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

The correct option is B $1$

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