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