Выразите дельта f/ дельта x через x0 и дельта x: (Номер 3)

Просто заменяем Δf на f(x₀+Δx)-f(x₀) получаем
Δf/Δx= $\frac{f(x_0+\triangle x)-f (x_0)}{\triangle x}$
a) $\frac{(x_0+\triangle x)^2-x_0-x_0^2+x_0}{\triangle x} = \frac{x_0^2+2x_0 \triangle x+\triangle x^2-x_0^2}{\triangle x} = \frac{2x_0 \triangle x+\triangle x^2}{\triangle x}= 2x_0 +\triangle x$
б)
$\frac{(x_0+\triangle x)^3+2-x_0^3-2}{\triangle x} = \frac{x_0^3+3x_0^2 \triangle x+3x_0\triangle x^2+\triangle x^3-x_0^3}{\triangle x}=\frac{3x_0^2 \triangle x+3x_0\triangle x^2+\triangle x^3}{\triangle x}= \\ =3x_0^2 +3x_0\triangle x+\triangle x^2$
в)
$\frac{3(x_0+\triangle x)-1-3 x_0+1}{\triangle x} = \frac{3x_0+3\triangle x-3 x_0}{\triangle x} = \frac{3\triangle x}{\triangle x} = 3$
г)
$\frac{ \frac{2}{x_0+\triangle x} - \frac{2}{x_0} }{\triangle x} = 2\frac{ \frac{x_0}{x_0 (x_0+\triangle x)} - \frac{x_0+\triangle x}{x_0(x_0+\triangle x)} }{\triangle x} =2 \frac{x_0-x_0-\triangle x}{x_0 \triangle x(x_0+\triangle x)}=2 \frac{-\triangle x}{x_0 \triangle x(x_0+\triangle x)}= \\ = -\frac{2}{x_0 (x_0+\triangle x)}=-\frac{2}{x_0 ^2+x_0\triangle x}$