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

Mar 16, 2015

The answer is: $- \sin 6$.

This is because, applying the chain rule:

$y ' = 2 \sin \left(3 - x\right) \cdot \cos \left(3 - x\right) \cdot \left(- 1\right)$ and

$y ' \left(0\right) = - 2 \sin \left(3 - 0\right) \cos \left(3 - 0\right) = - 2 \sin 3 \cos 3 =$

$= - \sin 6$.

Remembering that: $\sin 2 \alpha = 2 \sin \alpha \cos \alpha$.