Question

Differentiate

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

Answer 1

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