Question

Evaluate:

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

• A. $0$
• B. $1$
• C. $\frac{1}{2}$
• D. $2$

Answer 1

The correct option is A $0$

Given,

$(ta{n}^{-1}\frac{x}{1+x^2}+ta{n}^{-1}\frac{1+x^2}{x})$

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

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

$=\frac{1-x^2}{x^4+3x^2+1}+\frac{x^2-1}{x^4+3x^2+1}$

$=0$