Question

$\frac{d}{dx}\left({\tan }^{-1}\left(\frac{a-x}{1+ax}\right)\right)$ equals if ax > -1

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

Answer 1

The correct option is D $-\frac{1}{1+x^2}$

Let, $y={\tan }^{-1}\left(\frac{a-x}{1+ax}\right)$
$y={\tan }^{-1}\left(a\right)-{\tan }^{-1}\left(x\right)$
$\therefore \frac{dy}{dx}=-\frac{1}{1+x^2}$