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Lesson 51 — Derivative: definition via limit

Derivative as the limit of the average rate of change. Tangent line. Differentiability implies continuity, but not vice-versa. Calculation by definition for elementary functions.

Used in: 2nd year of High School (16–17 years) · Equiv. Japanese Math II (微分) · Equiv. German Klasse 11 (Analysis)

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

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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 the notations below represent the same object — the derivative of f. Leibniz (dy/dx), Lagrange (f'), Newton (f with dot) and the operator D are the most commonly used.

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

From Secant to Tangent — Geometry of the Limit

The secant line (orange) passes through 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 this limit.

Tangent Line and Normal Line

Given $f$ 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, if $f'(a) \neq 0$ ): $\quad y - f(a) = -\dfrac{1}{f'(a)}(x - a)$

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

Fundamental Derivatives via Definition

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

Solved Examples

Example— 1· Derivative of $x^2$ by Definition

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

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

Solution:

Calculate $f(a+h)$ : $f(a+h) = (a+h)^2 = a^2 + 2ah + h^2$
Set up the incremental 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 a slope of 6: $y = 6x - 9$ . Check point: $f(3{,}1) = 9{,}61$ 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 — CC-BY-SA 4.0 license.

Example— 2· Derivative of $1/x$ by Definition

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

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

Solution:

Incremental quotient: $\frac{f(a+h)-f(a)}{h} = \frac{\dfrac{1}{a+h} - \dfrac{1}{a}}{h}$
Combine the fractions in the numerator (least 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 a negative slope for all $x \neq 0$ — the function is always decreasing in 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 — CC-BY-NC-SA 4.0 license.

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:

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

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

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

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 — CC-BY-NC 4.0 license.

Example— 4· Non-Differentiability: The Absolute Value Cusp

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

Strategy: We calculate the left-hand and right-hand derivatives $f'_-(0)$ and $f'_+(0)$ separately. If they differ, $f$ is not differentiable at $0$ .

Solution:

Right-hand derivative ( $h \to 0^+$ , so $h > 0$ , thus $|h| = h$ ): $f'_+(0) = \lim_{h \to 0^+} \frac{|0+h| - |0|}{h} = \lim_{h \to 0^+} \frac{h}{h} = 1$
Left-hand derivative ( $h \to 0^-$ , so $h < 0$ , thus $|h| = -h$ ): $f'_-(0) = \lim_{h \to 0^-} \frac{|0+h| - |0|}{h} = \lim_{h \to 0^-} \frac{-h}{h} = -1$
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 cusp 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 a slope of $+1$ 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 — CC-BY-SA 4.0 license.

Example— 5· Marginal Cost — Economic Application

Problem: A company produces $q$ units of a product with a total cost $C(q) = q^2 + 30q + 500$ R $. Calculate the marginal cost at$ 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 calculate it by definition and evaluate it at $q = 50$ .

Solution:

Incremental 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 $q = 50$ : $C'(50) = 2(50) + 30 = 130 \text{ R\$}$

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

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

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

Exercise list

40 exercises · 10 with worked solution (25%)

Application 20Understanding 10Modeling 7Challenge 2Proof 1

Ex. 51.1Application
Calculate $f'(3)$ for $f(x) = x^2$ using the definition of the derivative. (Ans: $6$ .)
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Ex. 51.2Application
Calculate $f'(a)$ for $f(x) = x^3$ using the definition. (Ans: $3a^2$ .)
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Ex. 51.3Application
Calculate $f'(a)$ for $f(x) = c$ (real constant) by definition. (Ans: $0$ .)
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Ex. 51.4ApplicationAnswer key
Calculate $f'(a)$ for $f(x) = mx + b$ (affine function) by definition. (Ans: $m$ .)
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Ex. 51.5Application
Calculate $f'(2)$ for $f(x) = 2x^2 + 3x$ via definition. (Ans: $11$ .)
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Ex. 51.6Application
Calculate $f'(1)$ for $f(x) = 2x^2 - 5x + 1$ via definition. (Ans: $-1$ .)
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Ex. 51.7Application
Calculate the derivative function $f'(x)$ for $f(x) = 2x^2 - x + 3$ via definition. (Ans: $4x - 1$ .)
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Ex. 51.8Application
Use the definition to calculate $f'(a)$ for $f(x) = x^4$ . (Ans: $4a^3$ .)
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Ex. 51.9ApplicationAnswer key
Calculate $f'(0)$ for $f(x) = x^2 - x$ via definition. (Ans: $-1$ .)
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Ex. 51.10ApplicationAnswer key
Calculate $f'(a)$ for $f(x) = 2x^3 - 2x$ via definition. (Ans: $6a^2 - 2$ .)
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Ex. 51.11Application
Calculate $f'(2)$ for $f(x) = 1/x$ via definition. (Ans: $-1/4$ .)
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Ex. 51.12ApplicationAnswer key
Calculate $f'(4)$ for $f(x) = \sqrt{x}$ via definition. (Ans: $1/4$ .)
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Ex. 51.13ApplicationAnswer key
Calculate $f'(1)$ for $f(x) = 1/x$ via definition and write the equation of the tangent line at $x = 1$ . (Ans: $f'(1) = -1$ ; tangent $y = -x + 2$ .)
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Ex. 51.14Application
Determine the equation of the tangent line to $y = x^2$ at the point $x = 2$ .
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Ex. 51.15ApplicationAnswer key
Determine the equation of the tangent line to $y = 1/x$ at the point $x = 1$ .
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Ex. 51.16Application
For $f(x) = x^2 - 4x$ , at what value of $x$ is the tangent line horizontal? Also determine the point on the graph. (Ans: $x = 2$ ; point $(2, -4)$ .)
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Ex. 51.17Application
Calculate $f'(9)$ for $f(x) = \sqrt{x}$ via definition. (Ans: $1/6$ .)
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Ex. 51.18Application
Calculate $f'(a)$ for $f(x) = 1/x^2$ via definition. (Ans: $-2/a^3$ .)
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Ex. 51.19Application
Equation of the tangent line to $y = x^3$ at $x = 2$ .
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Ex. 51.20ApplicationAnswer key
Determine the equation of the normal line to $y = x^2$ at the point $x = 1$ . (Ans: $y = -\frac{1}{2}x + \frac{3}{2}$ .)
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Ex. 51.21Understanding
Is the function $f(x) = |x|$ differentiable at $x = 0$ ? Justify by calculating the one-sided derivatives.
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Ex. 51.22Understanding
Is the function $f(x) = x|x|$ differentiable at $x = 0$ ? (Ans: yes, $f'(0) = 0$ .)
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Ex. 51.23Understanding
Analyze $f(x) = \sqrt[3]{x}$ at $x = 0$ . Does the limit of the incremental quotient exist? (Ans: $+\infty$ — vertical tangent.)
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Ex. 51.24UnderstandingAnswer key
Let $f(x) = \begin{cases} x^2 & x \leq 1 \\ 3x - 2 & x > 1 \end{cases}$ . Is $f$ differentiable at $x = 1$ ? Calculate the one-sided derivatives. (Ans: not differentiable; $f'_-(1) = 2 \neq 3 = f'_+(1)$ .)
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Ex. 51.25Understanding
Let $f(x) = x^2\sin(1/x)$ for $x \neq 0$ and $f(0) = 0$ . Show that $f'(0) = 0$ . (Ans: use the Squeeze Theorem — $|h\sin(1/h)| \leq |h| \to 0$ .)
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Ex. 51.26Understanding
Let $f(x) = x\sin(1/x)$ for $x \neq 0$ and $f(0) = 0$ . Is the function differentiable at $x = 0$ ?
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Ex. 51.27Understanding
Let $f(x) = \begin{cases} x^2 & x \geq 0 \\ -x^2 & x < 0 \end{cases}$ . Calculate $f'(0)$ using one-sided derivatives. (Ans: $f'(0) = 0$ .)
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Ex. 51.28Understanding
Interpret geometrically: what do $f'(a) > 0$ , $f'(a) < 0$ , and $f'(a) = 0$ mean?
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Ex. 51.29Understanding
What is the correct relationship between differentiability and continuity?
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Ex. 51.30Understanding
Explain, with a numerical example, why the central difference $\frac{f(a+h)-f(a-h)}{2h}$ is numerically more accurate than the forward difference $\frac{f(a+h)-f(a)}{h}$ .
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Ex. 51.31ModelingAnswer key
An object moves with position $s(t) = 2t^2$ meters. What is its instantaneous velocity at $t = 2$ s?
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Ex. 51.32Modeling
Position $s(t) = t^2 + 5t$ meters. Calculate the instantaneous velocity at $t = 3$ s by the definition of the derivative. (Ans: $11$ m/s.)
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Ex. 51.33Modeling
Cost $C(q) = q^2 + 30q + 500$ R $. What is the marginal cost at$ q = 50$ units?
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Ex. 51.34Modeling
Population $P(t) = 100 + 5t^2$ individuals. Calculate the growth rate at $t = 4$ years by the definition of the derivative. (Ans: $40$ individuals/year.)
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Ex. 51.35Modeling
In machine learning, the loss function is $L(\theta) = (\theta - 3)^2$ . Calculate $L'(\theta)$ via definition and find the $\theta$ that minimizes $L$ . (Ans: $L'(\theta) = 2\theta - 6$ ; minimum at $\theta = 3$ .)
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Ex. 51.36Modeling
Electric charge $q(t) = t^2 + 2t$ coulombs. The current $i(t) = q'(t)$ . Calculate $i(2)$ .
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Ex. 51.37Modeling
Volume of a sphere $V(r) = \frac{4}{3}\pi r^3$ . Calculate the rate of change of volume with respect to the radius at $r = 2$ cm. (Ans: $16\pi$ cm³/cm. Bonus: relate the result to the surface area.)
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Ex. 51.38Challenge
Determine $k$ such that $f(x) = x^2 + kx$ has a horizontal tangent line at the point $x = -3/2$ . (Ans: $k = 3$ .)
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Ex. 51.39ChallengeAnswer key
Prove that if $f$ is an even function and differentiable at $x = 0$ , then $f'(0) = 0$ . (Hint: use the definition of one-sided derivatives and the property $f(-x) = f(x)$ .)
Ex. 51.40Proof
Let $h(x) = f(x) + g(x)$ , with $f$ and $g$ differentiable at $a$ . Use the definition of the derivative to demonstrate that $h'(a) = f'(a) + g'(a)$ (sum rule).
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Sources

Active Calculus 2.0 — Boelkins · 2024 · CC-BY-SA 4.0. Chapters §1.1 (instantaneous velocity), §1.3 (derivative at a point), §1.4 (derivative as a function), §1.7 (limits, continuity, and differentiability). Primary source. Guided activities on secant→tangent, graphical interpretation, and absolute value cusps.
Calculus, Volume 1 — OpenStax · Herman, Strang et al. · CC-BY-NC-SA 4.0. Chapters §3.1 (Defining the Derivative), §3.2 (The Derivative as a Function). Extensive exercises with calculation by definition and applications in physics, economics, and biology.
APEX Calculus — Hartman et al. · 5th ed. · CC-BY-NC 4.0. Chapter §2.1 (Instantaneous Rates of Change). Formal treatment with examples of tangent and normal lines, table of fundamental derivatives via definition.