Ответ(ы) на вопрос:
Гость[latex]f'(x)=(2\sqrt{2}sin^3x)'=2\sqrt{2}*3sin^2x*(sinx)'=\\=6\sqrt{2}sin^2x*cosx\\f'(\frac{\pi}{4})=6\sqrt{2}(sin\frac{\pi}{4})^2*(cos\frac{\pi}{4})=6\sqrt{2}*(\frac{\sqrt{2}}{2})^2*\frac{\sqrt{2}}{2}=\\=6\sqrt{2}*\frac{2}{4}*\frac{\sqrt{2}}{2}=3[/latex]
Гость[latex]f(x)=2\sqrt{2}sin^3 x[/latex] [latex]f'(x)=(2\sqrt{2}sin^3 x)'=2\sqrt{2}(sin^3 x)'=2\sqrt{2}*3sin^{3-1} x*(sin x)'=6\sqrt{2}sin^2 x cos x[/latex] [latex]f'(\frac{\pi}{4})=6\sqrt{2}*(sin \frac{\pi}{4})^2*cos \frac{\pi}{4}=6\sqrt{2}*(\frac{\sqrt{2}}{2})^2*\frac{\sqrt{2}}{2}=3[/latex]
Не нашли ответ?
Похожие вопросы