The Derivative Definition
Why Do We Need a New Limit?
You’ve learned how to compute limits. Now here’s a question: how steep is a curve at a single point?
A straight line has constant slope everywhere. But for curves like $y = x^2$, the steepness changes as you move along it. Near $x = 0$, the parabola is almost flat. Near $x = 3$, it’s quite steep. How do we capture this “instantaneous steepness” mathematically?
The answer is a special kind of limit called the derivative. It measures the slope at exactly one point by “zooming in” until the curve looks like a straight line.
Prerequisite Map
Quick Reference
| Property | Value |
|---|---|
| Section | Stewart 2.1 |
| Course | MATH161 |
| Difficulty | Intermediate |
| Time | ~20 minutes |
Key Concepts
The Difference Quotient
Before we can define the derivative, we need the difference quotient:
$$\frac{f(a+h) - f(a)}{h}$$
This measures the average rate of change of $f$ over the interval from $x = a$ to $x = a + h$.
Q = (a+h, f(a+h))
/
/ ← Secant line (slope = difference quotient)
/
─────────●─────────
/
/ P = (a, f(a))
/The Two Forms of the Derivative
The derivative of $f$ at the point $x = a$ is defined as:
Form 1 (h-form): $$\boxed{f'(a) = \lim_{h \to 0} \frac{f(a+h) - f(a)}{h}}$$
Form 2 (x-form): $$\boxed{f'(a) = \lim_{x \to a} \frac{f(x) - f(a)}{x - a}}$$
Both forms give the same answer. The h-form is often easier for computation; the x-form shows clearly that we’re taking a limit as a second point approaches our point.
Why These Forms Are Equivalent
If we let $x = a + h$, then:
- As $h \to 0$, we have $x \to a$
- The substitution gives $h = x - a$
So Form 1 transforms into Form 2 directly.
The Computation Process
To find $f'(a)$ using the definition:
Step 1: Write out the difference quotient $\frac{f(a+h) - f(a)}{h}$
Step 2: Substitute and expand $f(a+h)$
Step 3: Simplify algebraically (factor out $h$)
Step 4: Cancel the $h$ in numerator and denominator
Step 5: Evaluate the limit as $h \to 0$
Worked Example: Finding $f'(2)$ for $f(x) = x^2 - 3x$
Step 1: Set up the difference quotient: $$f'(2) = \lim_{h \to 0} \frac{f(2+h) - f(2)}{h}$$
Step 2: Compute $f(2+h)$ and $f(2)$:
- $f(2+h) = (2+h)^2 - 3(2+h) = 4 + 4h + h^2 - 6 - 3h = h^2 + h - 2$
- $f(2) = 4 - 6 = -2$
Step 3: Form the quotient: $$\frac{(h^2 + h - 2) - (-2)}{h} = \frac{h^2 + h}{h}$$
Step 4: Factor and cancel: $$\frac{h(h + 1)}{h} = h + 1$$
Step 5: Take the limit: $$f'(2) = \lim_{h \to 0}(h + 1) = 1$$
So the slope of $f(x) = x^2 - 3x$ at $x = 2$ is exactly $1$.
See It: Slide the Secant Into the Tangent
Move the slider to bring the second point toward the first on the graph of $x^2 - 3x$. The secant line rotates and its slope settles on one value. When the two points meet, the secant has become the tangent. Name the slope the secant slopes approached.
$f'(a) = f(a)$ -- the derivative value equals the function value.
This is the height-vs-slope error at the formal definition level. $f(a)$ is the HEIGHT of the graph at $x = a$. $f'(a)$ is the SLOPE of the tangent line at $x = a$. For $f(x) = x^2 - 3x$: $f(2) = 4 - 6 = -2$ (the height) and $f'(2) = 1$ (the slope, found by the limit). Both are found by evaluating something at $x = 2$, but one evaluates the function and the other evaluates the derivative. They are unrelated in general. When a problem asks for $f'(2)$, it asks for a slope, not a height.
you cannot compute the derivative at a point because the denominator is zero.
This is the limit-as-unreachable-barrier error applied to the formal definition. The expression $\frac{f(2+h) - f(2)}{h}$ has denominator $h$, which equals zero when $h = 0$. But the derivative is the LIMIT as $h \to 0$, not the value at $h = 0$. The key step is to simplify the numerator so that $h$ factors out and cancels: for $f(x) = x^2 - 3x$, the numerator $f(2+h) - f(2) = h^2 + h$ factors as $h(h+1)$. After canceling $h$, the expression is $h + 1$, whose limit as $h \to 0$ is $1$. The zero-denominator issue is resolved by simplification, not by avoiding the limit.
Practice Problems
Which of the following limits represents the derivative of $f(x) = x^3$ at $x = 2$?
(A) $\displaystyle\lim_{h \to 0} \frac{(2+h)^3 - 8}{h}$
(B) $\displaystyle\lim_{x \to 2} \frac{x^3 - 2}{x - 2}$
(C) $\displaystyle\lim_{h \to 0} \frac{h^3 - 8}{h}$
(D) $\displaystyle\lim_{x \to 0} \frac{x^3 - 8}{x - 2}$
Use the limit definition to find $f'(3)$ for $f(x) = 2x^2 + 1$.
Use the limit definition to find $f'(2)$ for $f(x) = \dfrac{4}{x}$.
Use the limit definition to find $f'(a)$ for $f(x) = x^2 - 5x + 2$, where $a$ is any number.
Then verify your answer by computing $f'(4)$ directly from your formula.
Use the limit definition to find $f'(a)$ for $f(x) = \sqrt{x}$, where $a > 0$.
Hint: You’ll need to rationalize the numerator.
Conceptual Questions (CCI-Style)
The limit $\displaystyle\lim_{h \to 0} \frac{(5+h)^4 - 625}{h}$ represents:
(A) The slope of the line through $(5, 625)$ and $(h, h^4)$
(B) The derivative of $f(x) = x^4$ at $x = 5$
(C) The derivative of $f(x) = x^4$ at $x = h$
(D) The average rate of change of $f(x) = x^4$ from $x = 5$ to $x = h$
Mastery Checklist
Mental Model
The Zoom Lens: Imagine pointing a camera at a curve and zooming in on a single point. The more you zoom, the more the curve looks like a straight line. The derivative is the slope of that line when you’ve zoomed in “infinitely.”
The difference quotient is the slope of the secant line when you’re zoomed out. The derivative is what that slope becomes as you zoom in until the two points merge into one.
Connections
Looking back:
- Limit laws from Chapter 1 make the final step (evaluating $\lim_{h \to 0}$) possible
- Algebraic factoring skills determine whether you can cancel the $h$
Looking ahead:
- This definition will be used to derive all the “shortcut” rules (power rule, product rule, etc.)
- The derivative at every point becomes the “derivative function” in Section 2.2
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| Continuity | Skills Index | Tangent Lines |
Last updated: 2026-01-22