Question

If $y={\tan }^{-1}\left\{\frac{\sqrt{1+x^2}+\sqrt{1-x^2}}{\sqrt{1+x^2}-\sqrt{1-x^2}}\right\},\mathrm{find}\frac{dy}{dx}$.

Answer 1

$$\begin{gathered} \mathrm{Here},y={\tan }^{-1}\left(\frac{\sqrt{1+x^2}+\sqrt{1-x^2}}{\sqrt{1+x^2}-\sqrt{1-x^2}}\right) \\ \mathrm{Put}{x}^2=\cos 2\theta \\ \Rightarrow y={\tan }^{-1}\left(\frac{\sqrt{1+\cos 2\theta }+\sqrt{1-\cos 2\theta }}{\sqrt{1+\cos 2\theta }-\sqrt{1-\cos 2\theta }}\right) \\ \Rightarrow y={\tan }^{-1}\left(\frac{\sqrt{2{\cos }^2\theta }+\sqrt{2{\sin }^2\theta }}{\sqrt{2{\cos }^2\theta }-\sqrt{2{\sin }^2\theta }}\right) \\ \Rightarrow y={\tan }^{-1}\left(\frac{\sqrt{2}\left(\cos \theta +sin\theta \right)}{\sqrt{2}\left(\cos \theta -sin\theta \right)}\right) \\ \Rightarrow y={\tan }^{-1}\left(\frac{{\displaystyle \frac{\cos \theta +sin\theta }{\cos \theta }}}{{\displaystyle \frac{\cos \theta -\sin \theta }{\cos \theta }}}\right)\left[\mathrm{Dividing}\mathrm{numerator}\mathrm{and}\mathrm{denominator}\mathrm{by}\cos \theta \right] \\ \Rightarrow y={\tan }^{-1}\left(\frac{{\displaystyle \frac{\cos \theta }{\cos \theta }}+{\displaystyle \frac{\sin \theta }{\cos \theta }}}{{\displaystyle \frac{\cos \theta }{\cos \theta }}-{\displaystyle \frac{\sin \theta }{\cos \theta }}}\right) \\ \Rightarrow y={\tan }^{-1}\left(\frac{1+\tan \theta }{1-\tan \theta }\right) \\ \Rightarrow y={\tan }^{-1}\left(\frac{\tan {\displaystyle \frac{\mathrm{\pi }}{4}}+\tan \theta }{1+\tan {\displaystyle \frac{\mathrm{\pi }}{4}}+\tan \theta }\right) \\ \Rightarrow y={\tan }^{-1}\left[\tan \left(\frac{\mathrm{\pi }}{4}+\theta \right)\right] \\ \Rightarrow y=\frac{\mathrm{\pi }}{4}+\theta \\ \Rightarrow y=\frac{\mathrm{\pi }}{4}+\frac{1}{2}{\cos }^{-1}{x}^2\left(\mathrm{Using}{x}^2=\cos 2\theta \right) \end{gathered}$$

Differentiate it with respect to x,

$$\begin{gathered} \frac{dy}{dx}=0+\frac{1}{2}\left[\frac{-1}{\sqrt{1-{\left(x^2\right)}^2}}\right]\times 2x \\ \frac{dy}{dx}=\frac{-x}{\sqrt{1-x^4}} \end{gathered}$$