Question

Let $\cup ={\sin }^{-1}\left(\frac{2x}{1+x^2}\right)\mathrm{and}V={\tan }^{-1}\left(\frac{2x}{1-x^2}\right),\mathrm{then}\frac{d\cup }{dV}=$
(a) 1/2
(b) x
(c) $\frac{1-x^2}{x^2-4}$
(d) 1

Answer 1

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

$\therefore \frac{du}{dv}=\frac{{\displaystyle \frac{du}{dx}}}{{\displaystyle \frac{dv}{dx}}}=\frac{2}{1+x^2}\times \frac{1+x^2}{2}=1$