Expressing $dx$ in terms of $d \theta$

Question

Given $x^2$ + $y^2$ = $r^2$ (right-angled triangle with angle $\theta$) and $dx$ as a small length of $x$,

we know that $x$ = $y$$cot \theta$.

However, the answer scheme proceeds to explain that:

$x$ = $y$$cot \theta$ => $dx$ = -$y$($csc^2 \theta$)$d \theta$

and I completely do not understand how they can derive the latter from the former. Can anyone help?

Answer 1

HINT: $$(\cot(x))'=\frac{-\sin^2(x)-\cos^2(x)}{\sin^2(x)}=-1-\cot^2(x)$$