How do you differentiate ${sin}^{2} (3 - x)$ $sin^2(3-x)$ ?

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Answer 1 · 2018-04-23T21:38:06

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

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

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

How do you differentiate sin^2(3-x)sin2(3−x)sin^2(3-x)?