Question

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

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

Answer 1

The correct option is D $1$

Let $u={\tan }^{-1}\frac{2x}{1-x^2}$

$u=2{\tan }^{-1}x$ ....(1)

and $v={\sin }^{-1}\frac{2x}{1+x^2}$

$v=2{\tan }^{-1}x$ .... (2)

From (1) and (2),

$u=v$

$\frac{du}{dv}=1$