How do you find $f'(0)$ if $f(x)=sin^2(3-x)$ ?

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Answer 1 · 2015-03-16T16:29:56

The answer is: $-sin6$ .

This is because, applying the chain rule:

$y'=2sin(3-x)* cos(3-x)* (-1)$ and

$y'(0)=-2sin(3-0)cos(3-0)=-2sin3cos3=$

$=-sin6$ .

Remembering that: $sin2alpha=2sinalphacosalpha$ .

How do you find f'(0)f'(0) if f(x)=sin^2(3-x)f(x)=sin^2(3-x)?