Question

If $y={\cot }^{-1}\left(\frac{1+x}{1-x}\right)$, then $\frac{dy}{dx}$ equals to?

• A. $\frac{1}{1+x^2}$
• B. $\frac{-1}{1+x^2}$
• C. $\frac{2}{1+x^2}$
• D. $\frac{-2}{1+x^2}$

Answer 1

The correct option is B

$\frac{-1}{1+x^2}$

Explanation for the correct option:

Finding the value of $\frac{dy}{dx}$:

Given function,

$y={\cot }^{-1}\left(\frac{1+x}{1-x}\right)$

Put $x=\tan \theta$ in the given function,

$\theta ={\tan }^{-1}x$

$\begin{array}{rcl}y& =& co{t}^{-1}\frac{\left(1+\tan \theta \right)}{\left(1-\tan \theta \right)}\\ & =& co{t}^{-1}\frac{\left(\tan \pi /4+\tan \theta \right)}{\left(1-\tan {\displaystyle \frac{\mathrm{\pi }}{4}}\tan \theta \right)}\left[\mathbf{\because }\mathbf{tan}\left(\frac{\mathbf{\pi }}{\mathbf{4}}\right)\mathbf{=}\mathbf{1}\right]\\ & =& co{t}^{-1}\tan \left(\frac{\mathrm{\pi }}{4}+\theta \right)\left[\because \mathit{}\tan \mathit{(}\mathit{A}\mathit{+}\mathit{B}\mathit{)}\mathit{=}\frac{\tan A+\tan B}{1-\tan A\tan B}\right]\\ & =& co{t}^{-1}cot\left(\frac{\mathrm{\pi }}{2}-\left(\frac{\mathrm{\pi }}{4}+\theta \right)\right)\left[\because \tan A=\cot \left(\frac{\pi }{2}-A\right)\right]\\ & =& \frac{\mathrm{\pi }}{4}-\theta \\ & =& \frac{\mathrm{\pi }}{4}–{\tan }^{-1}x\end{array}$

Differentiate the obtained function with respect to $x$

$\frac{dy}{dx}=\frac{-1}{1+x^2}$

Hence, option (B) is correct.