Let \(a\) and \(b\) be real numbers.

Let \(f\) be the linear function defined by \[f(t)=a\cdot t + b\] Then the derivative is a constant: \[f'(t)=a\]

For any real number \(t\) and \({\vartriangle}t\neq 0\), the difference quotient is \[\begin{aligned}\frac{f(t+{\vartriangle}t)-f(t)}{{\vartriangle}t} &= \frac{\bigl(a\cdot(t+{\vartriangle}t)+b\bigr)-\bigl(a\cdot t + b\bigr)}{{\vartriangle}t} \\ \\ &=\frac{a\cdot {\vartriangle}t}{{\vartriangle}t}\qquad \blue{\text{brackets expanded}}\\ \\ &=a \qquad\qquad\; \blue{\text{simplified with }{\vartriangle}t\neq0}\end{aligned}\] Because the difference quotient is equal to the constant \(a\), the derivative in any point \(t\) will also be equal to \(a\).

A special case is the constant \(f\) defined by \[f(t)= b\] Then the derivative is equal to: \[f'(t)=0\]

For each real number \(t\) and \({\vartriangle}t\neq 0\), the difference quotient is \[\frac{f(t+{\vartriangle}t)-f(t)}{{\vartriangle}t} = \frac{b-b}{{\vartriangle}t}=0\] Because the difference quotient is equal to zero, the derivative at any point \(t\) will also be equal to zero.