Упр.28.26 ГДЗ Мордкович 10-11 класс (Алгебра)

1. Для заданной функции $$f(x)$$ найдите значение её производной в указанной точке:
   а) $$f(x)=x^2\sin x,\quad f’\left(\frac{\pi}{2}\right)=?$$
   б) $$f(x)=\sqrt{3}\sin x+\frac{x^2}{\pi}+x\sin\left(\frac{\pi}{6}\right),\quad f’\left(\frac{\pi}{6}\right)=?$$
   в) $$f(x)=x(1+\cos x),\quad f'(\pi)=?$$
   г) $$f(x)=\sqrt{3}\cos x-x\cos\left(\frac{\pi}{6}\right)+\frac{x^2}{\pi},\quad f’\left(\frac{\pi}{3}\right)=?$$

а) $$f(x)=x^2\sin x$$

$$f'(x)=(x^2)’\sin x+x^2(\sin x)’=2x\sin x+x^2\cos x$$

$$f’\left(\frac{\pi}{2}\right)=2\cdot \frac{\pi}{2}\cdot \sin \frac{\pi}{2}+\left(\frac{\pi}{2}\right)^2\cos \frac{\pi}{2}=\pi\cdot 1+\frac{\pi^2}{4}\cdot 0=\pi$$

Ответ: $$\pi$$

б) $$f(x)=\sqrt{3}\sin x+\frac{x^2}{\pi}+x\sin \frac{\pi}{6}$$

$$f'(x)=\sqrt{3}\cos x+\frac{2x}{\pi}+\sin \frac{\pi}{6}$$

$$f’\left(\frac{\pi}{6}\right)=\sqrt{3}\cos \frac{\pi}{6}+\frac{2\cdot \frac{\pi}{6}}{\pi}+\sin \frac{\pi}{6}=\sqrt{3}\cdot \frac{\sqrt{3}}{2}+\frac{1}{3}+\frac{1}{2}=\frac{3}{2}+\frac{1}{3}+\frac{1}{2}=\frac{7}{3}$$

Ответ: $$\frac{7}{3}$$

в) $$f(x)=x(1+\cos x)$$

$$f'(x)=(x)'(1+\cos x)+x(1+\cos x)’=1+\cos x-x\sin x$$

$$f'(\pi)=1+\cos \pi-\pi\sin \pi=1-1-\pi\cdot 0=0$$

Ответ: $$0$$

г) $$f(x)=\sqrt{3}\cos x-x\cos \frac{\pi}{6}+\frac{x^2}{\pi}$$

$$f'(x)=\sqrt{3}(\cos x)’-\cos \frac{\pi}{6}\cdot (x)’+\frac{1}{\pi}(x^2)’=-\sqrt{3}\sin x-\cos \frac{\pi}{6}+\frac{2x}{\pi}$$

$$f’\left(\frac{\pi}{3}\right)=-\sqrt{3}\sin \frac{\pi}{3}-\cos \frac{\pi}{6}+\frac{2\cdot \frac{\pi}{3}}{\pi}=-\sqrt{3}\cdot \frac{\sqrt{3}}{2}-\frac{\sqrt{3}}{2}+\frac{2}{3}$$

$$=-\frac{3}{2}-\frac{\sqrt{3}}{2}+\frac{2}{3}=\frac{-9-3\sqrt{3}+4}{6}=-\frac{5+3\sqrt{3}}{6}$$

Ответ: $$-\frac{5+3\sqrt{3}}{6}$$