtain

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

Answer :

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

By product rule,

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

Answer:

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

By product rule,

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

Answer :

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

By product rule,

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

Answer :

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

By product rule,

$ \begin{aligned} f^{\prime}\left(x\right) & =\left(x+\cos x\right) \dfrac{d}{d x}\left(x-\tan x\right)+\left(x-\tan x\right) \dfrac{d}{d x}\left(x+\cos x\right) \\ \\ & =\left(x+\cos x\right)\left[\dfrac{d}{d x}\left(x\right)-\dfrac{d}{d x}\left(\tan x\right)\right]+\left(x-\tan x\right)\left(1-\sin x\right) \\ \\ & =\left(x+\cos x\right)\left[1-\dfrac{d}{d x} \tan x\right]+\left(x-\tan x\right)\left(1-\sin x\right)\qquad …(i) \\ \\ & \text{ Let } g\left(x\right)=\tan x \text{. Accordingly, } g\left(x+h\right)=\tan \left(x+h\right) \end{aligned} $

By first principle,

$ \begin{aligned} g^{\prime}\left(x\right) & =\lim _ {h \to 0} \dfrac{g\left(x+h\right)-g\left(x\right)}{h} \\ \\ & =\lim _ {h \to 0}\left(\dfrac{\tan \left(x+h\right)-\tan x}{h}\right) \\ \\ & =\lim _ {h \to 0} \dfrac{1}{h}\left[\dfrac{\sin \left(x+h\right)}{\cos \left(x+h\right)}-\dfrac{\sin x}{\cos x}\right] \\ \\ & =\lim _ {h \to 0} \dfrac{1}{h}\left[\dfrac{\sin \left(x+h\right) \cos x-\sin x \cos \left(x+h\right)}{\cos \left(x+h\right) \cos x}\right] \\ \\ & =\dfrac{1}{\cos x} \cdot \lim _ {h \to 0} \dfrac{1}{h}\left[\dfrac{\sin \left(x+h-x\right)}{\cos \left(x+h\right)}\right] \\ \\ & =\dfrac{1}{\cos x} \cdot \lim _ {h \to 0} \dfrac{1}{h}\left[\dfrac{\sin h}{\cos \left(x+h\right)}\right] \\ \\ & =\dfrac{1}{\cos {x}} \cdot\left(\lim _ {h \to 0} \dfrac{\sin h}{h}\right) \cdot\left(\lim _ {h \to 0} \dfrac{1}{\cos \left(x+h\right)}\right) \\ \\ & =\dfrac{1}{\cos x} \times 1 \times \dfrac{1}{\cos \left(x+0\right)} \\ \\ & =\dfrac{1}{\cos ^{2} x} \\ \\ & =\sec ^{2} x \qquad … {(ii)} \end{aligned} $

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

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

Answer :

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

By quotient rule,

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

Answer :

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

By quotient rule,

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

Answer :

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

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

$f^{\prime}\left(x\right)=\dfrac{\left(1+\tan x\right)-x \cdot \dfrac{d}{d x}\left(1+\tan x\right)}{\left(1+\tan x\right)^{2}}\qquad …(i)$

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

By first principle,

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

From (i) and (ii), we obtain

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

Answer :

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

By product rule,

$ \begin{aligned} f^{\prime}\left(x\right) & =\left(x+\sec x\right) \dfrac{d}{d x}\left(x-\tan x\right)+\left(x-\tan x\right) \dfrac{d}{d x}\left(x+\sec x\right) \\ \\ & =\left(x+\sec x\right)\left[\dfrac{d}{d x}\left(x\right)-\dfrac{d}{d x} \tan x\right]+\left(x-\tan x\right)\left[\dfrac{d}{d x}\left(x\right)+\dfrac{d}{d x} \sec x\right] \\ \\ & =\left(x+\sec x\right)\left[1-\dfrac{d}{d x} \tan x\right]+\left(x-\tan x\right)\left[1+\dfrac{d}{d x} \sec x\right] \qquad …{(i)} \end{aligned} $

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

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

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

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

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

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

Answer :

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

By quotient rule,

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

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}\left(x\right) & =\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\left(n \sin ^{n-1} x \cos x\right)}{\sin ^{2 n} x} \\ \\ & =\dfrac{\sin ^{n-1} x\left(\sin x-n x \cos x\right)}{\sin ^{2 n} x} \\ \\ & =\dfrac{\sin x-n x \cos x}{\sin ^{n+1} x} \end{aligned} $