Ответ:

[tex]\mathrm{B)} ~~ 0 \\\\ \Gamma) ~~ - \dfrac{5+ 3\sqrt{3} }{6}[/tex]

Пошаговое объяснение:

Вспомним

[tex](x^n)' = n \cdot x^{n-1} \\\\ (u \pm v)' = u 'v \pm uv'\\\\ (u\cdot v)' = u'v + uv' \\\\ (C\cdot x )' = C \\\\ (C\cdot f(x) ) ' = C \cdot f'(x) \\\\ (C-const)[/tex]

Тогда

[tex]\mathrm{B)} ~~ f(x) = x(1 + \cos x) ~~; ~~f(\pi ) = ?[/tex]

[tex]f'(x) =x' (1+ \cos x ) + x(1+\cos x)' = 1 + \cos x -x \cdot \sin x \\\\ f'(\pi ) = 1 + \cos \pi - \pi \cdot \sin \pi = 1- 1 -0 = 0[/tex]

[tex]\Gamma ) ~ f(x) = \sqrt{3} \cos x - x\cdot \cos\dfrac{\pi }{6} + \dfrac{x^2}{\pi } ~~ ; ~~ f'\left(\dfrac{\pi }{3} \right) = ?[/tex]

[tex]f'(x) = \sqrt{3} (\cos x) ' -x'\cdot \cos \dfrac{\pi }{6} + \left (\dfrac{x^2}{\pi } \right) ' = \\\\ -\sqrt{3} \sin x - \cos \dfrac{\pi }{6} + \dfrac{2x}{\pi } \\\\\\ f'\bigg(\dfrac{\pi }{3}\bigg ) = - \sqrt{3} \cdot \sin 60 - \cos 30 + \dfrac{2}{3} = \\\\\\- \dfrac{3}{2} - \dfrac{\sqrt{3} }{2 } + \dfrac{2}{3} = -\dfrac{5}{6} - \dfrac{\sqrt{3} }{2} = - \dfrac{5+ 3\sqrt{3} }{6}[/tex]