Question

If $y={\tan }^{-1}\left(\frac{2x}{1-x^2}\right)+{\sec }^{-1}\left(\frac{1+x^2}{1-x^2}\right),x>0$, prove that $\frac{dy}{dx}=\frac{4}{1+x^2}.$

Answer 1

$$\begin{gathered} \mathrm{Here},y={\tan }^{-1}\left(\frac{2x}{1-x^2}\right)+se{c}^{-1}\left(\frac{1+x^2}{1-x^2}\right) \\ \Rightarrow y={\tan }^{-1}\left(\frac{2x}{1-x^2}\right)+{\cos }^{-1}\left(\frac{1-x^2}{1+x^2}\right) \\ \mathrm{Put}x=\tan \theta \\ \therefore y={\tan }^{-1}\left(\frac{2\tan \theta }{1-{\tan }^2\theta }\right)+{\cos }^{-1}\left(\frac{1-{\tan }^2\theta }{1+{\tan }^2\theta }\right) \\ \Rightarrow y={\tan }^{-1}\left(\tan 2\theta \right)+{\cos }^{-1}\left(\cos 2\theta \right) \\ \Rightarrow \mathrm{y}=2\mathrm{\theta }+2\mathrm{\theta } \\ \Rightarrow \mathrm{y}=4\mathrm{\theta } \\ \Rightarrow \mathrm{y}=4{\tan }^{-1}\mathrm{x}\left[\mathrm{using},\mathrm{x}=\mathrm{tan\theta }\right] \end{gathered}$$

Differentiate it with respect to x,

$\therefore \frac{dy}{dx}=\frac{4}{1+x^2}$