Question

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

Answer 1

$$\begin{gathered} \mathrm{Here},y={\tan }^{-1}\left(\frac{\sqrt{1+x}-\sqrt{1-x}}{\sqrt{1+x}+\sqrt{1-x}}\right) \\ Putx=\cos 2\theta \\ \therefore y={\tan }^{-1}\left(\frac{\sqrt{1+\cos 2\theta }-\sqrt{1-\cos 2\theta }}{\sqrt{1+\cos 2\theta }+\sqrt{1-\cos 2\theta }}\right) \\ ={\tan }^{-1}\left(\frac{\sqrt{2{\cos }^2\theta }-\sqrt{2{\sin }^2\theta }}{\sqrt{2{\cos }^2\theta }+\sqrt{2{\sin }^2\theta }}\right) \\ ={\tan }^{-1}\left(\frac{\sqrt{2}\left(\cos \theta -sin\theta \right)}{\sqrt{2}\left(\cos \theta +sin\theta \right)}\right) \\ ={\tan }^{-1}\left(\frac{{\displaystyle \frac{\cos \theta -sin\theta }{\cos \theta }}}{{\displaystyle \frac{\cos \theta +\sin \theta }{\cos \theta }}}\right)\left[\mathrm{Dividing}\mathrm{numerator}\mathrm{and}\mathrm{denominator}\mathrm{by}\cos \theta \right] \\ ={\tan }^{-1}\left(\frac{{\displaystyle \frac{\cos \theta }{\cos \theta }}-{\displaystyle \frac{\sin \theta }{\cos \theta }}}{{\displaystyle \frac{\cos \theta }{\cos \theta }}+{\displaystyle \frac{\sin \theta }{\cos \theta }}}\right) \\ ={\tan }^{-1}\left(\frac{1-\tan \theta }{1+\tan \theta }\right) \end{gathered}$$

$$\begin{gathered} ={\tan }^{-1}\left(\frac{\tan {\displaystyle \frac{\mathrm{\pi }}{4}}-\tan \theta }{1+\tan {\displaystyle \frac{\mathrm{\pi }}{4}}\times \tan \theta }\right) \\ ={\tan }^{-1}\left[\tan \left(\frac{\mathrm{\pi }}{4}-\theta \right)\right] \\ =\frac{\mathrm{\pi }}{4}-\theta \\ =\frac{\mathrm{\pi }}{4}-\frac{1}{2}{\cos }^{-1}x\left(\mathrm{Using}x=\cos 2\theta \right) \end{gathered}$$

Differentiate it with respect to x,

$$\begin{gathered} \frac{dy}{dx}=0-\frac{1}{2}\left(\frac{-1}{\sqrt{1-x^2}}\right) \\ \therefore \frac{dy}{dx}=\frac{1}{2\sqrt{1-x^2}} \end{gathered}$$