The function is: $\displaystyle f(x)=2*cos^3x+cos^4x$
My derivative is:
$\displaystyle f'(x)=-sin(2x)cos(x)(3+2*cos(x))$
The back gives a totally different answer... I want to confirm mine.
Differentiate $\displaystyle 2\cos^3{x}$ and you get $\displaystyle -6\cos^2{x}\sin{x} = -3\sin{2x}\cos{x}$
Differentiate $\displaystyle \cos^4{x}$ and you get $\displaystyle -4\cos^3{x}\sin{x} = -2\sin{2x}\cos^2{x}$
Adding and factorising....
$\displaystyle f'(x) = -3\sin{2x}\cos{x} - 2\sin{2x}\cos^2{x} = -\sin{2x}\cos{x}\,(3+2\cos{x})$