Question

Решите производные......................

Answer 1

$1)\; \; f(x)= \frac{2x}{1-x^2}+\frac{1}{x}\\\\f'(x)= \frac{2(1-x^2)-2x(-2x)}{(1-x^2)^2}+\frac{-1}{x^2}= \frac{2(x^2+1)}{(1-x^2)^2}- \frac{1}{x^2} \\\\2)\; \; f(x)= \frac{2-x}{\sqrt{x}} + \frac{\sqrt{x}}{2-x}\\\\f'(x)=\frac{(2-x)'\cdot \sqrt{x}-(2-x)\cdot (\sqrt{x})'}{(\sqrt{x})^2}+\frac{(\sqrt{x})'\cdot (2-x)-\sqrt{x}\cdot (2-x)'}{(2-x)^2}=\\\\=\frac{-1\cdot \sqrt{x}-(2-x)\cdot \frac{1}{2\sqrt{x}}}{x}+\frac{\frac{1}{2\sqrt{x}}\cdot (2-x)-\sqrt{x}\cdot (-1)}{(2-x)^2} =\\\\= \frac{-\sqrt{x}-\frac{1}{\sqrt{x}}+\frac{\sqrt{x}}{2}}{x} +\frac{\frac{1}{\sqrt{x}}-\frac{\sqrt{x}}{2}+\sqrt{x}}{(2-x)^2} =(\frac{\sqrt{x}}{2}-\frac{1}{\sqrt{x}}-\sqrt{x})\cdot (\frac{1}{x}-\frac{1}{(2-x)^2})=$

$= \frac{x-2-2x}{2\sqrt{x}}\cdot \frac{x^2-5x+4}{x(2-x)^2}=\frac{-2-x}{2\sqrt{x}}\cdot \frac{x^2-5x+4}{x(2-x)^2}$