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

1 Answer
Y
May 23, 2017

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

Explanation:

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)#

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