Question

Differentiate cos^-1 ( 1 – x² / 1 + x² ) with respect to tan^-1 ( 3x – x³ / 1 – 3x² )
The answer: 2/3

Answer 1

$$\begin{gathered} \mathrm{Let}u={\cos }^{-1}\left[\frac{1-x^2}{1+x^2}\right] \\ \mathrm{put}x=\tan \theta ,weget \\ u={\cos }^{-1}\left(\frac{1-{\tan }^2\theta }{1+{\tan }^2\theta }\right) \\ \Rightarrow u={\cos }^{-1}\left(\cos 2\theta \right) \\ \Rightarrow u=2\theta \\ \Rightarrow u=2{\tan }^{-1}x \\ \mathrm{differentiating}\mathrm{both}\mathrm{sides}\mathrm{with}\mathrm{respect}\mathrm{to}x,\mathrm{we}\mathrm{get} \\ \frac{du}{dx}=2\times \frac{1}{1+x^2}=\frac{2}{1+x^2} \\ Letv={\tan }^{-1}\left[\frac{3x-x^3}{1-3x^2}\right] \\ putx=\tan \theta \\ v={\tan }^{-1}\left[\frac{3\tan \theta -{\tan }^3\theta }{1-3{\tan }^2\theta }\right] \\ \Rightarrow v={\tan }^{-1}\left(\tan 3\theta \right) \\ \Rightarrow v=3\theta \\ \Rightarrow v=3{\tan }^{-1}x \\ \mathrm{differentiating}\mathrm{both}\mathrm{sides}\mathrm{with}\mathrm{respect}\mathrm{to}x,\mathrm{we}\mathrm{get} \\ \frac{dv}{dx}=\frac{3}{1+x^2} \\ \mathrm{Now},\frac{du}{dv}=\frac{2}{1+x^2}\times \frac{1+x^2}{3} \\ \Rightarrow \boxed{\frac{du}{dv}=\frac{2}{3}} \end{gathered}$$