Question

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

• A. $0$
• B. $1$
• C. $\frac{1}{1-x^2}$
• D. $\frac{1}{1+x^2}$

Answer 1

Answer: (B)

Let $u=si{n}^{-1}\left(\frac{2x}{1+x^2}\right)=2ta{n}^{-1}x$
and $v=ta{n}^{-1}\left(\frac{2x}{1-x^2}\right)=2ta{n}^{-1}x$
$\Rightarrow u=v$
$\Rightarrow \frac{du}{dx}=\frac{dv}{dx}$
$\therefore \frac{du}{dv}=\frac{\left(du/dx\right)}{\left(dvdx/\right)}=1$