Найдите производные функций по определению

Решите задачу:

$\left(5-3x-x^2)'\right|_{x_0}\equiv\lim\limits_{h\to0}\dfrac{\left(5-3x-x^2)\right|_{x_0+h}-\left(5-3x-x^2)\right|_{x_0}}{h}=\\=\lim\limits_{h\to0}\dfrac{(5-3(x_0+h)-(x_0+h)^2)-(5-3x_0-x_0^2)}h=\\=\lim\limits_{h\to0}\dfrac{-3h-2x_0h-h^2}{h}=\lim\limits_{h\to0}(-3-2x_0-h)=-3-2x_0$

$\left.(\sqrt{3-2x})'\right|_{x_0}\equiv\lim\limits_{h\to0}\dfrac{\left.\sqrt{3-2x}\right|_{x_0+h}-\left.\sqrt{3-2x}\right|_{x_0}}{h}=\\=\lim\limits_{h\to0}\dfrac{\sqrt{3-2x_0-2h}-\sqrt{3-2x_0}}h=\lim\limits_{h\to0}\dfrac{(3-2x_0-2h)-(3-2x_0)}{h(\sqrt{3-2x_0-2h}+\sqrt{3-2x_0})}\\ =\lim\limits_{h\to0}\dfrac{-2h}{h(\sqrt{3-2x_0-2h}+\sqrt{3-2x_0})}=\\=-\lim\limits_{h\to0}\dfrac{2}{\sqrt{3-2x_0-2h}+\sqrt{3-2x_0}}=-\dfrac{2}{\sqrt{3-2x_0-0}+\sqrt{3-2x_0}}=\\=-\dfrac1{\sqrt{3-2x_0}}$