Answer
$f'(a)=\lim \limits_{h \to 0}(2a+h)$, which equals to $2a$.
$f'(1)=2\times1=2$
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, $f(a)=a^2-3$
By substituting, we get:
$\frac{(a+h)^2-3-(a^2-3)}{h}=\frac{a^2+2ah+h^2-3-a^2+3}{h}=\frac{2ah+h^2}{h}=2a+h$
$f'(a)=\lim \limits_{h \to 0}(2a+h)$, which equals to $2a$.
$f'(1)=2\times1=2$
