ПОМОГИТЕ!! [latex]f(x)=8sin ^{2} xcosx[/latex] Найти : f' (п/4) Ответ должен получиться : [latex]2 \sqrt{2} [/latex]

[latex]f'(x)=(8\sin^2x\cos x)'=8((\sin^2x)'\cos x+\sin^2x(\cos x)')=\\\\=8(2\sin x(\sin x)'\cos x+\sin^2x\bullet(-\sin x)=\\\\=16\sin x\cos^2 x-8\sin^3x=8\sin2x\cos x-8\sin^3x;\\\\\\f'(\frac{\pi}{4})=8\sin\frac{2\pi}{4}\cos \frac{\pi}{4}-8\sin^3(\frac{\pi}{4})=8\bullet \frac{\sqrt2}{2}-8(\frac{\sqrt2}{2}\bullet\frac{\sqrt2}{2}\bullet\frac{\sqrt2}{2})=\\\\=4\sqrt2-2\sqrt2 =2\sqrt2.[/latex]