turunan pertama dari f(x) = akar pangkat 3 dari sin kuadrat 3x adalah f'(x) = tolong dibantu ya kak please :)

f(x) = (sin^2 3x)^1/3
= (sin 3x)^2/3

f'(x) = (2/3) (sin 3x)^(-1/3) . cos 3x . 3
= 2 cos 3x / (sin 3x)^1/3
= 2 cos 3x / (sin 3x)^1/3 . (sin 3x)^2/3 / (sin 3x)^2/3
= 2 cos 3x . (sin 3x)^2/3 / (sin 3x)
= 2 (cos 3x)/(sin 3x) . (sin 3x)^2/3
= 2 cot 3x . (sin 3x)^2/3

^ = pangkat

[tex]\displaystyle f(x)=\sqrt[3]{\sin^23x}\\f(x)=\sin^{\frac23}3x\\\\f'(x)=\frac23\cdot3\sin^{-\frac13}3x\cos3x\\f'(x)=2\sin^{-\frac13}3x\cos3x\\f'(x)=\frac{2\cos3x}{\sqrt[3]{\sin3x}}\\f'(x)=\frac{2\cos3x}{\sqrt[3]{\sin3x}}\cdot\frac{\sqrt[3]{\sin^23x}}{\sqrt[3]{\sin^23x}}\\f'(x)=\frac{2\cos3x\sqrt[3]{\sin^23x}}{\sin3x}\\\boxed{\boxed{f'(x)=2\cot3x\sqrt[3]{\sin^23x}}}[/tex]