Упр.28.26 ГДЗ Мордкович 10-11 класс (Алгебра)
а) f(х) = х^2 sin х, f'(пи/2) = ?
б) f(x) = корень(3)sin x + x^2 / пи + x sin (пи/6), f'(пи/6) = ?
в) f(x) = x(1 + cos x), f'(пи) = ?
г) f(x) = корень(3)cos x — xcos пи/6 + x^2/пи, f'(пи/3) = ?
$$f(x)=x^2\sin x$$
$$f'(x)=(x^2)’\sin x+x^2(\sin x)’=2x\sin x+x^2\cos x$$
$$f’\left(\frac{\pi}{2}\right)=2\cdot \frac{\pi}{2}\cdot \sin \frac{\pi}{2}+\left(\frac{\pi}{2}\right)^2\cos \frac{\pi}{2}=\pi$$
$$f(x)=\sqrt{3}\sin x+\frac{x^2}{\pi}+x\sin \frac{\pi}{6}$$
$$f'(x)=\sqrt{3}\cos x+\frac{2x}{\pi}+\sin \frac{\pi}{6}$$
$$f’\left(\frac{\pi}{6}\right)=\sqrt{3}\cos \frac{\pi}{6}+\frac{2\cdot \frac{\pi}{6}}{\pi}+\sin \frac{\pi}{6}=\frac{3}{2}+\frac{1}{3}+\frac{1}{2}=\frac{7}{3}$$
$$f(x)=x(1+\cos x)$$
$$f'(x)=1\cdot(1+\cos x)+x(-\sin x)=1+\cos x-x\sin x$$
$$f'(\pi)=1+\cos \pi-\pi\sin \pi=1-1-0=0$$
$$f(x)=\sqrt{3}\cos x-x\cos \frac{\pi}{6}+\frac{x^2}{\pi}$$
$$f'(x)=-\sqrt{3}\sin x-\cos \frac{\pi}{6}+\frac{2x}{\pi}$$
$$f’\left(\frac{\pi}{3}\right)=-\sqrt{3}\sin \frac{\pi}{3}-\cos \frac{\pi}{6}+\frac{2\cdot \frac{\pi}{3}}{\pi}=-\frac{3}{2}-\frac{\sqrt{3}}{2}+\frac{2}{3}=-\frac{5+3\sqrt{3}}{6}$$
Ответ
а) $$\pi$$; б) $$\frac{7}{3}$$; в) $$0$$; г) $$-\frac{5+3\sqrt{3}}{6}$$.
