Question

Найдите производную функции:
3)$\frac{1}{x^{2} }$ -$\sqrt{x}$

Answer 1

$\frac{d}{dx}(\frac{1}{x^2}-\sqrt{x})=\\\\=\frac{d}{dx}(\frac{1}{x^2})-\frac{d}{dx}(\sqrt{x})=\\\\=\frac{d}{dx}(x^{-2})-\frac{d}{dx}(x^{\frac{1}{2}})=\\\\=-2x^{-3}-\frac{1}{2}x^{-\frac{1}{2}}=\\\\=-\frac{2}{x^3}-\frac{1}{2\sqrt{x}}=\\\\=-\frac{2\cdot \:2\sqrt{x}}{x^3\cdot \:2\sqrt{x}}-\frac{1\cdot \:x^{\frac{5}{2}}\sqrt{x}}{2\sqrt{x}x^{\frac{5}{2}}\sqrt{x}}=\\\\=-\frac{4\sqrt{x}}{2x^3\sqrt{x}}-\frac{x^{\frac{5}{2}}\sqrt{x}}{2x^3\sqrt{x}}=\\\\=\frac{-4\sqrt{x}-x^{\frac{5}{2}}\sqrt{x}}{2x^3\sqrt{x}}=$

$=-\frac{\sqrt{x}\left(4+x^{\frac{5}{2}}\right)}{2x^3\sqrt{x}}=\\\\=-\frac{x^{\frac{1}{2}}\left(x^{\frac{5}{2}}+4\right)}{2x^3\sqrt{x}}=\\\\=-\frac{x^{\frac{5}{2}}+4}{2\sqrt{x}x^{-\frac{1}{2}+3}}=\\\\=-\frac{x^{\frac{5}{2}}+4}{2x^{\frac{5}{2}}\sqrt{x}}$