Question

$\left(\frac{d}{dx}\right)\left[\frac{(1+cotx)}{(1-cotx)}\right]=$

Answer 1

Determine the derivative of the given function.

$\left(\frac{d}{dx}\right)\left[\frac{(1+cotx)}{(1-cotx)}\right]=$

Use formula:

$\frac{d}{dx}\frac{u}{v}=\frac{v\frac{du}{dx}-u\frac{dv}{dx}}{u^2}$

$$\begin{gathered} \left(\frac{d}{dx}\right)\left[\frac{(1+cotx)}{(1-cotx)}\right]=\frac{(1-cotx)\frac{d}{dx}(1+cotx)-(1+cotx)\frac{d}{dx}(1-cotx)}{{(1-cotx)}^2} \\ \Rightarrow =\frac{(1-cotx)\frac{d}{dx}1+\frac{d}{dx}cotx-(1+cotx)\frac{d}{dx}1-\frac{d}{dx}cotx}{{(1-cotx)}^2} \\ \Rightarrow =\frac{(1-cotx).0-\cos e{c}^{2}x-(1+cotx).0-(-\cos e{c}^{2}x)}{{(1-cotx)}^2}\because \frac{\mathbf{d}}{\mathbf{d}\mathbf{x}}\mathbf{}\mathit{c}\mathit{o}\mathit{t}\mathbf{}\mathit{x}=-\cos e{c}^{2}xand\frac{\mathbf{d}}{\mathbf{d}\mathbf{x}}\mathbf{1}=0 \\ \Rightarrow =\frac{-\cos e{c}^{2}x+(-\cos e{c}^{2}x)}{1^2+co{t}^{2}x-2.1.cotx}\because {\mathbf{(}\mathit{a}\mathbf{}\mathbf{-}\mathbf{}\mathit{b}\mathbf{)}}^{\mathbf{2}}=a^2+b^2-2.a.b \\ \Rightarrow =\frac{-\cos e{c}^{2}x-\cos e{c}^{2}x}{\cos e{c}^{2}x-2cotx}\because \cos e{c}^{2}x-co{t}^{2}x=1So,\mathit{c}\mathit{o}\mathit{s}\mathit{e}{\mathit{c}}^{\mathbf{2}\mathbf{}}\mathit{x}=1+co{t}^{2}x \\ \Rightarrow =\frac{-2\cos e{c}^{2}x}{({\displaystyle \frac{1}{{\sin }^{2}x}}-{\displaystyle \frac{2\cos x}{\sin x}})} \\ \Rightarrow =\frac{-2\cos e{c}^{2}x}{{\displaystyle \frac{1-2\sin x\cos x}{{\sin }^{2}x}}} \\ \Rightarrow =\frac{2}{{\displaystyle 2\sin x\cos x-1}}\mathbf{\therefore }\mathbf{}\mathbf{sin}\mathbf{}\mathbf{2}\mathit{x}\mathbf{}\mathbf{=}\mathbf{}2\sin x\cos x \\ \Rightarrow =\frac{2}{{\displaystyle \sin 2x-1}} \end{gathered}$$

Hence, the derivative of the given function $\left(\frac{d}{dx}\right)\left[\frac{(1+cotx)}{(1-cotx)}\right]$ is$"\frac{\mathbf{2}}{\mathbf{s}\mathbf{i}\mathbf{n}\mathbf{}\mathbf{2}\mathbf{x}\mathbf{}\mathbf{-}\mathbf{}\mathbf{1}}"$.