Question

The derivative of ${\sin }^{-1}\left(\frac{2x}{1+x^2}\right)$ with respect to ${\sin }^{-1}\left(\frac{1-x^2}{1+x^2}\right)$ is-

• A. $2$
• B. $\frac{1}{2}$
• C. $-1$
• D. $1$

Answer 1

The correct option is C $-1$

Let $u={\sin }^{-1}\left(\frac{2x}{1+x^2}\right)$
$\Rightarrow u=2{\tan }^{-1}x$ .... $[\because {\tan }^{-1}x={\sin }^{-1}\left(\frac{2x}{1+x^2}\right)]$
$⟹\frac{du}{dx}=\frac{2}{1+x^2}$ .... $\left(i\right)$
And $v={\sin }^{-1}\left(\frac{1-x^2}{1+x^2}\right)$
$⟹v=\frac{\pi }{2}-{\cos }^{-1}\left(\frac{1-x^2}{1+x^2}\right)$ ....... $[\because {\sin }^{-1}x+{\cos }^{-1}x=\frac{\pi }{2}]$
$⟹v=\frac{\pi }{2}-2{\tan }^{-1}x$ ..... $[\because {\tan }^{-1}x={\cos }^{-1}\left(\frac{1-x^2}{1+x^2}\right)]$
$⟹\frac{dv}{dx}=-\frac{2}{1+x^2}$ ..... $\left(ii\right)$
$\therefore \frac{du}{dv}=-1$ ..... From $\left(i\right)$ and $\left(ii\right)$