Answer
$f'(a)=\lim \limits_{h \to 0}(-1-6a^2-6ah-2h^2)$, which equals to $-1-6a^2$.
$f'(1)=-1-6\times 1^2=-7$
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, $f(a)=-a-2a^3$
By substituting, we get:
$\frac{-(a+h)-2(a+h)^3-(-a-2a^3)}{h}=\frac{-a-h-2a^3-6a^2h-6ah^2-2h^3+a+2a^3}{h}=\frac{-h-6a^2h-6ah^2-2h^3}{h}=-1-6a^2-6ah-2h^2$
$f'(a)=\lim \limits_{h \to 0}(-1-6a^2-6ah-2h^2)$, which equals to $-1-6a^2$.
$f'(1)=-1-6\times 1^2=-7$
