Question

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

• A. $-1$
• B. $1$
• C. $2$
• D. $4$

Answer 1

Answer: (B)

Let $p=si{n}^{-1}\left[\frac{2x}{1+x^2}\right]=2ta{n}^{-1}x$
and $q=co{s}^{-1}\left[\frac{1-x^2}{1+x^2}\right]=2ta{n}^{-1}x$
$\Rightarrow \frac{dp}{dx}=\frac{2}{1+x^2}$,
$\frac{dq}{dx}=\frac{2}{1+x^2}$
$\therefore \frac{dp}{dq}=\frac{\frac{dp}{dx}}{\frac{dq}{dx}}$
$=\frac{2}{\frac{\left(1+x^2\right)}{\frac{2}{\left(1+x^2\right)}}}=1$