Question

If $u={\sin }^{-1}\left(\frac{2x}{1+x^2}\right)\mathrm{and}v={\tan }^{-1}\left(\frac{2x}{1-x^2}\right)$, where $-1<x<1$, then write the value of $\frac{du}{dv}$.

Answer 1

$$\begin{gathered} \mathrm{We}\mathrm{have},u={\sin }^{-1}\left(\frac{2x}{1+x^2}\right)\mathrm{and}v={\tan }^{-1}\left(\frac{2x}{1-x^2}\right) \\ \Rightarrow \frac{du}{dx}=\frac{2}{1+x^2}\mathrm{and}\frac{dv}{dx}=\frac{2}{1+x^2} \\ \therefore \frac{du}{dv}=\frac{{\displaystyle \frac{du}{dx}}}{{\displaystyle \frac{dv}{dx}}}=\frac{2}{1+x^2}\times \frac{1+x^2}{2}=1 \end{gathered}$$