Answer
$\frac{-(a+h)-(a+h)^2-(-a-a^2)}{h}=\frac{-a-h-a^2-2ah-h^2+a+a^2}{h}=\frac{-h-2ah-h^2}{h}=-1-2a-h$
$f'(a)=\lim \limits_{h \to 0}(-1-2a-h)$, which equals to $-1-2a$.
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, $f(a)=-a-a^2$
By substituting, we get:
$\frac{-(a+h)-(a+h)^2-(-a-a^2)}{h}=\frac{-a-h-a^2-2ah-h^2+a+a^2}{h}=\frac{-h-2ah-h^2}{h}=-1-2a-h$
$f'(a)=\lim \limits_{h \to 0}(-1-2a-h)$, which equals to $-1-2a$.
