If f(x)=sin^2(3-x) then f'(0)=

Question

A. -2cos3
B. -2sin3cos3
C. 6cos3
D. 2sin3cos3
E. 6sin3cos3

I got B

Answer 1

Yes, B is correct.

f'(x)
d(sin^2(3-x))/dx
Apply chain rule
=2sin(3-x) dsin(3-x)/dx
=2sin(3-x)cos(3-x)d(3-x)/dx
=-2sin(3-x)cos(3-x)

Then calculate f'(0) from above.