Question

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

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

Answer 1

The correct option is A $\frac{1}{1+x^2}.$

Let $y={\tan }^{-1}x\therefore x=\tan y$
$y+\delta y={\tan }^{-1}(x+\delta x)$
$\therefore x+\delta x=\tan (y+\delta y)$
$\therefore \frac{dy}{dx}=\underset{\delta x\to 0}{lim}\frac{\delta y}{\delta x}$
$=\underset{\delta x\to 0}{lim}\frac{\delta y}{\tan (y+\delta y)-\tan y}=\frac{1}{{\sec }^{2}y}$
$=\frac{1}{1+{\tan }^{2}y}=\frac{1}{1+x^2}.$