Find the derivative of x³ln(cos²x) I just need the working and an explanation of how you got to your answer.

Try this option:

1) the rule 1: if f(x)=u(x)*v(x), then f'(x)=u'(x)*v(x)+u(x)*v'(x), where u(x)=x³, v(x)=ln(cos²x);

2) the rule 2: if f(x)=u(v(x)), then f'(x)=u'(v(x))*v'(x), where u(v(x))=ln(cos²x), v(x)=cos²x;

3) according to the rules above:

f'(x)=(x³)'*ln(cos²x)+x³*(ln(cos²x))';

3x²ln(cos²x)-2x³tanx.

answer: -8.2562234e+13