Question

If $y={\cot }^{-1}\left(x^2\right)$, then the value of $\frac{dy}{dx}$ 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

Answer: (C)

Let $y={\cot }^{-1}\left(x^2\right)$
$\Rightarrow \cot y=x^2$
Diff both side, w.r.t. '$x$' ;
$-cosec{}^{2}y\cdot \frac{dy}{dx}=2x$
$\frac{dy}{dx}=\frac{2x}{-cosec{}^{2}y}=\frac{2x}{-\left(1+{\cot }^{2}y\right)}=\frac{-2x}{\left(x^4+1\right)}$