Question

differentiate tan^-1(√1+x² -√1-x²/√1+x² +√1-x²) with respect to cos^​-1x²

Answer 1

$$\begin{gathered} \mathrm{Let}\mathrm{u}={\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{\theta } \\ \mathrm{Now},\mathrm{u}={\tan }^{-1}\left[\frac{\sqrt{1+\cos 2\mathrm{\theta }}-\sqrt{1-\cos 2\mathrm{\theta }}}{\sqrt{1+\cos 2\mathrm{\theta }}+\sqrt{1-\cos 2\mathrm{\theta }}}\right] \\ \Rightarrow \mathrm{u}={\tan }^{-1}\left[\frac{\sqrt{2{\cos }^2\mathrm{\theta }}-\sqrt{2{\sin }^2\mathrm{\theta }}}{\sqrt{2{\cos }^2\mathrm{\theta }}+\sqrt{2{\sin }^2\mathrm{\theta }}}\right] \\ \Rightarrow \mathrm{u}={\tan }^{-1}\left[\frac{\sqrt{2}\cos \mathrm{\theta }-\sqrt{2}\sin \mathrm{\theta }}{\sqrt{2}\cos \mathrm{\theta }+\sqrt{2}\sin \mathrm{\theta }}\right] \\ \Rightarrow \mathrm{u}={\tan }^{-1}\left[\frac{\cos \mathrm{\theta }-\sin \mathrm{\theta }}{\cos \mathrm{\theta }+\sin \mathrm{\theta }}\right] \\ \Rightarrow \mathrm{u}={\tan }^{-1}\left(\frac{1-\tan \mathrm{\theta }}{1+\tan \mathrm{\theta }}\right) \\ \Rightarrow \mathrm{u}={\tan }^{-1}\left[\frac{\tan \left(\mathrm{\pi }/4\right)-\tan \mathrm{\theta }}{1+\tan \left(\mathrm{\pi }/4\right)\times \tan \mathrm{\theta }}\right] \\ \Rightarrow \mathrm{u}={\tan }^{-1}\left[\tan \left(\frac{\mathrm{\pi }}{4}-\mathrm{\theta }\right)\right] \\ \Rightarrow \mathrm{u}=\frac{\mathrm{\pi }}{4}-\mathrm{\theta } \\ \Rightarrow \mathrm{u}=\frac{\mathrm{\pi }}{4}-\frac{{\cos }^{-1}\left({\mathrm{x}}^2\right)}{2} \\ \Rightarrow \frac{\mathrm{du}}{\mathrm{dx}}=\frac{1}{2}\times \frac{2\mathrm{x}}{\sqrt{1-{\mathrm{x}}^4}}=\frac{\mathrm{x}}{\sqrt{1-{\mathrm{x}}^4}} \\ \mathrm{Let}\mathrm{v}={\cos }^{-1}\left({\mathrm{x}}^2\right) \\ \Rightarrow \frac{\mathrm{dv}}{\mathrm{dx}}=-\frac{2\mathrm{x}}{\sqrt{1-{\mathrm{x}}^4}} \\ \mathrm{Now},\frac{\mathrm{du}}{\mathrm{dv}}=\frac{\mathrm{du}}{\mathrm{dx}}\times \frac{\mathrm{dx}}{\mathrm{dv}}=\frac{\mathrm{x}}{\sqrt{1-{\mathrm{x}}^4}}\times \left(-\frac{\sqrt{1-{\mathrm{x}}^4}}{2\mathrm{x}}\right)=-\frac{1}{2} \end{gathered}$$