Question

Differentiate

${\tan }^{-1}\left\{\frac{x}{a+\sqrt{a^2-x^2}}\right\},-a<x<a$

Answer 1

$$\begin{gathered} \mathrm{Let},y={\tan }^{-1}\left\{\frac{x}{a+\sqrt{a^2-x^2}}\right\} \\ \mathrm{Put}x=a\sin \theta \\ \Rightarrow y={\tan }^{-1}\left\{\frac{a\sin \theta }{a+\sqrt{a^2-a^2{\sin }^2\theta }}\right\} \\ \Rightarrow y={\tan }^{-1}\left(\frac{a\sin \theta }{a+\sqrt{a^2\left(1-{\sin }^2\theta \right)}}\right) \\ \Rightarrow y={\tan }^{-1}\left\{\frac{a\sin \theta }{{\displaystyle a+a\cos \theta }}\right\} \\ \Rightarrow y={\tan }^{-1}\left\{\frac{a\sin \theta }{a\left(1+\cos \theta \right)}\right\} \\ \Rightarrow y={\tan }^{-1}\left\{\frac{\sin \theta }{1+\cos \theta }\right\} \\ \Rightarrow y={\tan }^{-1}\left\{\frac{2\sin {\displaystyle \frac{\theta }{2}}\cos {\displaystyle \frac{\theta }{2}}}{2{\cos }^2{\displaystyle \frac{\theta }{2}}}\right\} \\ \Rightarrow y={\tan }^{-1}\left(\tan \frac{\theta }{2}\right)...\left(\mathrm{i}\right) \\ \mathrm{Here},-a<x<a \\ \Rightarrow -1<\frac{x}{a}<1 \\ \Rightarrow -1<\sin \theta <1 \\ \Rightarrow -\frac{\mathrm{\pi }}{2}<\theta <\frac{\mathrm{\pi }}{2} \\ \Rightarrow -\frac{\mathrm{\pi }}{4}<\frac{\theta }{2}<\frac{\mathrm{\pi }}{4} \\ \mathrm{So},\mathrm{from}\mathrm{equation}\left(\mathrm{i}\right), \\ y=\frac{\theta }{2}\left[\mathrm{Since},{\tan }^{-1}\left(\tan \theta \right)=\theta ,\mathrm{if}\theta \in \left[-\frac{\mathrm{\pi }}{2},\frac{\mathrm{\pi }}{2}\right]\right] \\ \Rightarrow y=\frac{1}{2}{\sin }^{-1}\left(\frac{x}{a}\right)\left[\mathrm{Since},x=a\sin \theta \right] \\ \mathrm{Differentiating}\mathrm{it}\mathrm{with}\mathrm{respect}\mathrm{to}x, \\ \frac{dy}{dx}=\frac{1}{2}\times \frac{1}{\sqrt{1-{\left({\displaystyle \frac{x}{a}}\right)}^2}}\frac{d}{dx}\left(\frac{x}{a}\right) \\ \Rightarrow \frac{dy}{dx}=\frac{a}{2\sqrt{a^2-x^2}}\times \left(\frac{1}{a}\right) \\ \therefore \frac{dy}{dx}=\frac{1}{2\sqrt{a^2-x^2}} \end{gathered}$$