Упр.41.43 ГДЗ Мордковича 10 класс профильный уровень (Алгебра)

Задача

а) $$f(x)=x^2\sin x,\quad f’\left(\frac{\pi}{2}\right)=?$$
б) $$f(x)=x(1+\cos x),\quad f'(\pi)=?$$
в) $$f(x)=\sqrt{3}\sin x+\frac{x^2}{\pi}+x\sin\frac{\pi}{6},\quad f’\left(\frac{\pi}{6}\right)=?$$
г) $$f(x)=\sqrt{3}\cos x-x\cos\frac{\pi}{6}+\frac{x^2}{\pi},\quad f’\left(\frac{\pi}{3}\right)=?$$

Подробный ответ

а) $$f(x)=x^2\sin x$$

Найдём производную:

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

Тогда

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

б) $$f(x)=x(1+\cos x)$$

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

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

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

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

Так как $$\sin\frac{\pi}{6}=\frac12,$$ то

$$f’\left(\frac{\pi}{6}\right)=\sqrt{3}\cos\frac{\pi}{6}+\frac{2\cdot\pi/6}{\pi}+\frac12=\sqrt{3}\cdot\frac{\sqrt{3}}{2}+\frac13+\frac12=\frac32+\frac13+\frac12=\frac{7}{3}.$$

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

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

Так как $$\cos\frac{\pi}{6}=\frac{\sqrt{3}}{2},$$ то

$$f’\left(\frac{\pi}{3}\right)=-\sqrt{3}\sin\frac{\pi}{3}-\frac{\sqrt{3}}{2}+\frac{2\cdot\pi/3}{\pi}=-\sqrt{3}\cdot\frac{\sqrt{3}}{2}-\frac{\sqrt{3}}{2}+\frac23=-\frac32-\frac{\sqrt{3}}{2}+\frac23=-\frac{5+3\sqrt{3}}{6}.$$