Question

The differential coefficient of the function ${\tan }^{-1}\frac{2x}{1-x^2}$ w.r.t. $x^2$ is-

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

Answer 1

The correct option is C $\frac{1}{x(1+x^2)}$

$\frac{d}{dx}\left({\tan }^{-1}\frac{2x}{1-x^2}\right)=\frac{\frac{d}{dx}\left(\frac{2x}{1-x^2}\right)}{1+{\left(\frac{2x}{1-x^2}\right)}^2}$

$=\frac{\frac{(1-x^2)\left(2\right)-\left(2x\right)-\left(2x\right)}{{(1-x^2)}^2}}{1+{\left(\frac{2x}{1-x^2}\right)}^2}=\frac{2-{2x}^2+{4x}^2}{{(1-x^2)}^2+{\left(2x\right)}^2}$

$=\frac{{2x}^2+2}{1+x^4-{2x}^2+{4x}^2}=\frac{2(x^2+1)}{(x^2+1)}=\frac{2}{x^2+1}$

$\frac{d}{dx}\left(x^2\right)=2x$

Therefore derivative of ${\tan }^{-1}\left(\frac{2x}{1-x^2}\right)$ w.r.t $x^2$ is

$=\frac{2}{x^2+1}\times \frac{1}{2x}=\frac{1}{x(x^2+1)}$