# How do you differentiate y=cot^2(sintheta)?

May 23, 2017

$y ' = - 2 {\csc}^{2} \left(\sin \left(\theta\right)\right) \cot \left(\sin \left(\theta\right)\right) \cos \left(\theta\right)$

#### Explanation:

Differentiate $y = {\cot}^{2} \left(\sin \theta\right)$

Chain rule:
For $h = f \left(g \left(x\right)\right)$,
$h ' = f ' \left(g \left(x\right)\right) \cdot g ' \left(x\right)$

First we note that the given equation can also be written as
$y = {\left(\cot \left(\sin \theta\right)\right)}^{2}$

We can apply the chain rule:
$y ' = 2 \left(\cot \left(\sin \left(\theta\right)\right)\right) \cdot - {\csc}^{2} \left(\sin \left(\theta\right)\right) \cdot \cos \left(\theta\right)$

Therefore,
$y ' = - 2 {\csc}^{2} \left(\sin \left(\theta\right)\right) \cot \left(\sin \left(\theta\right)\right) \cos \left(\theta\right)$