Question

Differentiate

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

Answer 1

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