Answer
$\frac{2(a+h)-(a+h)^2-(2a-a^2)}{h}=\frac{2a+2h-a^2-2ah-h^2-2a+a^2}{h}=\frac{2h-2ah-h^2}{h}=2-2a-h$
$f'(a)=\lim \limits_{h \to 0}(2-2a-h)$, which equals to $2-2a$.
Work Step by Step
The algebraic derivative of a function can be described as:
$f'(a)=\lim \limits_{h \to 0}\frac{f(a+h)-f(a)}{h}$
Here, $f(a)=2a-a^2$
By substituting, we get:
$\frac{2(a+h)-(a+h)^2-(2a-a^2)}{h}=\frac{2a+2h-a^2-2ah-h^2-2a+a^2}{h}=\frac{2h-2ah-h^2}{h}=2-2a-h$
$f'(a)=\lim \limits_{h \to 0}(2-2a-h)$, which equals to $2-2a$.
