Question

If $y={\tan }^{-1}\left(\frac{1-x}{1+x}\right),\mathrm{find}\frac{dy}{dx}$.

Answer 1

$\mathrm{We}\mathrm{have},y={\tan }^{-1}\left(\frac{1-x}{1+x}\right)$

$$\begin{gathered} \Rightarrow \frac{dy}{dx}=\frac{1}{1+{\left({\displaystyle \frac{1-x}{1+x}}\right)}^2}\frac{d}{dx}\left(\frac{1-x}{1+x}\right) \\ \Rightarrow \frac{dy}{dx}=\frac{{\left(1+x\right)}^2}{1+x^2+2x+1+x^2-2x}\left[\frac{\left(1+x\right){\displaystyle \frac{d}{dx}}\left(1-x\right)-\left(1-x\right){\displaystyle \frac{d}{dx}}\left(1+x\right)}{{\left(1+x\right)}^2}\right]\left[\mathrm{using}\mathrm{quotient}\mathrm{rule}\right] \\ \Rightarrow \frac{dy}{dx}=\frac{{\left(1+x\right)}^2}{2x^2+2}\left[\frac{\left(1+x\right)\left(-1\right)-\left(1-x\right)\left(1\right)}{\left(1+x\right)}\right] \\ \Rightarrow \frac{dy}{dx}=\frac{{\left(1+x\right)}^2}{2\left(x^2+1\right)}\left(\frac{-x-1-1+x}{{\left(1+x\right)}^2}\right) \\ \Rightarrow \frac{dy}{dx}=\frac{{\left(1+x\right)}^2}{2\left(x^2+1\right)}\times \frac{-2}{{\left(1+x\right)}^2} \\ \Rightarrow \frac{dy}{dx}=-\frac{1}{x^2+1} \end{gathered}$$