Question

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

• A. $\frac{2x}{1+x^4}$
• B. $\frac{2x}{\sqrt{1+4x}}$
• C. $\frac{-2x}{1+x^4}$
• D. $\frac{-2x}{\sqrt{1+x^2}}$

Answer 1

The correct option is C $\frac{-2x}{1+x^4}$

$y={\cot }^{-1}\left(x^2\right)$
$\Rightarrow \frac{dy}{dx}=\frac{-1}{1+{\left(x^2\right)}^2}\frac{d}{dx}\left(x^2\right)=\frac{-1}{1+x^4}\left(2x\right)=\frac{-2x}{1+x^4}$