Differentiation from First Principles

This lesson covers the formal definition of the derivative using limits, as required by the Edexcel A-Level Mathematics specification (9MA0). You need to understand the concept of differentiation from first principles and be able to apply it to polynomials, sin x, cos x, and eˣ.

The Definition of the Derivative

The derivative of a function f(x) is defined as:

f'(x) = lim[h→0] (f(x + h) - f(x)) / h

This is called differentiation from first principles. The expression (f(x + h) - f(x)) / h is the gradient of the chord between the points (x, f(x)) and (x + h, f(x + h)). As h → 0, the chord becomes the tangent, and the gradient of the chord becomes the gradient of the curve.

Term	Meaning
f(x + h) - f(x)	The change in y (the rise)
h	The change in x (the run)
(f(x + h) - f(x)) / h	The gradient of the chord (average rate of change)
lim[h→0]	Take the limit as h approaches zero
f'(x)	The gradient of the tangent (instantaneous rate of change)

First Principles for Polynomials

Worked Example 1: f(x) = x²

f'(x) = lim[h→0] ((x + h)² - x²) / h

Expand (x + h)² = x² + 2xh + h²:

= lim[h→0] (x² + 2xh + h² - x²) / h = lim[h→0] (2xh + h²) / h = lim[h→0] (2x + h) [divide numerator and denominator by h] = 2x + 0 = 2x

So d/dx(x²) = 2x. ✓

Worked Example 2: f(x) = x³

f'(x) = lim[h→0] ((x + h)³ - x³) / h

Expand (x + h)³ = x³ + 3x²h + 3xh² + h³:

= lim[h→0] (x³ + 3x²h + 3xh² + h³ - x³) / h = lim[h→0] (3x²h + 3xh² + h³) / h = lim[h→0] (3x² + 3xh + h²) = 3x² + 0 + 0 = 3x²

So d/dx(x³) = 3x². ✓

Worked Example 3: f(x) = 5x + 3

f'(x) = lim[h→0] (5(x + h) + 3 - (5x + 3)) / h = lim[h→0] (5x + 5h + 3 - 5x - 3) / h = lim[h→0] 5h / h = lim[h→0] 5 = 5

So d/dx(5x + 3) = 5. A linear function has a constant gradient. ✓

Exam Tip: In an exam, the first principles question will tell you which function to differentiate. Show every algebraic step, including the expansion, simplification, cancelling of h, and then substituting h = 0. Do not skip steps.

First Principles for sin x

To differentiate sin x from first principles, you need the limit result:

lim[h→0] sin h / h = 1 (where h is in radians)

and:

lim[h→0] (cos h - 1) / h = 0

Worked Example 4: f(x) = sin x

f'(x) = lim[h→0] (sin(x + h) - sin x) / h

Use the addition formula: sin(x + h) = sin x cos h + cos x sin h

= lim[h→0] (sin x cos h + cos x sin h - sin x) / h = lim[h→0] [sin x (cos h - 1) + cos x sin h] / h = lim[h→0] [sin x × (cos h - 1)/h + cos x × sin h / h] = sin x × 0 + cos x × 1 = cos x

So d/dx(sin x) = cos x. ✓

The Key Limit Results

Limit	Value	Condition
lim[h→0] sin h / h	1	h must be in radians
lim[h→0] (cos h - 1) / h	0	h must be in radians
lim[h→0] (eʰ - 1) / h	1	This defines the number e

Exam Tip: These limit results are given or expected knowledge at A-Level. You do not need to prove them, but you must use them correctly.

First Principles for cos x

Worked Example 5: f(x) = cos x

f'(x) = lim[h→0] (cos(x + h) - cos x) / h

Use the addition formula: cos(x + h) = cos x cos h - sin x sin h

= lim[h→0] (cos x cos h - sin x sin h - cos x) / h = lim[h→0] [cos x(cos h - 1) - sin x sin h] / h = cos x × 0 - sin x × 1 = -sin x

So d/dx(cos x) = -sin x. ✓

First Principles for eˣ

Worked Example 6: f(x) = eˣ

f'(x) = lim[h→0] (e^(x+h) - eˣ) / h

Factor out eˣ from the numerator: e^(x+h) = eˣ × eʰ

= lim[h→0] (eˣ eʰ - eˣ) / h = lim[h→0] eˣ(eʰ - 1) / h = eˣ × lim[h→0] (eʰ - 1) / h = eˣ × 1 = eˣ

So d/dx(eˣ) = eˣ. ✓

This is the defining property of e: it is the unique number for which lim[h→0] (eʰ - 1)/h = 1.

The Notation

There are several equivalent notations for the derivative:

Notation	Meaning
f'(x)	The derivative of f(x) — Lagrange notation
dy/dx	The derivative of y with respect to x — Leibniz notation
d/dx[f(x)]	Differentiate f(x) with respect to x — operator notation

All three are used interchangeably at A-Level. Leibniz notation (dy/dx) is particularly useful for the chain rule and related rates.

When First Principles Appears in Exams

First principles questions typically ask you to:

Write down the definition: f'(x) = lim[h→0] (f(x + h) - f(x)) / h
Substitute the given f(x)
Expand f(x + h)
Simplify the expression (cancel h from numerator and denominator)
Take the limit as h → 0

Worked Example 7: f(x) = 3x² + 2x

f'(x) = lim[h→0] [3(x + h)² + 2(x + h) - (3x² + 2x)] / h

Expand: 3(x² + 2xh + h²) + 2x + 2h - 3x² - 2x = 3x² + 6xh + 3h² + 2x + 2h - 3x² - 2x = 6xh + 3h² + 2h

Divide by h: 6x + 3h + 2

Take limit as h → 0: 6x + 2

Check: d/dx(3x² + 2x) = 6x + 2. ✓

Common Mistakes

Mistake	Correction
Forgetting lim[h→0] in working	Always write lim[h→0] until you actually substitute h = 0
Dividing by h before simplifying the numerator	Simplify the numerator first so h cancels cleanly
Using degrees instead of radians for trig	First principles for sin x and cos x only works in radians
Writing dy/dx = (f(x + h) - f(x)) / h (without the limit)	The derivative is the LIMIT of this expression, not the expression itself
Confusing f(x + h) with f(x) + h	f(x + h) means substitute (x + h) into f. For example, if f(x) = x², then f(x + h) = (x + h)²

Summary

The derivative is defined as f'(x) = lim[h→0] (f(x + h) - f(x)) / h.
For polynomials: expand f(x + h), subtract f(x), divide by h, then let h → 0.
For sin x: use the addition formula and the limit sin h / h → 1 to show d/dx(sin x) = cos x.
For cos x: use the addition formula and limits to show d/dx(cos x) = -sin x.
For eˣ: factor out eˣ and use the limit (eʰ - 1)/h → 1 to show d/dx(eˣ) = eˣ.
Always show the limit notation throughout your working.

A-Level Deep Dive: Differentiation from First Principles

Spec mapping

Edexcel 9MA0 specification section 9 — Differentiation, sub-strand 9.1 covers the derivative of $f(x)$ as the gradient of the tangent to the graph of $y = f(x)$ at a general point $(x, y)$ ; the gradient of the tangent as a limit; interpretation as a rate of change; differentiation from first principles for small positive integer powers of $x$ and for $\sin x$ and $\cos x$ (refer to the official specification document for exact wording). Synoptically the topic depends on section 7 (limits and continuity, the formal device behind $\lim_{h \to 0}$ ), section 5 (small-angle approximations $\sin h \approx h$ and $\cos h \approx 1 - h^2/2$ , used to differentiate $\sin x$ ), and section 2 (algebraic manipulation, especially binomial expansion of $(x+h)^n$ ). The definition of the derivative is the foundation on which every later rule — power, product, quotient, chain — is built; without it the rules are unmotivated recipes. Differentiation from first principles is examined explicitly on Paper 1 (Pure) and is one of the few topics that returns almost every cycle.

Worked example with full mark scheme

Question (8 marks): Given that $f(x) = x^3 - 2x$ , find $f'(x)$ from first principles.

Solution with mark scheme:

Step 1 — state the definition.

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

M1 — explicit statement of the limit definition. The mark is awarded for the correct quotient with the limit notation; writing $f'(x) = \frac{f(x+h) - f(x)}{h}$ without the $\lim$ symbol scores nothing because the equality is false until the limit is taken.

Step 2 — substitute $f(x+h)$ and $f(x)$ .

$f(x+h) = (x+h)^3 - 2(x+h)$ and $f(x) = x^3 - 2x$ , so:

$f'(x) = \lim_{h \to 0} \frac{(x+h)^3 - 2(x+h) - (x^3 - 2x)}{h}$

M1 — correct substitution into the definition. A common error: writing $f(x+h) = x^3 + h^3 - 2x - 2h$ , missing the cross terms in the binomial expansion.

Step 3 — expand $(x+h)^3$ .

$(x+h)^3 = x^3 + 3x^2 h + 3x h^2 + h^3$ . Substituting:

$f'(x) = \lim_{h \to 0} \frac{x^3 + 3x^2 h + 3x h^2 + h^3 - 2x - 2h - x^3 + 2x}{h}$

M1 — correct binomial expansion of $(x+h)^3$ .

Step 4 — cancel the $x^3$ and $-2x$ terms.

The $x^3$ and $-x^3$ cancel; the $-2x$ and $+2x$ cancel. What remains:

$f'(x) = \lim_{h \to 0} \frac{3x^2 h + 3x h^2 + h^3 - 2h}{h}$

A1 — correct numerator after cancellation, with no errors in sign tracking.

Step 5 — factor out $h$ and divide.

$f'(x) = \lim_{h \to 0} \frac{h(3x^2 + 3x h + h^2 - 2)}{h} = \lim_{h \to 0} (3x^2 + 3x h + h^2 - 2)$

M1 — factoring $h$ from every term and cancelling. Crucially, the cancellation $h/h = 1$ is valid because $h \neq 0$ (we are taking a limit as $h \to 0$ , never setting $h = 0$ ).

A1 — correct simplification before taking the limit.

Step 6 — take the limit.

As $h \to 0$ , $3xh \to 0$ and $h^2 \to 0$ , leaving:

$f'(x) = 3x^2 - 2$

M1 — correct evaluation of the limit, with the dependence on $h$ explicitly disappearing.

A1 — final answer $f'(x) = 3x^2 - 2$ .

Total: 8 marks (M5 A3, split as shown).

Specimen question modelled on the Edexcel 9MA0 Paper 1 format

Question (6 marks): The function $f$ is defined by $f(x) = 2x^2 + 5x$ .

(a) Use differentiation from first principles to show that $f'(x) = 4x + 5$ . (5)

(b) Hence find the gradient of the curve $y = f(x)$ at the point where $x = 3$ . (1)

Mark scheme decomposition by AO:

(a)

M1 (AO1.1a) — stating $f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}$ correctly with the limit notation.
M1 (AO1.1b) — substituting $f(x+h) = 2(x+h)^2 + 5(x+h) = 2x^2 + 4xh + 2h^2 + 5x + 5h$ correctly.
M1 (AO1.1b) — forming the difference quotient $\frac{4xh + 2h^2 + 5h}{h}$ after cancelling $2x^2$ and $5x$ .
A1 (AO1.1b) — simplifying to $4x + 2h + 5$ after dividing by $h$ .
A1 (AO2.1) — taking the limit as $h \to 0$ explicitly, justifying the $2h \to 0$ step, to obtain the printed result $4x + 5$ .

(b)

B1 (AO1.1b) — substituting $x = 3$ into $f'(x) = 4x + 5$ to obtain gradient $= 17$ .

Total: 6 marks split AO1 = 5, AO2 = 1. This mirrors the typical Edexcel pattern: the bulk of marks reward procedural fluency (AO1), with a single AO2 mark held back for explicit limit reasoning. Skipping the "as $h \to 0$ " justification — even with all the algebra correct — costs the AO2 mark.

Synoptic links

Connects to:

Section 7 — Limits and continuity: the very definition of $f'(x)$ rests on the limit $\lim_{h \to 0}$ . Continuity at $x$ is a necessary (but not sufficient) condition for differentiability at $x$ . Functions like $|x|$ are continuous at $0$ but not differentiable there because the left- and right-hand limits of the difference quotient disagree.
Section 9 — gradient of secant becoming tangent: geometrically, $\frac{f(x+h) - f(x)}{h}$ is the gradient of the secant through $(x, f(x))$ and $(x+h, f(x+h))$ . Letting $h \to 0$ slides the second point onto the first, and the secant rotates into the tangent. The derivative is the limit of secant gradients.
Section 5 — small-angle approximations: to differentiate $\sin x$ from first principles you need $\lim_{h \to 0} \frac{\sin h}{h} = 1$ and $\lim_{h \to 0} \frac{\cos h - 1}{h} = 0$ . These follow from the small-angle results $\sin h \approx h$ and $\cos h \approx 1 - h^2/2$ for small $h$ in radians.
Section 9 — chain, product, quotient rules: every standard rule is derived from first principles. The product rule $(uv)' = u'v + uv'$ falls out of expanding $\frac{u(x+h)v(x+h) - u(x)v(x)}{h}$ by adding and subtracting $u(x)v(x+h)$ . Treat the rules as theorems, not axioms.
Section 8 — integration as inverse: the Fundamental Theorem of Calculus says that if $F'(x) = f(x)$ , then $\int_a^b f(x)\,dx = F(b) - F(a)$ . Without the first-principles definition of the derivative, integration has no rigorous link to area.

Mark-scheme literacy

First-principles questions on 9MA0 split AO marks as follows:

AO	Typical share	Earned by
AO1 (knowledge / procedure)	60–70%	Stating the definition, expanding $(x+h)^n$ , cancelling and simplifying the difference quotient, applying small-angle approximations
AO2 (reasoning / interpretation)	25–35%	Justifying the cancellation $h/h = 1$ for $h \neq 0$ , justifying the limit step "as $h \to 0$ ", explaining why the $h$ -dependent terms vanish
AO3 (problem-solving)	0–10%	Rare for this topic; appears occasionally when first principles is combined with a tangent or rate-of-change application

Examiner-rewarded phrasing: "as $h$ tends to $0$ "; "in the limit"; "since $h \neq 0$ , we may divide by $h$ "; "using the small-angle approximation $\sin h \approx h$ for $h$ near $0$ "; "the secant gradient approaches the tangent gradient". Phrases that lose marks: writing "let $h = 0$ " (which produces $0/0$ before cancellation); omitting the $\lim$ symbol on every line where $h$ still appears; using " $\approx$ " loosely without invoking a small-angle result; cancelling $h$ without acknowledging $h \neq 0$ .

A specific Edexcel pattern to watch: questions phrased "use differentiation from first principles" forbid the use of the power rule. Writing " $\frac{d}{dx}(x^3) = 3x^2$ by the power rule" inside a first-principles question scores zero — the question is testing the derivation, not the result.

Grade-band model answers

3-mark question

Question: Use the definition of the derivative as a limit to find $f'(x)$ when $f(x) = 5x + 2$ .

Grade C response (~210 words):

The definition is $f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}$ .

Substituting: $f(x+h) = 5(x+h) + 2 = 5x + 5h + 2$ , so:

$\frac{f(x+h) - f(x)}{h} = \frac{5x + 5h + 2 - 5x - 2}{h} = \frac{5h}{h} = 5$

Taking the limit, $f'(x) = 5$ .

Examiner commentary: Full marks (3/3). The candidate states the definition with $\lim$ , substitutes correctly, cancels and simplifies, and takes the limit. The working is brief but every required step is present. For a linear function the limit is constant — $5$ does not depend on $h$ — which the candidate handles smoothly. A weaker answer would either omit the limit notation or skip the substitution step, jumping straight to "the gradient is the coefficient of $x$ , so it's $5$ " — which is the right answer obtained from the wrong method. Examiners explicitly check that the question's instruction "from the definition" is honoured.

Grade A response (~250 words):*

By the limit definition of the derivative:

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

Substituting $f(x+h) = 5(x+h) + 2$ and $f(x) = 5x + 2$ :

$f'(x) = \lim_{h \to 0} \frac{5(x+h) + 2 - (5x + 2)}{h} = \lim_{h \to 0} \frac{5x + 5h + 2 - 5x - 2}{h}$

The $5x$ and $+2$ terms cancel, leaving:

$f'(x) = \lim_{h \to 0} \frac{5h}{h}$

Since $h \neq 0$ throughout the limiting process, we may divide:

$f'(x) = \lim_{h \to 0} 5 = 5$

So $f'(x) = 5$ , consistent with the geometric fact that $y = 5x + 2$ is a straight line of constant gradient $5$ .

Examiner commentary: Full marks (3/3). Beyond procedural correctness the candidate explicitly justifies the cancellation $h/h = 1$ by invoking $h \neq 0$ , uses display-math notation, and closes with a geometric sanity-check linking the calculus result to the line's known gradient. The "consistent with" phrase is the kind of self-checking remark that flags examiner-aware sophistication. No mark is lost, but on a longer question the same habits earn the AO2 reasoning marks.

6-mark question

Question: Given that $f(x) = x^2 + 4x$ , use differentiation from first principles to find $f'(x)$ .

Grade B response (~290 words):

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

$f(x+h) = (x+h)^2 + 4(x+h) = x^2 + 2xh + h^2 + 4x + 4h$

$f(x+h) - f(x) = x^2 + 2xh + h^2 + 4x + 4h - x^2 - 4x = 2xh + h^2 + 4h$

Divide by $h$ : $\frac{2xh + h^2 + 4h}{h} = 2x + h + 4$

Let $h = 0$ : $f'(x) = 2x + 4$ .

Examiner commentary: Method correct, answer correct, but the candidate writes "Let $h = 0$ " at the final step rather than "as $h \to 0$ ". This is the canonical Grade B error — the maths is right but the language betrays a misunderstanding of what a limit is. Setting $h = 0$ in the original quotient $\frac{f(x+h) - f(x)}{h}$ gives $0/0$ , which is undefined. The reason the cancellation works is that we never set $h = 0$ — we cancel first, then take the limit. Examiners may award full marks here (the algebra is correct), but on a 9-mark version of this question the AO2 mark for limit reasoning would be lost. Score: 5/6 or 6/6 depending on examiner strictness.

Grade A response (~340 words):*

By the limit definition of the derivative:

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

Compute $f(x+h)$ by binomial expansion:

$f(x+h) = (x+h)^2 + 4(x+h) = x^2 + 2xh + h^2 + 4x + 4h$

The difference $f(x+h) - f(x)$ is:

$(x^2 + 2xh + h^2 + 4x + 4h) - (x^2 + 4x) = 2xh + h^2 + 4h$

Substituting into the difference quotient and factoring $h$ :

$\frac{f(x+h) - f(x)}{h} = \frac{h(2x + h + 4)}{h}$

Since $h \neq 0$ throughout the limiting process (the limit considers $h$ approaching but never equal to $0$ ), we may cancel $h$ :

$\frac{f(x+h) - f(x)}{h} = 2x + h + 4$

Taking the limit as $h \to 0$ , the $h$ term vanishes:

$f'(x) = \lim_{h \to 0} (2x + h + 4) = 2x + 4$

Examiner commentary: Full marks (6/6). The candidate uses the precise phrase "since $h \neq 0$ throughout the limiting process" — exactly the AO2 justification examiners reward. The factoring step $h(2x + h + 4)$ is shown explicitly before cancellation. The closing line states "as $h \to 0$ , the $h$ term vanishes" rather than the lazy "let $h = 0$ ". Every line keeps the $\lim$ symbol present until the limit is actually taken. This is precisely the discipline that distinguishes a Grade A* script: the mathematics has not changed but the language of limits has been used correctly throughout.

9-mark question

Question: (a) Show that $\lim_{h \to 0} \frac{\sin h}{h} = 1$ may be assumed (state but do not prove). (1) (b) Hence use differentiation from first principles to prove that $\frac{d}{dx}(\sin x) = \cos x$ . (8)

Grade A response (~410 words):*

(a) For $h$ measured in radians, the standard limit $\lim_{h \to 0} \frac{\sin h}{h} = 1$ is taken as known. We will also use $\lim_{h \to 0} \frac{\cos h - 1}{h} = 0$ , which follows.

(b) By the limit definition of the derivative:

$\frac{d}{dx}(\sin x) = \lim_{h \to 0} \frac{\sin(x+h) - \sin x}{h}$

Apply the addition formula $\sin(x+h) = \sin x \cos h + \cos x \sin h$ :

$= \lim_{h \to 0} \frac{\sin x \cos h + \cos x \sin h - \sin x}{h}$

Factor $\sin x$ from the first and third terms:

$= \lim_{h \to 0} \frac{\sin x (\cos h - 1) + \cos x \sin h}{h}$

Split the limit:

$= \sin x \cdot \lim_{h \to 0} \frac{\cos h - 1}{h} + \cos x \cdot \lim_{h \to 0} \frac{\sin h}{h}$

( $\sin x$ and $\cos x$ are constants with respect to $h$ and may be taken outside the limit.)

Using the assumed limits:

$= \sin x \cdot 0 + \cos x \cdot 1 = \cos x$

So $\frac{d}{dx}(\sin x) = \cos x$ , as required. The key intuition is that the small-angle approximations $\sin h \approx h$ and $\cos h \approx 1 - h^2/2$ make the first limit zero (the $-h^2/2$ vanishes faster than $h$ ) and the second limit $1$ (the dominant term is $h$ itself).

Examiner commentary: Full marks (9/9). (a) is a B1 for stating the limit cleanly. (b) deploys M1 for the definition, M1 for the addition formula, M1 for factoring, M1 for splitting the limit, A1 for taking constants outside, A1 for substituting the assumed limits, A1 for the final result, A1 (AO2) for the closing intuition linking small-angle approximations to the limits. The candidate's discipline of splitting the limit only after factoring is crucial — splitting prematurely (before the $\cos h - 1$ structure emerges) produces $\lim \frac{\sin x \cos h}{h}$ , which diverges. This is examination craft of the highest level.

A-Level-depth misconceptions

The errors that distinguish A from A* on first-principles questions:

Setting $h = 0$ immediately rather than taking the limit. The single most common error: candidates write "let $h = 0$ in the difference quotient" and obtain $0/0$ . The correct approach is to cancel $h$ first (justified by $h \neq 0$ during the limiting process) and then take the limit. The distinction is the conceptual heart of calculus.
Mishandling cancellation in $\frac{(x+h)^3 - x^3}{h}$ . Candidates expand $(x+h)^3 = x^3 + h^3$ (forgetting the cross terms $3x^2 h$ and $3xh^2$ ), or expand correctly but then fail to factor $h$ from every term. The full expansion is $x^3 + 3x^2 h + 3xh^2 + h^3$ ; subtracting $x^3$ leaves $3x^2 h + 3xh^2 + h^3$ , every term divisible by $h$ .
Using $\lim_{h \to 0}$ notation incorrectly. Writing $f'(x) = \frac{f(x+h) - f(x)}{h}$ without the $\lim$ is a category error — the equation is false until the limit is taken. Equally bad is dropping the $\lim$ midway through the working, then re-introducing it at the end. The $\lim$ symbol should appear on every line until the limit is actually evaluated.
Confusing $f(x+h)$ with $f(x) + h$ . $f(x+h)$ requires substituting $(x+h)$ everywhere $x$ appears in $f$ . For $f(x) = x^2$ , $f(x+h) = (x+h)^2 = x^2 + 2xh + h^2$ , not $x^2 + h$ . This is the foundational substitution error.
Treating $\sin(x+h)$ as $\sin x + \sin h$ . The addition formula $\sin(x+h) = \sin x \cos h + \cos x \sin h$ is essential for differentiating $\sin x$ from first principles. Distributing $\sin$ over addition is a classic A-Level howler that destroys the proof.
Forgetting that small-angle approximations require radians. $\sin h \approx h$ holds only when $h$ is in radians. In degrees, $\sin h \approx \frac{\pi h}{180}$ . Edexcel A-Level differentiation is always in radians; failing to flag this is a conceptual error even if the numbers come out right.
Cancelling $h$ inside a limit as if it were ordinary algebra. The cancellation $h/h = 1$ is justified by $h \neq 0$ , which is true throughout the limiting process (we approach $0$ but never reach it). Stating " $h \neq 0$ , so we cancel" earns the AO2 mark; silently cancelling does not.

Common errors and mark-loss patterns on first-principles proofs

Three patterns repeatedly cost candidates marks on Paper 1 first-principles questions. They are about discipline of notation, not technique.

Dropping the limit symbol mid-proof. Candidates state $f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}$ , do the algebra without the $\lim$ , then re-attach the limit at the end. This loses the AO2 mark because the equality $\frac{\text{something}}{h} = 2x + h + 4$ is true for all $h \neq 0$ but $f'(x)$ is the limit of that quantity. The cure: write $\lim_{h \to 0}$ on every line where $h$ still appears.
Substituting $h = 0$ rather than taking $h \to 0$ . "Let $h = 0$ " produces $0/0$ in the original quotient and is the language of someone who hasn't internalised what a limit is. The cure: rehearse the phrase "as $h \to 0$ " until it replaces "let $h = 0$ " automatically.
Failure to justify cancellation. Cancelling $h$ from numerator and denominator looks innocent but rests on $h \neq 0$ , which is the definition-side property of the limit. State "since $h \neq 0$ during the limiting process, we may divide by $h$ " once explicitly — examiners reward this with the AO2 mark.

This pattern is endemic to Paper 1 first-principles questions: candidates know the algebra, lose marks on the language of limits.

Going further — university and academic signposting

Differentiation from first principles points directly toward several undergraduate trajectories:

Real analysis (Year 1): the formal $\varepsilon$ - $\delta$ definition of a limit makes "as $h \to 0$ " precise. $\lim_{h \to 0} g(h) = L$ means: for every $\varepsilon > 0$ there exists $\delta > 0$ such that $0 < |h| < \delta$ implies $|g(h) - L| < \varepsilon$ . Differentiability is then defined as the existence of this limit, and theorems like the Mean Value Theorem and Taylor's Theorem rest on it.
Non-standard analysis (Year 3 / postgraduate): Abraham Robinson's framework rehabilitates Leibniz's infinitesimals — quantities $dx$ smaller than every positive real — as rigorous objects in a hyperreal field. The derivative becomes the standard part of the difference quotient $\frac{f(x + dx) - f(x)}{dx}$ , exactly the way physicists informally compute it.
Multivariable calculus (Year 2): the Frechet derivative generalises $f'(x)$ to functions $f : \mathbb{R}^n \to \mathbb{R}^m$ as the linear map $Df(x)$ such that $\lim_{h \to 0} \frac{\|f(x+h) - f(x) - Df(x)h\|}{\|h\|} = 0$ . The single-variable definition is the case $n = m = 1$ .
Complex analysis (Year 2/3): for $f : \mathbb{C} \to \mathbb{C}$ , the same limit definition applies but $h$ is now complex. The requirement that the limit exist regardless of how $h$ approaches $0$ in the complex plane yields the Cauchy-Riemann equations $\frac{\partial u}{\partial x} = \frac{\partial v}{\partial y}$ , $\frac{\partial u}{\partial y} = -\frac{\partial v}{\partial x}$ , which characterise holomorphic functions.

Oxbridge interview prompt: "Differentiate $f(x) = |x|$ at $x = 0$ from first principles. What goes wrong? Now define a function that is differentiable everywhere but whose derivative is not continuous."

Additional A* practice: derivative of $\sin x$ from first principles

A common A* question on 9MA0 Paper 1 is to derive $\frac{d}{dx}(\sin x) = \cos x$ from first principles using the addition formula and small-angle approximations.

Worked example: Show from first principles that $\frac{d}{dx}(\sin x) = \cos x$ .

By the limit definition:

$\frac{d}{dx}(\sin x) = \lim_{h \to 0} \frac{\sin(x+h) - \sin x}{h}$

Apply the addition formula $\sin(x+h) = \sin x \cos h + \cos x \sin h$ :

$= \lim_{h \to 0} \frac{\sin x \cos h + \cos x \sin h - \sin x}{h} = \lim_{h \to 0} \frac{\sin x (\cos h - 1) + \cos x \sin h}{h}$

Split into two limits (valid since both individual limits exist):

$= \sin x \cdot \lim_{h \to 0} \frac{\cos h - 1}{h} + \cos x \cdot \lim_{h \to 0} \frac{\sin h}{h}$

For small $h$ in radians, $\sin h \approx h$ and $\cos h \approx 1 - \frac{h^2}{2}$ . Therefore:

$\frac{\sin h}{h} \approx \frac{h}{h} = 1, \qquad \frac{\cos h - 1}{h} \approx \frac{-h^2/2}{h} = -\frac{h}{2}$

As $h \to 0$ , $\frac{\sin h}{h} \to 1$ and $\frac{\cos h - 1}{h} \to 0$ . Substituting:

$\frac{d}{dx}(\sin x) = \sin x \cdot 0 + \cos x \cdot 1 = \cos x$

Why A candidates spot this immediately:* the structure " $\sin$ of a sum" inside a limit is the trigger for the addition formula, and the resulting $\cos h - 1$ structure is the trigger for the small-angle approximation $\cos h \approx 1 - h^2/2$ (not $\cos h \approx 1$ , which gives $0/h = 0$ — correct here, but accidentally so). The same technique derives $\frac{d}{dx}(\cos x) = -\sin x$ via $\cos(x+h) = \cos x \cos h - \sin x \sin h$ .

A subtlety: small-angle approximations are only valid in radians. Stating "for small $h$ in radians" once explicitly is examiner-rewarded — it is the AO2 reasoning that distinguishes mechanical recall from genuine understanding.

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This content is aligned with the Pearson Edexcel GCE A Level Mathematics (9MA0) specification, Paper 1 — Pure Mathematics, Section 9: Differentiation. For the most accurate and up-to-date information, please refer to the official Pearson Edexcel specification document.

Visual summary

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    B --> C{"Type of f?"}
    C -->|"Polynomial"| D["Expand (x+h)ⁿ<br/>using binomial"]
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    C -->|"eˣ"| F["Factor eˣ<br/>eˣ⁺ʰ = eˣ · eʰ"]
    D --> G["Cancel f(x) terms<br/>factor h from numerator"]
    E --> H["Use small-angle limits:<br/>sin h / h → 1<br/>(cos h − 1) / h → 0"]
    F --> I["Use limit:<br/>(eʰ − 1) / h → 1"]
    G --> J["Divide by h<br/>(valid since h ≠ 0)"]
    H --> K["Take limit as h → 0"]
    I --> K
    J --> K
    K --> L["f’(x) — the derivative"]

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Differentiation from First Principles

Differentiation from First Principles

The Definition of the Derivative

First Principles for Polynomials

Worked Example 1: f(x) = x²

Worked Example 2: f(x) = x³

Worked Example 3: f(x) = 5x + 3

First Principles for sin x

Worked Example 4: f(x) = sin x

The Key Limit Results

First Principles for cos x

Worked Example 5: f(x) = cos x

First Principles for eˣ

Worked Example 6: f(x) = eˣ

The Notation

When First Principles Appears in Exams

Worked Example 7: f(x) = 3x² + 2x

Common Mistakes

Summary

A-Level Deep Dive: Differentiation from First Principles

Spec mapping

Worked example with full mark scheme

Specimen question modelled on the Edexcel 9MA0 Paper 1 format

Synoptic links

Mark-scheme literacy

Grade-band model answers

3-mark question

6-mark question

9-mark question

A-Level-depth misconceptions

Common errors and mark-loss patterns on first-principles proofs

Going further — university and academic signposting

Additional A* practice: derivative of sin⁡x\sin xsinx from first principles

Edexcel A-Level alignment footer

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Additional A* practice: derivative of $\sin x$ from first principles