could somebody please check my answer if it is correct.

Question

the question is: Find the value of f'(-2) if f(x)=(x^2+2) sin(4-3x)

Answer 1

f(x)=(x^2+2) sin(4-3x)

f'(x)=(x^2+2)(-3)cos(4-3x)+2xsin(4-3x)

        =-3(x^2+2)cos(4-3x)+2xsin(4-3x)

substitute x=-2

f'(-2)=-3((-2)^2+2)cos(4-3(-2))+2(-2)sin(4-3(-2))

           =-18cos(10)-4sin(10)

          =-18cos(10*180/pi)-4sin(10*180/pi)

          =-18*-0.84-4*-0.545

          =15.12+2.18

           =17.3