Question

If the ans is negative, fill the absolute value

$y={\cot }^{-1}x$, find $\left|\frac{dy}{dx}\right|$ at $x=0$.

Answer 1

Let $y={\cot }^{-1}x\Rightarrow \cot y=x$

Differentiating both side w.r.t. $x$ we get,

$-cose{c}^{2}y\frac{dy}{dx}=1\Rightarrow \frac{dy}{dx}=-\frac{1}{1+{\cot }^{2}y}$

$\therefore \frac{dy}{dx}=-\frac{1}{1+x^2}.$

Clearly at $x=0$, $\left|\frac{dy}{dx}\right|=|-1|=1$