Question

How do you do this question?

Answer 1

Answer:

Step-by-step explanation:

$f(x)=sin^2(3-x)\\f'(x)=2 sin(3-x) cos(3-x)(-1)\\f'(0)=-2 sin 3cos 3\\B$

Answer 2

Answer:

B

Step-by-step explanation:

f(x) = sin^2(3 - x)

f'(x) = 2*sin(3 - x)*cos(3 - x)(-1)

This comes from d(sin^2(3 - x))/dx which is 2*sin^2(3 - x)

d(sin(3 - x)/dx which is cos(3 - x)

d(3 - x) = - 1

f'(0) = (-1) * 2*sin(3)*cos(3)

Looks like B