How do you differentiate #sin^2(3-x)#?

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Answer 1 · 2018-04-23T21:38:06

#2sin(x-3)cos(x-3)#

We use the chain rule here

#d/dx[f(x)]^n=[f(x)]^(n-1)f'(x)#

We also use these facts about #sin# and #cos#

#sin(-x)=-sinx#

#cos(-x)=cosx#

So

#d/dxsin^2(3-x)=-2sin(3-x)cos(3-x)=-2(-sin(x-3))cos(x-3)=2sin(x-3)cos(x-3)#