Question

Срочно!! Помогите пожалуйста.

Answer 1

Ответ:

Вложение

Объяснение:

Answer 2

$1)\; \; f(x)=x^{-1}\cdot \sqrt{x^{-1}}\cdot \sqrt[3]{x^2}=\frac{1}{x}\cdot \frac{1}{\sqrt{x}}\cdot x^{2/3}=x^{-\frac{5}{6}}\\\\f'(x)=-\frac{5}{6}\cdot x^{-\frac{11}{6}}=-\frac{5}{6\sqrt[6]{x^{11}}}\\\\\\2)\; \; f(x)=\sqrt[3]{x}-\frac{2}{\sqrt{x}}+\frac{3}{x^2}-\frac{1}{5x^3}+4\\\\f'(x)=\frac{1}{3}x^{-\frac{2}{3}}-2\cdot (-\frac{1}{2})x^{-\frac{3}{2}}+3\cdot (-2)x^{-3}-\frac{1}{5}\cdot (-3)x^{-4}=\\\\=-\frac{1}{3\sqrt[3]{x^2}}+\frac{1}{\sqrt{x^3}}-\frac{6}{x^3}+\frac{3}{5x^4}$

$3)\; \; y=\frac{1}{1+\sqrt{t}}-\frac{1}{1-\sqrt{t}}\\\\y'=-\frac{\frac{1}{2\sqrt{t}}}{(1+\sqrt{t})^2}+\frac{\frac{-1}{2\sqrt{t}}}{(1-\sqrt{t})^2}=-\frac{1}{2\sqrt{t}(1+\sqrt{t})^2}-\frac{1}{2\sqrt{t}(1-\sqrt{t})^2}=-\frac{1+t}{\sqrt{t}(1-t)^2}\\\\\\4)\; \; y=(x^2+6)(x^2-3)^2\\\\y'=2x(x^2-3)^2+2(x^2-3)\cdot 2x(x^2+6)=\\\\=2x(x^2-3)\cdot (x^2-3+2x^2+12)=6x(x^2-3)(x^2+3)$

$5)\; \; y=2\sqrt{x}+\frac{3}{2\sqrt[3]{x^2}}-\frac{2}{\sqrt{x}}-\frac{1}{x}+1\\\\y'=\frac{2}{2\sqrt{x}}+\frac{3}{2}\cdot (-\frac{2}{3})x^{-\frac{5}{3}}-2\cdot (-\frac{1}{2})x^{-\frac{3}{2}}+x^{-2}=\\\\=\frac{1}{\sqrt{x}}-\frac{1}{\sqrt[3]{x^5}}+\frac{1}{\sqrt{x^3}}+\frac{1}{x^2}$