Применяем правило нахождения производной произведения:
[m]f`(x)=(\sqrt{x}-2)`\cdot (5-6\sqrt{x})+(\sqrt{x}-2)\cdot (5-6\sqrt{x})`=[/m]
[m]=(\frac{1}{2\sqrt{x}}-0)\cdot (5-6\sqrt{x})+(\sqrt{x}-2)\cdot (0-6\frac{1}{2\sqrt{x}})=(\frac{1}{2\sqrt{x}})\cdot (5-6\sqrt{x})+(\sqrt{x}-2)\cdot (\frac{3}{\sqrt{x}})=[/m]
[m]=\frac{5}{2\sqrt{x}}-\frac{1}{2\sqrt{x}}\cdot 6\sqrt{x}-\sqrt{x}\cdot\frac{3}{\sqrt{x}} +2\cdot \frac{6}{2\sqrt{x}}=\frac{17}{2\sqrt{x}}-3-3 - \frac{12}{2\sqrt{x}}=-6-\frac{7}{2\sqrt{x}}[/m]
2 cпособ
Упрощаем:
[m]f(x)=5\sqrt{x}-10-6x+12\sqrt{x}[/m]
[m]f(x)=17\sqrt{x}-10-6x[/m]
[m]f(x)=(17\sqrt{x}-10-6x)`[/m]
[m]f(x)=17(\sqrt{x})`-(10)`-(6x)`[/m]
[m]f`(x)=17\cdot \frac{1}{2\sqrt{x}}-0-6[/m]