Question

The derivative of ${\sin }^{-1}\frac{2x}{1+x^2}$ with respect to ${\cos }^{-1}\frac{1-x^2}{1+x^2}$ is

• A. -1
• B. 1
• C. 2
• D. 4

Answer 1

Answer: (B)

Let $P={\sin }^{-1}\frac{2x}{1+x^2}=2{\tan }^{-1}x$
And $q={\cos }^{-1}\frac{1-x^2}{1+x^2}=2{\tan }^{-1}x$
$\therefore \frac{dp}{dx}=\frac{2}{1+x^2}=$ and $\frac{dq}{dx}=\frac{2}{1+x^2}$
$\Rightarrow \frac{dp}{dq}=\frac{\frac{dp}{dx}}{\frac{2}{dq}}=\frac{\frac{2}{1+x^2}}{\frac{2}{1+x^2}}=1$