Question

Differentiate

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

Answer 1

$$\begin{gathered} \mathrm{Let}y=\sqrt{\frac{1-x^2}{1+x^2}} \\ \Rightarrow y={\left(\frac{1-x^2}{1+x^2}\right)}^{\frac{1}{2}} \\ \mathrm{Differentiate}\mathrm{it}\mathrm{with}\mathrm{respect}\mathrm{to}x\mathrm{we}\mathrm{get}, \\ \frac{\mathit{d}\mathit{y}}{\mathit{d}\mathit{x}}=\frac{d}{dx}{\left(\frac{1-x^2}{1+x^2}\right)}^{\frac{1}{2}} \\ =\frac{1}{2}{\left(\frac{1-x^2}{1+x^2}\right)}^{\frac{1}{2}-1}\times \frac{d}{dx}\left(\frac{1-x^2}{1+x^2}\right)\left[\mathrm{Using}\mathrm{chain}\mathrm{rule}\right] \\ =\frac{1}{2}{\left(\frac{1-x^2}{1+x^2}\right)}^{\frac{-1}{2}}\times \left\{\frac{\left(1+x^2\right){\displaystyle \frac{d}{dx}}\left(1-x^2\right)-\left(1-x^2\right){\displaystyle \frac{d}{dx}}\left(1+x^2\right)}{{\left(1+x^2\right)}^2}\right\} \\ =\frac{1}{2}{\left(\frac{1+x^2}{1-x^2}\right)}^{\frac{1}{2}}\left\{\frac{-2x\left(1+x^2\right)-2x\left(1-x^2\right)}{{\left(1+x^2\right)}^2}\right\} \\ =\frac{1}{2}{\left(\frac{1+x^2}{1-x^2}\right)}^{\frac{1}{2}}\left\{\frac{-2x-2x^3-2x+2x^3}{{\left(1+x^2\right)}^2}\right\} \\ =\frac{1}{2}{\left(\frac{1+x^2}{1-x^2}\right)}^{\frac{1}{2}}\left\{\frac{-4x}{{\left(1+x^2\right)}^2}\right\} \\ =\frac{-2x}{\sqrt{1-x^2}{\left(1+x^2\right)}^{\frac{3}{2}}} \end{gathered}$$