Preview of the Derivative
Define the derivative as the limit of a difference quotient: the instantaneous rate of change and the slope of the tangent line at a point. Compute simple derivatives from the definition for f(x) = x² and f(x) = x³.
A lesson of Pre-Calculus, in Limits: An Introduction to Calculus, about 10 minutes.
By the end of this lesson, you will be able to compute average rates of change, understand the derivative as the limit of a difference quotient (the instantaneous rate of change), compute derivatives of simple functions from the definition, and interpret derivatives as slopes of tangent lines.
The Big Idea
The derivative is the central concept of differential calculus. It measures the instantaneous rate of change of a function — the slope of the tangent line to the curve at a point. Before calculus, you could only find the slope of a line (which has constant slope). The derivative extends this to curves of any shape.
Average Rate of Change
The average rate of change of $f$ from $x = a$ to $x = b$:
$$\frac{\Delta f}{\Delta x} = \frac{f(b) - f(a)}{b - a}$$
This is the slope of the secant line connecting $(a, f(a))$ and $(b, f(b))$.
Instantaneous Rate of Change: The Derivative
As $b$ approaches $a$ (or using $h$ for the increment $h = b - a$):
$$f'(a) = \lim_{h \to 0} \frac{f(a+h) - f(a)}{h}$$
This is the derivative of $f$ at $x = a$ — the limit of average rates of change as the interval shrinks to zero.
Geometrically: $f'(a)$ is the slope of the tangent line to $y = f(x)$ at the point $(a, f(a))$.
The Difference Quotient
The expression $\dfrac{f(x+h) - f(x)}{h}$ is called the difference quotient.
Computing $f'(x)$ means evaluating the limit of this as $h \to 0$.
Example 1 — Derivative from Definition
Find $f'(x)$ for $f(x) = x^2$.
$$\frac{f(x+h)-f(x)}{h} = \frac{(x+h)^2 - x^2}{h} = \frac{x^2 + 2xh + h^2 - x^2}{h} = \frac{2xh + h^2}{h} = 2x + h$$
$$f'(x) = \lim_{h \to 0}(2x + h) = 2x$$
So $f'(x) = 2x$: the slope of the tangent to $y = x^2$ at any point $x$ is $2x$.
Example 2 — Derivative at a Specific Point
Find the slope of the tangent to $f(x) = x^3$ at $x = 2$.
$$f'(2) = \lim_{h \to 0}\frac{(2+h)^3 - 8}{h} = \lim_{h \to 0}\frac{8 + 12h + 6h^2 + h^3 - 8}{h}$$
$$= \lim_{h \to 0}(12 + 6h + h^2) = 12$$
Tangent slope = 12 at $(2, 8)$.
Example 3 — Derivative of $f(x) = 1/x$
$$\frac{f(x+h)-f(x)}{h} = \frac{\frac{1}{x+h} - \frac{1}{x}}{h} = \frac{\frac{x-(x+h)}{x(x+h)}}{h} = \frac{-1}{x(x+h)}$$