if y = tan^-1 [(root(1+x^2) - root (1-x^2))/root(1+x^2) + root (1-x^2))], find dy/dx

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Please find below the solution to the asked query:

$$\begin{gathered} \mathrm{We} \mathrm{have} \\ \mathrm{y}={\tan }^{-1}\left[\frac{\sqrt{1+{\mathrm{x}}^2}-\sqrt{1-{\mathrm{x}}^2}}{\sqrt{1+{\mathrm{x}}^2}+\sqrt{1-{\mathrm{x}}^2}}\right] \\ \mathrm{Put} {\mathrm{x}}^2=\cos 2\mathrm{A} \left[\Rightarrow \mathrm{A}=\frac{1}{2}{\cos }^{-1}\left({\mathrm{x}}^2\right)\right] \\ \Rightarrow \mathrm{y}={\tan }^{-1}\left[\frac{\sqrt{1+\cos 2\mathrm{A}}-\sqrt{1-\cos 2\mathrm{A}}}{\sqrt{1+\cos 2\mathrm{A}}+\sqrt{1-\cos 2\mathrm{A}}}\right] \\ ={\tan }^{-1}\left[\frac{\sqrt{1+2{\cos }^2\mathrm{A}-1}-\sqrt{1-1+2{\sin }^2\mathrm{A}}}{\sqrt{1+2{\cos }^2\mathrm{A}-1}-\sqrt{1-1+2{\sin }^2\mathrm{A}}}\right] \\ ={\tan }^{-1}\left[\frac{\sqrt{2}\mathrm{cosA}-\sqrt{2}\mathrm{sinA}}{\sqrt{2}\mathrm{cosA}+\sqrt{2}\mathrm{sinA}}\right] \\ ={\tan }^{-1}\left[\frac{\mathrm{cosA}-\mathrm{sinA}}{\mathrm{cosA}+\mathrm{sinA}}\right] \\ ={\tan }^{-1}\left[\frac{1-\mathrm{tanA}}{1+\mathrm{tanA}}\right] \left[\mathrm{Divide} \mathrm{by} \mathrm{cosA}\right] \\ ={\tan }^{-1}\left[\tan \left(\frac{\mathrm{\pi }}{4}-\mathrm{A}\right)\right] \\ =\frac{\mathrm{\pi }}{4}-\mathrm{A} \\ \mathrm{y}=\frac{\mathrm{\pi }}{4}-\frac{1}{2}{\cos }^{-1}\left({\mathrm{x}}^2\right) \\ \mathrm{Differentiate} \mathrm{with} \mathrm{respect} \mathrm{to} \mathrm{x} \\ \frac{\mathrm{dy}}{\mathrm{dx}}=-\frac{1}{2}\left[-\frac{2\mathrm{x}}{\sqrt{1-{\mathrm{x}}^4}}\right]=\frac{\mathrm{x}}{\sqrt{1-{\mathrm{x}}^4}}\left[\mathrm{Answer}\right] \end{gathered}$$

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