Question

Найти f ' (π/4), если f(x)=2√2sin^3x

Answer 1

$f'(x)=(2sqrt{2}sin^3x)'=2sqrt{2}*3sin^2x*(sinx)'=\=6sqrt{2}sin^2x*cosx\f'(frac{pi}{4})=6sqrt{2}(sinfrac{pi}{4})^2*(cosfrac{pi}{4})=6sqrt{2}*(frac{sqrt{2}}{2})^2*frac{sqrt{2}}{2}=\=6sqrt{2}*frac{2}{4}*frac{sqrt{2}}{2}=3$