A simple derivative

Differentiation, derivatives and Taylor approximations: Tangent line

A simple derivative

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.