# mmu

find dy/dx of y=1/2 cot^7(-x^2)

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1. Which of these is your school SUBJECT?

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2. use the chain rule. let

v = -x^2
u = cot v
y = 1/2 u^7

y' = 7/2 u^6 u'
u' = -csc^2 v v'
v' = -2x

So,

dy/dx = 7/2 cot^6(-x^2) * -csc^2(-x^2) * -2x
= 7x cot^2(-x^2) csc^2(-x^2)
or, since
cot(-x) = -cotx
csc(-x) = -csc(x)

7x cot^2(x^2) csc^2(x^2)

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posted by Steve

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