Answer
$f'(a)=\lim \limits_{h \to 0}(6a+3h+1)$, which equals to $6a+1$.
$f'(1)=6\times1+1=7$
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, $f(a)=3a^2+a$
By substituting, we get:
$\frac{3(a+h)^2+(a+h)-(3a^2+a)}{h}=\frac{3(a^2+2ah+h^2)+a+h-3a^2-a}{h}=\frac{3a^2+6ah+3h^2+a+h-3a^2-a}{h}=\frac{6ah+3h^2+h}{h}=6a+3h+1$
$f'(a)=\lim \limits_{h \to 0}(6a+3h+1)$, which equals to $6a+1$.
$f'(1)=6\times1+1=7$
