Question

Помогите решить 30 баллов

Answer 1

$1)\; \; f(x)= \frac{ 4\, \sqrt[3]{x+6} }{x-1}\\\\f'(x)=4\cdot \frac{\frac{1}{3}(x+6)^{-\frac{2}{3}}\cdot (x-1)-\sqrt[3]{x+6}\cdot 1}{(x-1)^2}=4\cdot \frac{\frac{x-1}{3\sqrt[3]{(x+6)^2}}-\sqrt[3]{x+6}}{(x-1)^2}=\\\\=4\cdot \frac{x-1-3(x+6)}{3\sqrt[3]{x+6}\cdot (x-1)^2} =4\cdot \frac{-2x-19}{3\, \sqrt[3]{x+6}(x-1)^2 }\\\\f'(2)=4\cdot \frac{-4-19}{3\cdot 2\cdot 1} = \frac{-92}{6} =-15 \frac{1}{3}\\\\2)\; \; fx)=ln \sqrt{tg3x}\\\\f'(x)= \frac{1}{\sqrt{tg3x}}\cdot \frac{1}{2\sqrt{tg3x}}\cdot \frac{3}{cos^23x}$

$f'(\frac{\pi}{12})= \frac{1}{\sqrt{tg\frac{\pi}{4}}}\cdot \frac{1}{2\sqrt{tg\frac{\pi}{4}}} \cdot \frac{3}{cos^2\frac{\pi}{4}}= \frac{3}{2\cdot \frac{\sqrt2}{2}} = \frac{3\sqrt2}{2}$