Definition of the derivative

IB Mathematics: Analysis and Approaches HL· Topic 5: Calculus, Subtopic 5.3· 15 min read

1. Formal Limit Definition from First Principles★★☆☆☆HL only⏱ 5 min

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📘 Definition

Derivative from First Principles

Two equivalent forms:

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

The derivative of a function $f$ at a point $x$ is the limit of the average rate of change over an interval, as the length of the interval approaches zero. The derivative exists only if this limit is finite.

Example:

For $f (x) = x^{2}$ , the derivative at $x = 3$ is $lim_{h \to 0} \frac{( 3 + h ) ^{2} - 9}{h} = 6$

📐 Worked Example

Find $f^{'} (x)$ for $f (x) = 2 x^{2} + 3 x - 1$ using first principles.

1
Start with the formal limit definition:
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$f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}$
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Expand and simplify $f (x + h)$ :
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$f(x+h) = 2(x+h)^2 + 3(x+h) - 1 = 2x^2 + 4xh + 2h^2 + 3x + 3h - 1$
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Subtract $f (x)$ and factor the numerator:
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$f(x+h) - f(x) = (2x^2 + 4xh + 2h^2 + 3x + 3h - 1) - (2x^2 + 3x - 1) = h(4x + 2h + 3)$
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Cancel $h$ (valid since $h \neq = 0$ when taking the limit) then evaluate the limit:
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$f'(x) = \lim_{h \to 0} (4x + 2h + 3) = 4x + 3$

Exam tip:

If an exam question explicitly says 'differentiate from first principles', you must use the limit definition. Using standard differentiation rules will earn you zero marks.

2. Geometric Interpretation of the Derivative★★☆☆☆HL only⏱ 6 min

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The derivative $f^{'} (a)$ is exactly equal to the gradient of the tangent line to the curve $y = f (x)$ at the point $(a, f (a))$ . The expression $\frac{f ( a + h ) - f ( a )}{h}$ gives the gradient of the secant line connecting $(a, f (a))$ and $(a + h, f (a + h))$ . As $h \to 0$ , the secant line approaches the tangent line, so its gradient approaches the derivative.

📐 Worked Example

Find the equation of the tangent line to $f (x) = x^{3}$ at $x = 2$ using the definition of the derivative.

1
Calculate the derivative (gradient) at $x = 2$ :
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$f'(2) = \lim_{h \to 0} \frac{f(2+h) - f(2)}{h}$
3
Expand and simplify the expression:
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$\frac{(2+h)^3 - 8}{h} = \frac{8 + 12h + 6h^2 + h^3 - 8}{h} = 12 + 6h + h^2$
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Evaluate the limit to get the gradient:
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$f'(2) = \lim_{h \to 0} (12 + 6h + h^2) = 12$
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Use point-gradient form for the tangent line (passes through $(2, 8)$ ):
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$y - 8 = 12(x - 2) \implies y = 12x - 16$

3. Differentiability and Continuity★★★☆☆HL only⏱ 7 min

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📘 Definition

Differentiability at a point

Left-hand derivative: $f_{-}^{'} (a) = lim_{h \to 0^{-}} \frac{f ( a + h ) - f ( a )}{h}$
Right-hand derivative: $f_{+}^{'} (a) = lim_{h \to 0^{+}} \frac{f ( a + h ) - f ( a )}{h}$

A function is differentiable at $x = a$ if and only if $f_{-}^{'} (a) = f_{+}^{'} (a)$ and both are finite.

A key theorem: If a function is differentiable at a point, it must be continuous at that point. The converse is not true: a function can be continuous but not differentiable. The classic example is $f (x) = ∣ x ∣$ at $x = 0$ , which is continuous but the left and right derivatives do not match.

📐 Worked Example

Prove that the piecewise function $f (x) = {x^{2} 2 x - 1 x \leq 1 x > 1$ is differentiable at $x = 1$ .

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First confirm continuity at $x = 1$ (required for differentiability):
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Left limit: $lim_{x \to 1^{-}} x^{2} = 1$ , Right limit: $lim_{x \to 1^{+}} (2 x - 1) = 1$ , $f (1) = 1$ , so continuous.
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Calculate the left-hand derivative:
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$f'_-(1) = \lim_{h \to 0^-} \frac{(1+h)^2 - 1}{h} = \lim_{h \to 0^-} \frac{2h + h^2}{h} = 2$
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Calculate the right-hand derivative:
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$f'_+(1) = \lim_{h \to 0^+} \frac{(2(1+h) - 1) - 1}{h} = \lim_{h \to 0^+} \frac{2h}{h} = 2$
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Since $f_{-}^{'} (1) = f_{+}^{'} (1) = 2$ , the limit exists, so $f$ is differentiable at $x = 1$ .

✓ Quick check

Test your understanding:

If a function is continuous at $x = a$ , it must be differentiable at $x = a$ . True or false?
- True
- False
Reveal answer
1 —
Continuous functions can have corners, cusps, or vertical tangents where the derivative does not exist. The classic example is $f (x) = ∣ x ∣$ at $x = 0$ .

4. Common Pitfalls

Wrong move:

Evaluating the limit at $h = 0$ before canceling $h$ , leading to division by zero.

Why:

When taking the limit as $h \to 0$ , $h$ approaches 0 but is never equal to 0, so cancellation is allowed before substitution.

Correct move:

Simplify the expression fully to remove $h$ from the denominator before evaluating the limit.

Wrong move:

Claiming a piecewise function is differentiable just because it is continuous at the junction point.

Why:

Continuity is a necessary condition for differentiability, not a sufficient one. Left and right derivatives must also match.

Correct move:

Always calculate both left-hand and right-hand derivatives at junction points to confirm they are equal.

Wrong move:

Not expanding $f (x + h)$ fully before canceling $h$ , leading to an incorrect limit.

Why:

Canceling $h$ incorrectly before factoring leaves an erroneous expression for the limit.

Correct move:

Expand $f (x + h)$ fully, subtract $f (x)$ , combine like terms, then factor out $h$ before canceling.

Wrong move:

Using standard differentiation rules when asked to differentiate from first principles.

Why:

Exam questions explicitly test your understanding of the limit definition, so alternative methods earn no marks.

Correct move:

Always start from the limit definition and show all steps when asked for first principles.

5. Quick Reference Cheatsheet

Concept	Formula
Derivative at $x$	f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}
Derivative at $x = a$ (alternative)	f'(a) = \lim_{x \to a} \frac{f(x) - f(a)}{x - a}
Left-hand derivative	f'-(a) = \lim{h \to 0^-} \frac{f(a+h) - f(a)}{h}
Right-hand derivative	f'+(a) = \lim{h \to 0^+} \frac{f(a+h) - f(a)}{h}
Tangent line at $(a, f (a))$	y - f(a) = f'(a)(x - a)
Key relationship	Differentiable $⟹$ Continuous, not vice versa

When this came up on past exams

AI-estimated based on syllabus patterns — cross-check with official past papers for accuracy. Use only as revision-focus signals.

2021 · Paper 1
Differentiate quadratic from first principles
2023 · Paper 1
Check differentiability of piecewise function

What's Next

Now that you understand the definition of the derivative from first principles, you can move on to standard differentiation rules for polynomials, trigonometric, exponential, and logarithmic functions. These rules let you compute derivatives far faster than first principles, and are used in almost all subsequent calculus topics. This foundational definition is the basis for implicit differentiation, related rates, integration, and all applications of calculus to optimization, kinematics, and differential equations.