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

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Answer 1 · 2017-05-23T13:14:17

$y'=-2csc^2(sin(theta))cot(sin(theta))cos(theta)$

Differentiate $y=cot^2(sintheta)$

Chain rule:
For $h=f(g(x))$ ,
$h'=f'(g(x))*g'(x)$

First we note that the given equation can also be written as
$y=(cot(sintheta))^2$

We can apply the chain rule:
$y'=2(cot(sin(theta)))*-csc^2(sin(theta))*cos(theta)$

Therefore,
$y'=-2csc^2(sin(theta))cot(sin(theta))cos(theta)$

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