Question

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

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

Answer 1

The correct option is A $0$

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

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

using ${\tan }^{-1}A+{\tan }^{-1}B={\tan }^{-1}[\frac{A+B}{}1-AB]$

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

$=\frac{d}{dx}\left\{{\tan }^{-1}\infty \right\}$

This term is content term & differentiation of way is zero

$=0$