The Derivative at a Point
Textbook Reference
| Primary source | OpenStax Calculus Volume 1, Section 3.1: “Defining the Derivative” |
| Book URL | https://openstax.org/details/books/calculus-volume-1 |
Freely available and openly licensed.
Key idea
You already know that the slope of a secant line through $(a, f(a))$ and $(a+h, f(a+h))$ is \[ \frac{f(a+h) - f(a)}{h}. \] This is the average rate of change over the interval $[a, a+h]$. The derivative at $a$ is the limit of this ratio as $h \to 0$ -- the instantaneous rate of change, the slope of the tangent line.
Every derivative computation from the definition is an algebraic-limit problem of type $\frac{0}{0}$. The $h$ in the numerator, after simplification, cancels the $h$ in the denominator, and you get a finite answer. If you cannot cancel the $h$, you have made an algebra error.
Prerequisite Check
Before this lesson, make sure you can do all of the following:
Quick Reference
Definition of the derivative at $a$. \[ f'(a) = \lim_{h \to 0} \frac{f(a+h) - f(a)}{h}, \] provided this limit exists.
Equivalent form (using $x \to a$): \[ f'(a) = \lim_{x \to a} \frac{f(x) - f(a)}{x - a}. \]
Geometric meaning. $f'(a)$ is the slope of the tangent line to the graph of $f$ at the point $(a, f(a))$.
Physical meaning. If $s = f(t)$ is position, then $f'(a)$ is the instantaneous velocity at time $t = a$.
Key Concepts
1. Setting Up the Difference Quotient
To compute $f'(a)$ from the definition:
- Write $f(a+h)$ -- substitute $a+h$ for every $x$ in the formula for $f$.
- Form the difference $f(a+h) - f(a)$.
- Divide by $h$ to get the difference quotient.
- Simplify: factor or rationalize so that $h$ cancels from the denominator.
- Take $\lim_{h \to 0}$ of the simplified expression.
Example 1. Find $f'(3)$ for $f(x) = x^2$.
See It: Slide the Secant Into the Tangent
Move the slider to bring the second point toward the first on the graph of $x^2$. 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.
Step 1: $f(3+h) = (3+h)^2 = 9 + 6h + h^2$.
Step 2: $f(3+h) - f(3) = 9 + 6h + h^2 - 9 = 6h + h^2$.
Step 3: $\dfrac{f(3+h)-f(3)}{h} = \dfrac{6h + h^2}{h} = \dfrac{h(6+h)}{h} = 6 + h$.
Step 4: $\lim_{h \to 0}(6 + h) = 6$.
So $f'(3) = 6$.
Example 2. Find $f'(a)$ for $f(x) = x^2$ at a general point $a$.
$f(a+h) = a^2 + 2ah + h^2$.
$f(a+h) - f(a) = 2ah + h^2 = h(2a + h)$.
$\dfrac{f(a+h)-f(a)}{h} = 2a + h$.
$\lim_{h \to 0}(2a + h) = 2a$.
So $f'(a) = 2a$. This means the derivative at any point $a$ is $2a$: the slope of $y = x^2$ at $x = a$ is $2a$.
2. Polynomial Examples
Example 3. Find $f'(1)$ for $f(x) = x^3 - 2x$.
$f(1+h) = (1+h)^3 - 2(1+h) = 1 + 3h + 3h^2 + h^3 - 2 - 2h = -1 + h + 3h^2 + h^3$.
$f(1+h) - f(1) = (-1 + h + 3h^2 + h^3) - (1 - 2) = (-1 + h + 3h^2 + h^3) - (-1) = h + 3h^2 + h^3$.
$\dfrac{f(1+h)-f(1)}{h} = 1 + 3h + h^2$.
$\lim_{h \to 0}(1 + 3h + h^2) = 1$.
So $f'(1) = 1$.
3. Rational and Root Functions
Example 4. Find $f'(a)$ for $f(x) = \dfrac{1}{x}$.
$f(a+h) = \dfrac{1}{a+h}$.
$f(a+h) - f(a) = \dfrac{1}{a+h} - \dfrac{1}{a} = \dfrac{a - (a+h)}{a(a+h)} = \dfrac{-h}{a(a+h)}$.
$\dfrac{f(a+h)-f(a)}{h} = \dfrac{-h}{h \cdot a(a+h)} = \dfrac{-1}{a(a+h)}$.
$\lim_{h \to 0} \dfrac{-1}{a(a+h)} = \dfrac{-1}{a^2}$ (for $a \neq 0$).
So $f'(a) = -\dfrac{1}{a^2}$.
Example 5. Find $f'(a)$ for $f(x) = \sqrt{x}$ at $a > 0$.
$f(a+h) - f(a) = \sqrt{a+h} - \sqrt{a}$.
Rationalize by multiplying by $\dfrac{\sqrt{a+h}+\sqrt{a}}{\sqrt{a+h}+\sqrt{a}}$:
\[ \frac{\sqrt{a+h}-\sqrt{a}}{h} \cdot \frac{\sqrt{a+h}+\sqrt{a}}{\sqrt{a+h}+\sqrt{a}} = \frac{(a+h)-a}{h(\sqrt{a+h}+\sqrt{a})} = \frac{h}{h(\sqrt{a+h}+\sqrt{a})} = \frac{1}{\sqrt{a+h}+\sqrt{a}}. \]
$\lim_{h \to 0} \dfrac{1}{\sqrt{a+h}+\sqrt{a}} = \dfrac{1}{2\sqrt{a}}$.
So $f'(a) = \dfrac{1}{2\sqrt{a}}$ for $a > 0$.
4. The Alternate Definition
Instead of $h \to 0$, you can use $x \to a$: \[ f'(a) = \lim_{x \to a} \frac{f(x) - f(a)}{x - a}. \]
Both definitions are equivalent and give the same answer. The $x \to a$ form is sometimes cleaner when the function is given as a specific expression near $a$.
Example 6. Find $f'(4)$ for $f(x) = \sqrt{x}$ using the alternate form.
\[ f'(4) = \lim_{x \to 4} \frac{\sqrt{x} - 2}{x - 4}. \]
Write $x - 4 = (\sqrt{x}-2)(\sqrt{x}+2)$ and cancel:
\[ \lim_{x \to 4} \frac{\sqrt{x}-2}{(\sqrt{x}-2)(\sqrt{x}+2)} = \lim_{x \to 4} \frac{1}{\sqrt{x}+2} = \frac{1}{4}. \]
This matches $f'(a) = 1/(2\sqrt{a})$ at $a = 4$.
a function value that is large means the function is changing rapidly there.
This is the height-vs-slope error. The function $f(x) = 1000$ is a horizontal line -- its value is large everywhere, but its derivative is zero everywhere. The function $g(x) = 0.01x^2$ has values near zero for small $x$, but its derivative $g'(x) = 0.02x$ grows without bound. Height (the output value) and slope (the rate of change) are separate quantities. On a graph, height is how far up the curve sits; slope is how steeply it is rising or falling. A statement like “at $x = 5$, the function is 100” tells you where the graph is, not how fast it is changing. To know the rate of change, you need the derivative.
the derivative is only an approximation because the limit “never quite reaches” $h = 0$.
This is the limit-as-unreachable-barrier error. The expression $2a + h$ as $h \to 0$ has the limit exactly $2a$, not “approximately $2a$.” A limit is a precise mathematical statement about what value an expression approaches; it is not an approximation technique. Once the $h$ cancels from the difference quotient and the limit is evaluated, the result is exact. The derivative $f'(a)$ is a number, computed exactly by the limit process.
Common Errors
| Error | Example | Correction |
|---|---|---|
| Not expanding $f(a+h)$ fully | Writing $(a+h)^2 = a^2 + h^2$ | $(a+h)^2 = a^2 + 2ah + h^2$; the middle term $2ah$ is essential |
| Canceling $h$ before simplifying | Dividing $\frac{6h+h^2}{h}$ and writing $6$ without carrying through | Factor $h(6+h)$ then cancel; the limit of $6+h$ as $h \to 0$ is 6, not just 6 |
| Forgetting to take the limit | Stopping at the difference quotient and writing “$f'(a) = 2a + h$” | The derivative is the limit of the quotient, not the quotient itself |
| Confusing $f'(a)$ (a number) with $f'(x)$ (a function) | “The derivative is $2x$” when asked for $f'(3)$ | Evaluate at the specific point: $f'(3) = 2(3) = 6$ |
Leveled Practice
Level 1 -- Direct Computation
Problem 1. Find $f'(2)$ for $f(x) = 3x^2 - 1$ using the definition.
Show answer
$f(2+h) = 3(2+h)^2 - 1 = 3(4 + 4h + h^2) - 1 = 11 + 12h + 3h^2$.
$f(2+h) - f(2) = 11 + 12h + 3h^2 - 11 = 12h + 3h^2 = h(12 + 3h)$.
$\dfrac{f(2+h)-f(2)}{h} = 12 + 3h$.
$\lim_{h \to 0}(12 + 3h) = 12$.
$f'(2) = 12$.
Problem 2. Find $f'(a)$ for $f(x) = x^3$ at a general point $a$.
Show answer
$(a+h)^3 = a^3 + 3a^2h + 3ah^2 + h^3$.
$f(a+h) - f(a) = 3a^2h + 3ah^2 + h^3 = h(3a^2 + 3ah + h^2)$.
Difference quotient: $3a^2 + 3ah + h^2$.
$\lim_{h \to 0}(3a^2 + 3ah + h^2) = 3a^2$.
$f'(a) = 3a^2$.
Level 2 -- Rational and Root Functions
Problem 3. Find $f'(4)$ for $f(x) = \dfrac{1}{x-2}$.
Show answer
$f(4+h) = \dfrac{1}{4+h-2} = \dfrac{1}{2+h}$.
$f(4+h) - f(4) = \dfrac{1}{2+h} - \dfrac{1}{2} = \dfrac{2-(2+h)}{2(2+h)} = \dfrac{-h}{2(2+h)}$.
Difference quotient: $\dfrac{-1}{2(2+h)}$.
$\lim_{h \to 0} \dfrac{-1}{2(2+h)} = -\dfrac{1}{4}$.
$f'(4) = -\dfrac{1}{4}$.
Problem 4. Find $f'(9)$ for $f(x) = \sqrt{x}$.
Show answer
Using the formula derived in Example 5: $f'(a) = 1/(2\sqrt{a})$.
$f'(9) = \dfrac{1}{2\sqrt{9}} = \dfrac{1}{6}$.
(Or verify directly: $\dfrac{\sqrt{9+h}-3}{h} \cdot \dfrac{\sqrt{9+h}+3}{\sqrt{9+h}+3} = \dfrac{h}{h(\sqrt{9+h}+3)} = \dfrac{1}{\sqrt{9+h}+3} \to \dfrac{1}{6}$.)
Level 3 -- Analysis
Problem 5. Use the alternate form $f'(a) = \lim_{x \to a} \frac{f(x)-f(a)}{x-a}$ to find $f'(2)$ for $f(x) = x^3 - 8$.
Show answer
$f'(2) = \lim_{x \to 2} \dfrac{x^3 - 8}{x - 2}$.
Factor: $x^3 - 8 = (x-2)(x^2+2x+4)$.
$\lim_{x \to 2}(x^2 + 2x + 4) = 4 + 4 + 4 = 12$.
$f'(2) = 12$.
Check using $f'(x) = 3x^2$: $f'(2) = 3(4) = 12$. Consistent.
Problem 6. Show that $f(x) = |x|$ is not differentiable at $x = 0$.
Show answer
Using the alternate form: \[ \lim_{x \to 0^+} \frac{|x| - 0}{x - 0} = \lim_{x \to 0^+} \frac{x}{x} = 1. \] \[ \lim_{x \to 0^-} \frac{|x| - 0}{x - 0} = \lim_{x \to 0^-} \frac{-x}{x} = -1. \]
The left and right one-sided limits of the difference quotient are $-1$ and $1$. They disagree, so the limit does not exist. $f$ is not differentiable at $x = 0$.
(Geometrically, the graph of $|x|$ has a corner at the origin; a unique tangent line does not exist.)
Mastery Checklist
You have mastered this skill when you can do all of the following without referring to notes:
Mental Model
The derivative measures the slope of the tangent line. The tangent line at $a$ is what the secant line through $(a, f(a))$ and $(a+h, f(a+h))$ approaches as $h \to 0$. The slope of the secant is $[f(a+h)-f(a)]/h$; the slope of the tangent is the limit.
Every difference quotient starts as $\frac{0}{0}$ -- both numerator and denominator go to zero as $h \to 0$, because $f(a+h) \to f(a)$ when $f$ is continuous. The algebra you do to simplify the numerator is always in service of canceling the $h$ from the denominator. Once canceled, the limit is straightforward.
Connections
Within Calculus I (MATH161)
- Tangent lines: $f'(a)$ is the slope of the tangent to the graph at $(a, f(a))$. The tangent line equation is $y - f(a) = f'(a)(x - a)$.
- Rates of change: $f'(a)$ is the instantaneous rate of change at $a$. If $f$ is position, $f'$ is velocity; if $f$ is a cost function, $f'$ is marginal cost.
- Derivative as a function: Repeating the calculation for general $a$ produces $f'(x)$, the derivative function. Differentiation rules (power rule, product rule, chain rule) give shortcuts to avoid the definition every time.
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