What is differentiation in the context of Edexcel International A Levels Pure Mathematics 1?

Differentiation is a fundamental concept in calculus that deals with finding the derivative of a function. In the context of Edexcel International A Levels Pure Mathematics 1, it involves understanding how to calculate the rate at which a function is changing at any given point. This is crucial for solving problems related to slopes of curves and rates of change.

How do you differentiate a polynomial function in Pure Mathematics 1?

To differentiate a polynomial function in Pure Mathematics 1, apply the power rule: for any term ax^n, the derivative is anx^(n-1). For example, if you have f(x) = 3x^4 + 2x^3 - x + 7, the derivative f'(x) would be 12x^3 + 6x^2 - 1. This method is a key part of the Differentiation 1 subtopic.

What are the applications of differentiation in real-world problems?

Differentiation has numerous real-world applications, such as calculating velocity and acceleration in physics, determining the slope of a curve in economics, and optimizing functions in engineering. In the Edexcel International A Levels curriculum, students learn to apply these concepts to solve practical problems, enhancing their analytical skills.

What is the significance of the derivative in understanding the behavior of functions?

The derivative of a function provides critical information about the function's behavior. It indicates where the function is increasing or decreasing and helps identify local maxima and minima. In Pure Mathematics 1, understanding these concepts is essential for analyzing and sketching graphs, which is a key skill in the Differentiation 1 subtopic.

How does one find the derivative of trigonometric functions in Differentiation 1?

To find the derivative of trigonometric functions in Differentiation 1, use specific rules: the derivative of sin(x) is cos(x), and the derivative of cos(x) is -sin(x). These rules are part of the foundational techniques taught in the Edexcel International A Levels Pure Mathematics 1 course, enabling students to handle more complex calculus problems.

Why is "Using the power rule when 'first principles' is demanded" a common mistake in Differentiation 1?

Why it happens: Rushing. How to avoid it: 'From first principles' requires the LIMIT definition. Power rule scores zero marks. Always start with f'(x) = \lim_h 0 (f(x + h) - f(x))/h. Source: WMA11/01 examiner reports — flagged when this question appears

Why is "Differentiating $1/x^{3}$ as $1/(3x^{2})$" a common mistake in Differentiation 1?

Why it happens: Treating the fraction as a special case. How to avoid it: Rewrite first: 1/x^3 = x^-3, then power rule: -3x^-4 = -3/x^4.

Why is "Using the wrong $y$-coordinate for the tangent (e.g. the $x$-value substituted into the derivative instead of the function)" a common mistake in Differentiation 1?

Why it happens: Confusing 'gradient' with 'point'. How to avoid it: Two separate substitutions: y_0 = f(x_0) (point) and m = f'(x_0) (gradient). Don't mix.

Why is "Concluding 'point of inflection' just because $f''(x) = 0$" a common mistake in Differentiation 1?

Why it happens: Half-remembered rule. How to avoid it: f''(x_0) = 0 is INCONCLUSIVE. It could be a max, min, or inflection. Use a sign-change test on f'(x) around x_0 to confirm. E.g. for y = x^4 at x = 0: f''(0) = 0 but it's a MINIMUM.

Why is "Normal gradient = $1/f'$ instead of $-1/f'$" a common mistake in Differentiation 1?

Why it happens: Forgetting the minus. How to avoid it: Perpendicular means m_1 m_2 = -1. So m_normal = -1/m_tangent. Verify: m_tan · m_nor should equal -1. Source: WMA11/01 examiner reports

Pure Mathematics 1Edexcel International A Levels Mathematics (XMA01-YMA01)

Differentiation 1

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Differentiation 1 Study Notes — Edexcel IAL Mathematics P1 (WMA11/01, 2018 spec — 2026 onwards)

First principles, the power rule, sum rule, tangents, normals, stationary points and increasing/decreasing functions. The most heavily-weighted topic in P1.

What you’ll learn

Mapped to the Cambridge IGCSE XMA01-YMA01 syllabus (2026 onwards).

8.1 — Understand differentiation from first principles; differentiate $x^{n}$ , sums, and constant multiples.
8.2 — Use the derivative to find the gradient of a curve at a point.
8.3 — Find equations of tangents and normals to a curve.
8.4 — Use the second derivative to determine the nature of stationary points (max, min, inflection).
8.5 — Determine intervals on which a function is increasing or decreasing.

Differentiation from first principles

The limit definition. The 'why' behind the power rule.

Definition. The derivative of $f$ at $x$ is

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

Geometrically, this is the gradient of the tangent to $y = f(x)$ at $x$ . The numerator is a secant chord's $y$ -change; the denominator is the $x$ -change. As $h \to 0$ the secant approaches the tangent.

Standard exam example: differentiate $x^{2}$ .

$f'(x) = \lim_{h \to 0} \dfrac{(x + h)^{2} - x^{2}}{h} = \lim_{h \to 0} \dfrac{2xh + h^{2}}{h} = \lim_{h \to 0} (2x + h) = 2x.$

Another example: $f(x) = x^{3}$ .

$(x + h)^{3} = x^{3} + 3x^{2}h + 3xh^{2} + h^{3},$

so $f(x + h) - f(x) = 3x^{2}h + 3xh^{2} + h^{3}$ . Divide by $h$ : $3x^{2} + 3xh + h^{2}$ . Take limit: $3x^{2}$ .

Mark scheme. Always:

Write down the LIMIT DEFINITION (B1).
Substitute $f(x + h)$ correctly (M1).
Expand and simplify (M1).
Cancel $h$ (A1 — the key algebraic step).
Take the limit (A1).

When to use first principles. ONLY when the question explicitly says 'from first principles'. Otherwise, use the power rule.

Definition: limit of secant gradient as $h \to 0$ .
Step 1: substitute $f(x + h)$ .
Step 2: expand, cancel $h$ .
Step 3: take the limit (set $h = 0$ ).
Only required when explicitly asked.

Power rule and sum rule

Bring the index down, subtract one. Differentiate sums term by term.

The power rule.

$\dfrac{d}{dx}(x^{n}) = n x^{n - 1} \quad \text{for any real } n.$

This works for:

Positive integers: $\dfrac{d}{dx}(x^{5}) = 5x^{4}$ .
Negative integers: $\dfrac{d}{dx}(x^{-2}) = -2x^{-3}$ (equivalently: $\dfrac{d}{dx}(1/x^{2}) = -2/x^{3}$ ).
Fractional: $\dfrac{d}{dx}(x^{1/2}) = \tfrac{1}{2}x^{-1/2}$ (equivalently: $\dfrac{d}{dx}(\sqrt{x}) = 1/(2\sqrt{x})$ ).

Sum rule. Differentiate term by term:

$\dfrac{d}{dx}(f + g) = f' + g'.$

Constant-multiple rule. Constants slide through:

$\dfrac{d}{dx}(c f) = c f'.$

Constants vanish. $\dfrac{d}{dx}(c) = 0$ for any constant $c$ (a horizontal line has zero gradient).

Full example.

$y = 3x^{4} - 5x^{2} + 7\sqrt{x} - \dfrac{2}{x^{3}}.$

Rewrite using powers: $y = 3x^{4} - 5x^{2} + 7x^{1/2} - 2x^{-3}$ .

Differentiate:

$\dfrac{dy}{dx} = 12x^{3} - 10x + \tfrac{7}{2}x^{-1/2} + 6x^{-4} = 12x^{3} - 10x + \dfrac{7}{2\sqrt{x}} + \dfrac{6}{x^{4}}.$

Key technique: rewrite roots and reciprocals as powers. ALWAYS do this BEFORE applying the power rule.

Expression	Rewrite	Derivative
$\sqrt{x}$	$x^{1/2}$	$\tfrac{1}{2}x^{-1/2}$
$1/x$	$x^{-1}$	$-x^{-2}$
$1/x^{2}$	$x^{-2}$	$-2x^{-3}$
$\sqrt[3]{x^{2}}$	$x^{2/3}$	$\tfrac{2}{3}x^{-1/3}$

$\dfrac{d}{dx}(x^{n}) = nx^{n-1}$ for any real $n$ .
Rewrite roots and reciprocals as powers first.
Differentiate term by term.
Constants → 0; constant multiples pass through.

Tangents and normals

Tangent: same gradient as the curve. Normal: perpendicular (negative reciprocal).

Tangent at the point $(x_{0}, y_{0})$ on $y = f(x)$ .

Compute $y_{0} = f(x_{0})$ .
Compute $m = f'(x_{0})$ .
Use point-gradient: $y - y_{0} = m(x - x_{0})$ .

Example. Tangent to $y = 2x^{3} - 3x^{2} + 1$ at $x = 2$ .

$y_{0} = 2(8) - 3(4) + 1 = 5$ , so point is $(2, 5)$ .
$f'(x) = 6x^{2} - 6x$ . $f'(2) = 24 - 12 = 12$ .
Tangent: $y - 5 = 12(x - 2) \Rightarrow y = 12x - 19$ .

Normal at $(x_{0}, y_{0})$ .

Same point and same derivative.
Normal gradient is the NEGATIVE RECIPROCAL: $m_{\perp} = -1/f'(x_{0})$ .
Use point-gradient with $m_{\perp}$ .

Example. Normal to $y = \sqrt{x}$ at $(4, 2)$ .

$f'(x) = 1/(2\sqrt{x})$ . $f'(4) = 1/4$ .
Normal gradient: $-4$ .
Normal: $y - 2 = -4(x - 4) \Rightarrow y = -4x + 18$ .

Special cases.

Tangent gradient	Normal gradient
Finite, non-zero	Negative reciprocal
Zero (horizontal tangent)	Undefined (vertical normal: $x = x_{0}$ )
Undefined (vertical tangent)	Zero (horizontal normal: $y = y_{0}$ )

Edexcel tip. Always state the tangent/normal gradient EXPLICITLY before writing the equation — it's the M-mark step.

Tangent gradient = $f'(x_{0})$ .
Normal gradient = $-1/f'(x_{0})$ .
Point-gradient form for both.
Compute $y_{0}$ and $m$ separately.

Stationary points — find, then classify

$f'(x) = 0$ for the location. $f''(x)$ sign for the nature.

Stationary point. A point where $f'(x) = 0$ . At this point the tangent is horizontal.

Three types:

Local maximum: $f$ is greater here than nearby.
Local minimum: $f$ is less here than nearby.
Point of inflection (horizontal): $f$ is neither a max nor min, but the curve momentarily flattens.

Method.

Compute $f'(x)$ .
Solve $f'(x) = 0$ for the $x$ -coordinates of stationary points.
Substitute each $x$ into $f$ to get the $y$ -coordinate.
Use a second-derivative test OR a sign-change test to classify each.

Second-derivative test.

$f''(x_{0})$	Nature
$> 0$	Local minimum (curve concave up)
$< 0$	Local maximum (curve concave down)
$= 0$	INCONCLUSIVE — use sign-change test

Sign-change test. Evaluate $f'(x)$ slightly to the left and right of $x_{0}$ :

$f'$ around $x_{0}$	Nature
$-$ to $+$	Local minimum
$+$ to $-$	Local maximum
Same sign on both sides	Point of inflection

Worked example. $y = x^{3} - 6x^{2} + 9x + 2$ .

$f'(x) = 3x^{2} - 12x + 9 = 3(x - 1)(x - 3)$ .
Stationary at $x = 1$ and $x = 3$ .
$y$ -coords: $f(1) = 6$ , $f(3) = 2$ .
$f''(x) = 6x - 12$ .
$f''(1) = -6 < 0 \Rightarrow$ max at $(1, 6)$ .
$f''(3) = +6 > 0 \Rightarrow$ min at $(3, 2)$ .

Edexcel tip. Always WRITE OUT $f''$ when classifying. Don't just say 'maximum because it's the bigger $y$ -value'.

$f'(x) = 0$ locates stationary points.
Substitute to find $y$ -coordinates.
$f''(x_{0}) > 0$ → min; $< 0$ → max.
$f''(x_{0}) = 0$ → use sign-change test.

Increasing and decreasing functions

$f' > 0$ : increasing. $f' < 0$ : decreasing. Solve the inequality.

Definitions.

$f$ is strictly increasing on an interval if $f'(x) > 0$ throughout.
$f$ is strictly decreasing on an interval if $f'(x) < 0$ throughout.
$f$ is non-increasing/decreasing allows the inequality to be non-strict.

Method. Find where $f'(x)$ is positive (or negative).

Example. Find values of $x$ where $f(x) = x^{3} - 12x + 5$ is decreasing.

$f'(x) = 3x^{2} - 12$ .
Decreasing: $3x^{2} - 12 < 0 \Rightarrow x^{2} < 4 \Rightarrow -2 < x < 2$ .

Where stationary points fit. $f'(x) = 0$ at $x = \pm 2$ . The function is decreasing strictly between them, increasing outside.

Geometric picture. Increasing intervals correspond to ascending parts of the graph; decreasing to descending parts. Stationary points mark transitions.

Worked example. Determine the increasing and decreasing intervals of $y = 2x^{3} - 9x^{2} + 12x + 1$ .

$f'(x) = 6x^{2} - 18x + 12 = 6(x - 1)(x - 2)$ .
Stationary at $x = 1$ and $x = 2$ .
Sign chart of $f'$ : parabola opens up, roots at $1$ and $2$ .
$f' > 0$ for $x < 1$ OR $x > 2$ (outside roots).
$f' < 0$ for $1 < x < 2$ (between roots).
So: INCREASING on $(-\infty, 1) \cup (2, \infty)$ ; DECREASING on $(1, 2)$ .

Edexcel tip. When the question says 'find the set of values', use set or interval notation. Mark schemes are picky about strict vs non-strict — read 'increasing' as $f' > 0$ (strict) by default.

Increasing $\Leftrightarrow f' > 0$ .
Decreasing $\Leftrightarrow f' < 0$ .
Solve the inequality.
Stationary points mark transitions.

How it’s examined

Differentiation is the highest-weighted topic in WMA11/01 P1, often 12-18 marks across multiple questions. Typical content: a power-rule question with mixed surd/reciprocal terms (4-5 marks); a tangent OR normal question (5-7 marks); a stationary-points question with second-derivative classification (7-10 marks); occasionally a first-principles question (5 marks, appears every 2-3 sittings). A* candidates aim for full marks on the algebra and pay close attention to (1) rewriting roots before differentiating, (2) using $f''$ NOT inspection to classify, (3) correctly negating-and-reciprocating for normals.

Sources: Pearson Edexcel International Advanced Level Mathematics specification (2018 — current for 2026 onwards); Edexcel IAL Mathematical Formulae and Statistical Tables (2018); WMA11/01 Examiner Reports — January 2023, June 2023, January 2024, June 2024. Last reviewed 2026-05-12.

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Worked examples, formulae, definitions and the mistakes examiners flag — everything you need to push from a pass to an A*.

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Step-by-step worked examples — Differentiation 1

Step-by-step solutions to past-paper-style questions on differentiation 1, written exactly the way a tutor would explain them at the board.

1Differentiate $x^{2}$ from first principles (5 marks, P1)
Extended• Style: WMA11/01 January 2024 Q6• first principles
Question
Differentiate $f(x) = x^{2}$ from first principles.
Step-by-step solution
1. Step 1
  Apply the definition:
  $f'(x) = \lim_{h \to 0} \dfrac{f(x + h) - f(x)}{h}$
2. Step 2
  Substitute $f(x) = x^{2}$ :
  $f'(x) = \lim_{h \to 0} \dfrac{(x + h)^{2} - x^{2}}{h}$
3. Step 3
  Expand the numerator: $(x + h)^{2} - x^{2} = x^{2} + 2xh + h^{2} - x^{2} = 2xh + h^{2}$ .
4. Step 4
  Simplify:
  $\dfrac{2xh + h^{2}}{h} = 2x + h$
5. Step 5
  Take the limit as $h \to 0$ :
  $f'(x) = \lim_{h \to 0} (2x + h) = 2x$
Answer
$f'(x) = 2x$ .
Examiner tip
B1 for stating the limit definition; M1 for substituting $f(x + h)$ ; M1 for expansion and simplification; A1 for $2x + h$ ; A1 for the limit. 'From first principles' demands the LIMIT NOTATION explicitly — using the power rule short-cut earns no marks.
2Differentiate using the power rule (4 marks, P1)
Core• Style: WMA11/01 June 2024 Q5• power rule
Question
Find $\dfrac{dy}{dx}$ when $y = 3x^{4} - 5x^{2} + 7\sqrt{x} - \dfrac{2}{x^{3}}$ .
Step-by-step solution
1. Step 1
  Rewrite as powers of $x$ :
  $y = 3x^{4} - 5x^{2} + 7x^{1/2} - 2x^{-3}$
2. Step 2
  Apply $\frac{d}{dx}(x^{n}) = nx^{n-1}$ term by term:
  $\dfrac{dy}{dx} = 12x^{3} - 10x + \tfrac{7}{2}x^{-1/2} + 6x^{-4}$
3. Step 3
  Optional: rewrite in surd/fractional form: $12x^{3} - 10x + \dfrac{7}{2\sqrt{x}} + \dfrac{6}{x^{4}}$ .
Answer
$\dfrac{dy}{dx} = 12x^{3} - 10x + \dfrac{7}{2\sqrt{x}} + \dfrac{6}{x^{4}}$ .
Examiner tip
M1 for rewriting roots and reciprocals as fractional/negative powers; M1 for power-rule applied to at least three terms; A1 A1 for the correct derivative. The $-2/x^{3}$ becomes $-2x^{-3}$ , with derivative $+6x^{-4}$ — a common sign-error site.
3Equation of a tangent (5 marks, P1)
Core• Style: WMA11/01 January 2023 Q8• tangent
Question
Find the equation of the tangent to the curve $y = 2x^{3} - 3x^{2} + 1$ at the point where $x = 2$ .
Step-by-step solution
1. Step 1
  Find the $y$ -coordinate: $y = 2(8) - 3(4) + 1 = 16 - 12 + 1 = 5$ . Point is $(2, 5)$ .
2. Step 2
  Differentiate: $\dfrac{dy}{dx} = 6x^{2} - 6x$ .
3. Step 3
  Gradient at $x = 2$ : $6(4) - 6(2) = 24 - 12 = 12$ .
4. Step 4
  Use point-gradient form: $y - 5 = 12(x - 2)$ , i.e. $y = 12x - 19$ .
Answer
$y = 12x - 19$ .
Examiner tip
B1 for $y$ -coordinate; M1 for differentiating; A1 for the derivative; M1 for substituting $x = 2$ to find gradient; A1 for the tangent equation. The gradient and the $y$ -coordinate must BOTH be computed at the same $x$ -value — a common error is computing gradient at one $x$ and $y$ at another.
4Stationary points and nature (7 marks, P1)
Extended• Style: WMA11/01 June 2023 Q10• stationary points, second derivative
Question
Find the stationary points of $y = x^{3} - 6x^{2} + 9x + 2$ and determine the nature of each.
Step-by-step solution
1. Step 1
  Differentiate: $\dfrac{dy}{dx} = 3x^{2} - 12x + 9$ .
2. Step 2
  Set $\dfrac{dy}{dx} = 0$ : $3x^{2} - 12x + 9 = 0 \Rightarrow x^{2} - 4x + 3 = 0 \Rightarrow (x - 1)(x - 3) = 0$ . So $x = 1$ or $x = 3$ .
3. Step 3
  Find $y$ -coordinates: $x = 1 \Rightarrow y = 1 - 6 + 9 + 2 = 6$ ; $x = 3 \Rightarrow y = 27 - 54 + 27 + 2 = 2$ .
4. Step 4
  Second derivative: $\dfrac{d^{2}y}{dx^{2}} = 6x - 12$ .
5. Step 5
  At $x = 1$ : $6(1) - 12 = -6 < 0$ . Concave down → maximum. So $(1, 6)$ is a local max.
6. Step 6
  At $x = 3$ : $6(3) - 12 = 6 > 0$ . Concave up → minimum. So $(3, 2)$ is a local min.
Answer
Maximum at $(1, 6)$ ; minimum at $(3, 2)$ .
Examiner tip
M1 for differentiating; A1 for derivative; M1 for setting $= 0$ and solving; A1 for both $x$ -values; A1 for both $y$ -values; M1 for using $\dfrac{d^{2}y}{dx^{2}}$ to test nature; A1 for naming both correctly. Mark scheme accepts sign-change test as an alternative to second derivative.
5Increasing and decreasing intervals (5 marks, P1)
Extended• Style: WMA11/01 January 2024 Q9• increasing, decreasing
Question
Find the values of $x$ for which the function $f(x) = x^{3} - 12x + 5$ is decreasing.
Step-by-step solution
1. Step 1
  $f'(x) = 3x^{2} - 12$ .
2. Step 2
  Decreasing $\Leftrightarrow f'(x) < 0$ : $3x^{2} - 12 < 0 \Rightarrow x^{2} < 4$ .
3. Step 3
  Solve: $-2 < x < 2$ .
Answer
$-2 < x < 2$ , or $\{x : -2 < x < 2\}$ .
Examiner tip
M1 for $f'(x)$ ; A1 for the derivative; M1 for setting up the inequality $f' < 0$ ; A1 for the quadratic step; A1 for the final solution. 'Decreasing' is strictly $f' < 0$ . 'Non-increasing' would be $f' \le 0$ . Read the wording carefully.
6Equation of a normal (5 marks, P1)
Extended• Style: WMA11/01 June 2024 Q9• normal
Question
Find the equation of the normal to the curve $y = \sqrt{x}$ at the point $(4, 2)$ .
Step-by-step solution
1. Step 1
  Rewrite: $y = x^{1/2}$ . Differentiate: $\dfrac{dy}{dx} = \tfrac{1}{2}x^{-1/2} = \dfrac{1}{2\sqrt{x}}$ .
2. Step 2
  Gradient of tangent at $x = 4$ : $\dfrac{1}{2\sqrt{4}} = \dfrac{1}{4}$ .
3. Step 3
  Gradient of normal: $-1/(1/4) = -4$ .
4. Step 4
  Equation: $y - 2 = -4(x - 4) \Rightarrow y = -4x + 18$ .
Answer
$y = -4x + 18$ .
Examiner tip
M1 A1 for derivative; A1 for tangent gradient; B1 for normal gradient (negative reciprocal); A1 for the equation. The normal is PERPENDICULAR to the tangent, so its gradient is the negative reciprocal.

Key Formulae — Differentiation 1

The formulae you need to memorise for differentiation 1 on the Pearson Edexcel IAL XMA01 / YMA01 paper, with every variable defined in plain English and a note on when to use it.

Derivative from first principles
$f'(x) = \lim_{h \to 0} \dfrac{f(x + h) - f(x)}{h}$
$f(x)$
the function being differentiated
$h$
small increment in $x$
When to use
Only when 'from first principles' is explicitly demanded. NOT in the formula booklet — must be memorised.
Example
For $f(x) = x^{2}$ : $f'(x) = \lim_{h \to 0}((x + h)^{2} - x^{2})/h = 2x$ .
Power rule
$\dfrac{d}{dx}(x^{n}) = n x^{n - 1}, \text{ for any real } n$
$n$
any real exponent (positive, negative, fractional)
When to use
The workhorse of differentiation. Applies to any power of $x$ . NOT in the formula booklet.
Example
$\dfrac{d}{dx}(x^{3}) = 3x^{2}$ . $\dfrac{d}{dx}(x^{-2}) = -2x^{-3}$ . $\dfrac{d}{dx}(\sqrt{x}) = \tfrac{1}{2}x^{-1/2}$ .
Sum and constant-multiple rules
$\dfrac{d}{dx}(af + bg) = af' + bg'; \quad \dfrac{d}{dx}(c) = 0$
$a, b, c$
constants
$f, g$
functions of $x$
When to use
Differentiating polynomials term by term.
Example
$\dfrac{d}{dx}(3x^{2} - 5x + 7) = 6x - 5$ .
Tangent and normal lines
$\text{Tangent: } y - y_{0} = f'(x_{0})(x - x_{0}); \quad \text{Normal: } y - y_{0} = -\dfrac{1}{f'(x_{0})}(x - x_{0})$
$(x_0, y_0)$
point on the curve
$f'(x_0)$
gradient at that point
When to use
Tangent: line touching the curve at $(x_{0}, y_{0})$ . Normal: perpendicular to the tangent at that point.
Example
At $(2, 5)$ on $y = 2x^{3} - 3x^{2} + 1$ with $f'(2) = 12$ : tangent $y - 5 = 12(x - 2)$ .
Stationary points and second derivative test
$f'(x) = 0 \text{ at stationary points; } f''(x) > 0 \Rightarrow \min, \; f''(x) < 0 \Rightarrow \max, \; f''(x) = 0 \Rightarrow \text{test required}$
$f', f''$
first and second derivatives
When to use
Finding and classifying local maxima, minima, and (possible) points of inflection.
Example
$y = x^{3} - 6x^{2} + 9x + 2$ : stationary at $x = 1, 3$ . Max at $(1, 6)$ , min at $(3, 2)$ .

Key Definitions and Keywords — Differentiation 1

Definitions to memorise and the exact keywords mark schemes credit for differentiation 1 answers — sharpened from recent examiner reports for the 2026 Pearson Edexcel IAL XMA01 / YMA01 sitting.

Derivative
Examiner keyword
The instantaneous rate of change of a function with respect to its variable. Notations: $\dfrac{dy}{dx}$ , $f'(x)$ , $y'$ .
Gradient function
The derivative $f'(x)$ viewed as a function of $x$ . Gives the gradient of the tangent at any $x$ .
Tangent
Examiner keyword
A straight line that touches the curve at a single point and has the same gradient as the curve there.
Normal
Examiner keyword
A straight line perpendicular to the tangent at the point of contact. Gradient = $-1/f'(x_{0})$ .
Stationary point
Examiner keyword
A point where $f'(x) = 0$ . Can be a local max, local min, or point of inflection.
Increasing / decreasing function
Examiner keyword
Increasing on an interval: $f'(x) > 0$ there. Decreasing: $f'(x) < 0$ . Strictly increasing/decreasing if the inequality is strict.

Common Mistakes and Misconceptions — Differentiation 1

The traps other students keep falling into on differentiation 1 questions — taken from recent Pearson Edexcel IAL XMA01 / YMA01 examiner reports and mark schemes — and how to avoid them.

✕Using the power rule when 'first principles' is demanded
WMA11/01 examiner reports — flagged when this question appears
Why it happens
Rushing.
How to avoid it
'From first principles' requires the LIMIT definition. Power rule scores zero marks. Always start with $f'(x) = \lim_{h \to 0} (f(x + h) - f(x))/h$ .
✕Differentiating $1/x^{3}$ as $1/(3x^{2})$
Why it happens
Treating the fraction as a special case.
How to avoid it
Rewrite first: $1/x^{3} = x^{-3}$ , then power rule: $-3x^{-4} = -3/x^{4}$ .
✕Using the wrong $y$ -coordinate for the tangent (e.g. the $x$ -value substituted into the derivative instead of the function)
Why it happens
Confusing 'gradient' with 'point'.
How to avoid it
Two separate substitutions: $y_{0} = f(x_{0})$ (point) and $m = f'(x_{0})$ (gradient). Don't mix.
✕Concluding 'point of inflection' just because $f''(x) = 0$
Why it happens
Half-remembered rule.
How to avoid it
$f''(x_{0}) = 0$ is INCONCLUSIVE. It could be a max, min, or inflection. Use a sign-change test on $f'(x)$ around $x_{0}$ to confirm. E.g. for $y = x^{4}$ at $x = 0$ : $f''(0) = 0$ but it's a MINIMUM.
✕Normal gradient = $1/f'$ instead of $-1/f'$
WMA11/01 examiner reports
Why it happens
Forgetting the minus.
How to avoid it
Perpendicular means $m_{1} m_{2} = -1$ . So $m_{\text{normal}} = -1/m_{\text{tangent}}$ . Verify: $m_{\text{tan}} \cdot m_{\text{nor}}$ should equal $-1$ .

Practice questions

Exam-style questions with step-by-step worked solutions. Try one before checking the method.

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Video lesson

Short walkthrough of the concepts students most often get stuck on.

Differentiation 1 — frequently asked questions

The things students keep getting wrong in this sub-topic, answered.

Differentiation 1

Short Study Notes in the form of Slides

Detailed Study Notes

What you’ll learn

How it’s examined

1Differentiate x2x^{2}x2 from first principles (5 marks, P1)

2Differentiate using the power rule (4 marks, P1)

3Equation of a tangent (5 marks, P1)

4Stationary points and nature (7 marks, P1)

5Increasing and decreasing intervals (5 marks, P1)

6Equation of a normal (5 marks, P1)

Derivative from first principles

Power rule

Sum and constant-multiple rules

Tangent and normal lines

Stationary points and second derivative test

Derivative

Gradient function

Tangent

Normal

Stationary point

Increasing / decreasing function

✕Using the power rule when 'first principles' is demanded

✕Differentiating 1/x31/x^{3}1/x3 as 1/(3x2)1/(3x^{2})1/(3x2)

✕Using the wrong yyy-coordinate for the tangent (e.g. the xxx-value substituted into the derivative instead of the function)

✕Concluding 'point of inflection' just because f′′(x)=0f''(x) = 0f′′(x)=0

✕Normal gradient = 1/f′1/f'1/f′ instead of −1/f′-1/f'−1/f′

Practice questions

Past paper style quiz

Assess your understanding

Video lesson

Differentiation 1 — frequently asked questions

1Differentiate $x^{2}$ from first principles (5 marks, P1)

✕Differentiating $1/x^{3}$ as $1/(3x^{2})$

✕Using the wrong $y$ -coordinate for the tangent (e.g. the $x$ -value substituted into the derivative instead of the function)

✕Concluding 'point of inflection' just because $f''(x) = 0$

✕Normal gradient = $1/f'$ instead of $-1/f'$