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