Question

Помогите с решением, просто решение без рисунков. ​

Answer 1

$1)\; \; f(x)=2x^7-\frac{x^3}{12}+\frac{2x}{3}-\frac{1}{x}-\frac{\sqrt3}{2}\\\\f'(x)=14x^6-\frac{1}{12}\cdot 3x^2+\frac{2}{3}+\frac{1}{x^2} \; ,\\\\f'(-1)=14-\frac{1}{4}+\frac{2}{3}+1=15\frac{5}{12}\\\\2)\; \; f(x)=\frac{6}{5\sqrt[6]{x^5}}=\frac{6}{5}\cdot x^{-\frac{5}{6}}\\\\f'(x)=\frac{6}{5}\cdot (-\frac{5}{6})\cdot x^{-\frac{11}{6}}=-\frac{1}{\sqrt[6]{x^{11}}}\\\\3)\; \; f(x)=6\, tgx-sinx\; \; ,\; \; f'(x)=6\cdot \frac{1}{cos^2x}-cosx\\\\f'(\frac{\pi }{3})=\frac{6}{1/4}-\frac{1}{2}=24-\frac{1}{2}=23,5$

$4)\; \; f(x)=0,5\, lnx-4^{x}+2e^{x}\\\\f'(x)=0,5\cdot \frac{1}{x}-4^{x}\, ln4+2e^{x}\\\\5)\; \; f(x)=\frac{log_2x}{3}-4\sqrt{x}\; \; ,\; \; f'(x)=\frac{1}{3x\cdot ln2}-\frac{2}{\sqrt{x}}$

$6)\; \; f(x)=cosx\cdot \sqrt[4]{x}\\\\f'(x)=-sinx\cdot \sqrt[4]{x}+cosx\cdot \frac{1}{4}\cdot x^{-\frac{3}{4}}=-sinx\cdot \sqrt[4]{x}+\frac{cosx}{4\sqrt[4]{x^3}}$