Question

If $y=(1+x^2){\tan }^{-1}x-x,$ then $\frac{dy}{dx}$ is equal to

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

Answer 1

The correct option is B $2x{\tan }^{-1}x$

$y=(1+x^2){\tan }^{-1}x-x$
By product rule
$\Rightarrow \frac{dy}{dx}=(1+x^2)\frac{d}{dx}\left({\tan }^{-1}x\right)+\left({\tan }^{-1}x\right)\frac{d}{dx}(1+x^2)-\frac{d}{dx}\left(x\right)$
$\Rightarrow \frac{dy}{dx}=(1+x^2)\cdot \frac{1}{(1+x^2)}+{\tan }^{-1}x\left(2x\right)-1$
$\Rightarrow \frac{dy}{dx}=2x{\tan }^{-1}x$