v1 · padrão canônico

Lesson 51 — The derivative: definition via limits

The derivative as the limit of the average rate of change. Tangent line. Differentiability implies continuity, but not conversely. Computing derivatives from the definition for elementary functions.

Used in: Year 2 of high school (ages 16–17) · Equiv. Japanese Math II (微分) · Equiv. German Klasse 11 (Analysis)

f'(a) = \lim_{h \to 0} \frac{f(a+h) - f(a)}{h}

The derivative of f at the point a: the limit of the average rate of change as the increment $h$ shrinks to zero. Geometrically, it is the slope of the tangent line to the graph of f at the point $(a,\, f(a))$ . Physically, it is the instantaneous rate of change of f at a.

Choose your door

Rigorous notation, full derivation, hypotheses

Rigorous definition and theorems

Definition of the derivative

"We say that a function $f$ is differentiable at $x = a$ whenever $f'(a)$ exists. […] The derivative measures the instantaneous rate of change of the function, as well as the slope of the tangent line to the function at the given point." — Boelkins, Active Calculus §1.3

"The derivative of a function $f(x)$ at a point $a$ in its domain, if it exists, is $f'(a) = \lim_{h \to 0}\frac{f(a+h)-f(a)}{h}$ ." — OpenStax Calculus Vol. 1, §3.1

Equivalent notations

f'(x) \;=\; \frac{df}{dx} \;=\; \frac{dy}{dx} \;=\; Df(x) \;=\; \dot{f}(x)

what this means · All notations below represent the same object — the derivative of f. Leibniz (dy/dx), Lagrange (f'), Newton (f with a dot), and the operator D are the most common.

The expression $\frac{df}{dx}\Big|_{x=a}$ denotes the derivative evaluated at the point $a$ .

From secant to tangent — geometry of the limit

The secant line (orange) passes through the points (a, f(a)) and (a+h, f(a+h)). As h → 0, the secant rotates until it coincides with the tangent line (gold). The derivative is the slope of that limiting line.

Tangent line and normal line

If $f$ is differentiable at $a$ :

Tangent line at $(a, f(a))$ : $\quad y - f(a) = f'(a)(x - a)$
Normal line at $(a, f(a))$ (perpendicular to the tangent, when $f'(a) \neq 0$ ): $\quad y - f(a) = -\dfrac{1}{f'(a)}(x - a)$

Fundamental theorem of differentiability

"If $f$ is differentiable at $a$ , then $f$ is continuous at $a$ . […] The converse is not true, and a function can be continuous but fail to be differentiable at a point." — OpenStax Calculus Vol. 1, §3.2

Points of non-differentiability

Basic derivatives via the definition

Function $f(x)$	$f'(x)$
$c$ (constant)	$0$
$x$	$1$
$x^2$	$2x$
$x^3$	$3x^2$
$x^n$ ( $n \in \mathbb{Z}$ )	$n x^{n-1}$
$\dfrac{1}{x}$	$-\dfrac{1}{x^2}$
$\sqrt{x}$	$\dfrac{1}{2\sqrt{x}}$

Worked examples

Example— 1· Derivative of $x^2$ from the definition

Problem: Use the definition of the derivative to compute $f'(x)$ for $f(x) = x^2$ .

Strategy: Apply $f'(a) = \lim_{h \to 0} \frac{f(a+h)-f(a)}{h}$ directly and simplify algebraically to cancel $h$ from the denominator before taking the limit.

Solution:

Compute $f(a+h)$ : $f(a+h) = (a+h)^2 = a^2 + 2ah + h^2$
Form the difference quotient: $\frac{f(a+h)-f(a)}{h} = \frac{a^2 + 2ah + h^2 - a^2}{h} = \frac{2ah + h^2}{h}$
Cancel $h$ (valid because $h \neq 0$ in the limit process): $\frac{2ah + h^2}{h} = \frac{h(2a + h)}{h} = 2a + h$
Take the limit: $f'(a) = \lim_{h \to 0}(2a + h) = 2a$

Since $a$ is arbitrary, the derivative function is $f'(x) = 2x$ .

Verification: At $x = 3$ , $f'(3) = 6$ . The tangent line passes through $(3, 9)$ with slope 6: $y = 6x - 9$ . Check: $f(3.1) = 9.61$ and $y(3.1) = 6(3.1) - 9 = 9.6$ — the tangent approximates the graph well for small $h$ . $\checkmark$

Source. Active Calculus, §1.3, Activity 1.3.2 — license CC-BY-SA 4.0.

Example— 2· Derivative of $1/x$ from the definition

Problem: Compute $f'(a)$ for $f(x) = \dfrac{1}{x}$ , with $a \neq 0$ , using the definition.

Strategy: The critical step is combining the fractions in the numerator into a single fraction, which allows canceling $h$ from the denominator.

Solution:

Difference quotient: $\frac{f(a+h)-f(a)}{h} = \frac{\dfrac{1}{a+h} - \dfrac{1}{a}}{h}$
Combine the fractions in the numerator (common denominator $a(a+h)$ ): $= \frac{\dfrac{a - (a+h)}{a(a+h)}}{h} = \frac{-h}{h \cdot a(a+h)}$
Cancel $h$ : $= \frac{-1}{a(a+h)}$
Take the limit $h \to 0$ : $f'(a) = \lim_{h \to 0} \frac{-1}{a(a+h)} = \frac{-1}{a \cdot a} = -\frac{1}{a^2}$

Result: $(1/x)' = -1/x^2$ . Geometrically: the hyperbola $y = 1/x$ has a tangent with negative slope for every $x \neq 0$ — the function is always decreasing on the half-planes $x > 0$ and $x < 0$ .

Verification: At $a = 2$ : $f'(2) = -1/4$ . Tangent line: $y - 1/2 = -\tfrac{1}{4}(x-2) \Rightarrow y = -x/4 + 1$ . At $x = 2.1$ : tangent gives $0.475$ ; $f(2.1) = 1/2.1 \approx 0.476$ . $\checkmark$

Source. OpenStax Calculus Volume 1, §3.1, Example 3.6 — license CC-BY-NC-SA 4.0.

Example— 3· Equation of the tangent line to $x^3$

Problem: Find the equation of the tangent line to the graph of $f(x) = x^3$ at the point $x = 2$ .

Strategy: The tangent line has equation $y - f(a) = f'(a)(x - a)$ . We need $f(a)$ and $f'(a)$ .

Solution:

Compute $f(2) = 2^3 = 8$ . The point of tangency is $(2, 8)$ .
Compute $f'(a)$ from the definition: $\frac{(a+h)^3 - a^3}{h} = \frac{a^3 + 3a^2h + 3ah^2 + h^3 - a^3}{h} = \frac{3a^2h + 3ah^2 + h^3}{h} = 3a^2 + 3ah + h^2$

$f'(a) = \lim_{h \to 0}(3a^2 + 3ah + h^2) = 3a^2$

Therefore $f'(2) = 3 \cdot 4 = 12$ .

Equation of the tangent line: $y - 8 = 12(x - 2) \implies y = 12x - 16$

Verification: At $x = 2.1$ : $f(2.1) = (2.1)^3 = 9.261$ ; tangent: $y = 12(2.1) - 16 = 9.2$ . Absolute error $0.061$ — grows with $h^2$ as expected for a linear approximation. $\checkmark$

Source. APEX Calculus, §2.1, Example 2.1.3 — license CC-BY-NC 4.0.

Example— 4· Non-differentiability: the corner of the absolute value

Problem: Investigate whether $f(x) = |x|$ is differentiable at $x = 0$ .

Strategy: Compute the one-sided derivatives $f'_-(0)$ and $f'_+(0)$ separately. If they differ, $f$ is not differentiable at $0$ .

Solution:

Right-hand derivative ( $h \to 0^+$ , so $h > 0$ , hence $|h| = h$ ): $f'_+(0) = \lim_{h \to 0^+} \frac{|0+h| - |0|}{h} = \lim_{h \to 0^+} \frac{h}{h} = 1$
Left-hand derivative ( $h \to 0^-$ , so $h < 0$ , hence $|h| = -h$ ): $f'_-(0) = \lim_{h \to 0^-} \frac{|0+h| - |0|}{h} = \lim_{h \to 0^-} \frac{-h}{h} = -1$
Conclusion: $f'_+(0) = 1 \neq -1 = f'_-(0)$ .

Since the one-sided derivatives differ, $f(x) = |x|$ is not differentiable at $x = 0$ — there is a corner in the graph. Note that $f$ is continuous at $0$ (since $\lim_{x\to 0}|x| = 0 = |0|$ ), confirming that continuity does not imply differentiability.

Verification: The graph of $|x|$ has slope $+1$ for $x > 0$ and $-1$ for $x < 0$ . There is no unique tangent line at $x = 0$ . $\checkmark$

Source. Active Calculus, §1.7, Activity 1.7.3 — license CC-BY-SA 4.0.

Example— 5· Marginal cost — economic application

Problem: A company produces $q$ units of a product with total cost $C(q) = q^2 + 30q + 500$ (in dollars). Compute the marginal cost at $q = 50$ units using the definition of the derivative and interpret the result.

Strategy: Marginal cost = $C'(q)$ , the instantaneous rate of change of cost with respect to quantity. We compute it from the definition and evaluate at $q = 50$ .

Solution:

Difference quotient: $\frac{C(q+h) - C(q)}{h} = \frac{(q+h)^2 + 30(q+h) + 500 - (q^2 + 30q + 500)}{h}$

$= \frac{q^2 + 2qh + h^2 + 30q + 30h + 500 - q^2 - 30q - 500}{h} = \frac{2qh + h^2 + 30h}{h}$

$= 2q + h + 30$

Take the limit: $C'(q) = \lim_{h \to 0}(2q + h + 30) = 2q + 30$
Evaluate at $q = 50$ : $C'(50) = 2(50) + 30 = 130 \text{ dollars}$

Interpretation: When producing 50 units, each additional unit costs approximately $130. Note that$ C(51) - C(50) = (2601 + 1530 + 500) - (2500 + 1500 + 500) = 4631 - 4500 = 131 $dollars — the marginal cost underestimates by$ 1 (a second-order difference in $h$ ).

Verification: $C'(q) = 2q + 30 > 0$ for all $q > 0$ : costs are increasing, as expected. At $q = 0$ : $C'(0) = 30$ — the marginal cost of the first unit is $30.$ \checkmark$

Source. OpenStax Calculus Volume 1, §3.1, Applied example — marginal cost — license CC-BY-NC-SA 4.0.

Exercise list

40 exercises · 10 with worked solution (25%)

Application 20Understanding 10Modeling 7Challenge 2Proof 1