Question

If $y={\sin }^{-1}\left(\frac{2x}{1+x^2}\right)$, write the value of $\frac{dy}{dx}\mathrm{for}x>1$.

Answer 1

$$\begin{gathered} \mathrm{We}\mathrm{have},y={\sin }^{-1}\left(\frac{2x}{1+x^2}\right) \\ \mathrm{Putting}x=\tan \theta \\ \Rightarrow 1<\tan \theta <\infty \\ \Rightarrow \frac{\mathrm{\pi }}{4}<\theta <\frac{\mathrm{\pi }}{2} \\ \Rightarrow \frac{\mathrm{\pi }}{2}<2\theta <\mathrm{\pi } \\ \therefore \mathrm{y}={\sin }^{-1}\left(\sin 2\mathrm{\theta }\right) \\ \Rightarrow \mathrm{y}={\sin }^{-1}\left\{\sin \left(\mathrm{\pi }-2\mathrm{\theta }\right)\right\} \\ \Rightarrow \mathrm{y}=\mathrm{\pi }-2\mathrm{\theta } \\ \Rightarrow \mathrm{y}=\mathrm{\pi }-2{\tan }^{-1}x \\ \Rightarrow \frac{dy}{dx}=0-\frac{2}{1+x^2} \\ \Rightarrow \frac{dy}{dx}=-\frac{2}{1+x^2} \end{gathered}$$