What is the chain rule in Differentiation 2 and how is it applied?

The chain rule is a fundamental technique in Differentiation 2 used to differentiate composite functions. It states that if you have a function y = f(g(x)), the derivative dy/dx is found by multiplying the derivative of the outer function f with respect to the inner function g by the derivative of the inner function g with respect to x. In formula terms, dy/dx = f'(g(x)) * g'(x). This rule is crucial for solving complex differentiation problems in Edexcel International A Levels Pure Mathematics 2.

How do you differentiate trigonometric functions in Differentiation 2?

In Differentiation 2, differentiating trigonometric functions involves using specific derivative formulas. For example, the derivative of sin(x) is cos(x), the derivative of cos(x) is -sin(x), and the derivative of tan(x) is sec^2(x). These derivatives are essential for solving problems involving trigonometric functions in the Edexcel International A Levels Pure Mathematics 2 curriculum.

What is implicit differentiation and when is it used in Differentiation 2?

Implicit differentiation is a technique used in Differentiation 2 when dealing with equations where y is not explicitly solved in terms of x. Instead of solving for y, you differentiate both sides of the equation with respect to x, treating y as a function of x. This method is particularly useful for finding the derivative of equations that define y implicitly, a common requirement in Edexcel International A Levels Pure Mathematics 2.

Can you explain the concept of higher-order derivatives in Differentiation 2?

Higher-order derivatives refer to the derivatives of a function taken multiple times. In Differentiation 2, the first derivative represents the rate of change or slope of the function. The second derivative provides information about the concavity and acceleration of the function. Higher-order derivatives are calculated by differentiating the function repeatedly and are important for analyzing the behavior of functions in Edexcel International A Levels Pure Mathematics 2.

How is the product rule used in Differentiation 2?

The product rule is a differentiation technique used in Differentiation 2 to find the derivative of the product of two functions. If you have two functions u(x) and v(x), the derivative of their product is given by (uv)' = u'v + uv'. This rule is essential for solving problems involving the multiplication of functions, a common task in the Edexcel International A Levels Pure Mathematics 2 syllabus.

Why is "Differentiating $(3 x^{2} - 1)^{5}$ as $5 (3 x^{2} - 1)^{4}$" a common mistake in Differentiation 2?

Why it happens: Forgetting to multiply by the derivative of the inside (chain rule's 'extra' factor). How to avoid it: Always two factors: derivative-of-outside (evaluated at the inside) MULTIPLIED BY derivative-of-inside. Here: 5 (3 x^2 - 1)^4 · 6 x = 30 x (3 x^2 - 1)^4. Source: WMA12/01 examiner reports — flagged annually

Why is "Writing $\dfrac{u' v + u v'}{v^{2}}$ in the quotient rule" a common mistake in Differentiation 2?

Why it happens: Borrowing the plus sign from the product rule. How to avoid it: Quotient rule has a MINUS in the numerator. Memory aid: 'low-D-high MINUS high-D-low, square the bottom and away we go.'

Why is "Writing $\dfrac{d}{dx}(\cos x) = \sin x$ (missing minus)" a common mistake in Differentiation 2?

Why it happens: Mirror confusion with . How to avoid it: d/dx( x) = x (NO sign); d/dx( x) = - x (MINUS). Memory rule: cosine 'falls'.

Why is "Differentiating $\sin x°$ as $\cos x°$" a common mistake in Differentiation 2?

Why it happens: Not realising the derivative formulae for trig functions are valid ONLY in radians. How to avoid it: Convert degrees to radians first: x° = ( x / 180). Then d/dx[( x / 180)] = /180 ( x / 180).

Why is "Differentiating $\ln(-x)$ and reporting $\dfrac{1}{-x}$ (or worrying that the log is undefined)" a common mistake in Differentiation 2?

Why it happens: Mixing up the domain restriction with the derivative formula. How to avoid it: (-x) is defined for x < 0, and by chain rule d/dx[(-x)] = -1/-x = 1/x. The general rule for |x|: d/dx[|x|] = 1/x for all x ≠ 0.

Pure Mathematics 2Edexcel International A Levels Mathematics (XMA01-YMA01)

Differentiation 2

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Differentiation 2 Study Notes — Edexcel IAL Mathematics P2 (WMA12/01, 2018 spec — 2026 onwards)

Derivatives of $\sin x$ , $\cos x$ , $e^{x}$ , $\ln x$ . The chain rule for composite functions, the product rule and the quotient rule. The engine of P2 calculus, paired naturally with the Integration 2 topic.

What you’ll learn

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

P2 7.1 — Differentiate $\sin kx$ , $\cos kx$ , $e^{kx}$ , $\ln(kx)$ and related sums, differences and constant multiples.
P2 7.2 — Apply the chain rule to functions of the form $f(g(x))$ for differentiable $f$ and $g$ .
P2 7.3 — Apply the product rule and the quotient rule.
P2 7.4 — Apply differentiation to find equations of tangents and normals, stationary points, increasing/decreasing functions, and rates of change.

Standard derivatives

Memorise the four standard results. Trig needs radians.

Function	Derivative
$x^{n}$	$n x^{n - 1}$
$\sin x$	$\cos x$
$\cos x$	$-\sin x$
$e^{x}$	$e^{x}$
$\ln x$ ( $x > 0$ )	$1/x$
$a^{x}$	$a^{x} \ln a$

Radians. The trig derivatives $(\sin x)' = \cos x$ etc. are valid ONLY when $x$ is in radians. If a problem is in degrees, convert first: $\sin x° = \sin(\pi x/180)$ , so $(\sin x°)' = (\pi/180) \cos x°$ — but this conversion almost never appears in P2.

$a^{x}$ via $\ln$ . Since $a^{x} = e^{x \ln a}$ , the chain rule gives $(a^{x})' = e^{x \ln a} \cdot \ln a = a^{x} \ln a$ .

Constant multiples and sums. $(c f)' = c f'$ and $(f + g)' = f' + g'$ . So $\dfrac{d}{dx}[3 \sin x - 4 \cos x + 2 e^{x}] = 3 \cos x + 4 \sin x + 2 e^{x}$ .

Multiple appearances. When the SAME variable appears multiple times — e.g. $y = \sin x \cos x$ — you need product rule or trig identities, not just the standard derivatives.

$(\sin x)' = \cos x$ (no sign change).
$(\cos x)' = -\sin x$ (MINUS).
$(e^{x})' = e^{x}$ — its own derivative.
$(\ln x)' = 1/x$ .

Chain rule

Differentiate outside, leave inside. Then multiply by derivative of inside.

Formula. If $y$ is a function of $u$ , and $u$ is a function of $x$ :

$\dfrac{dy}{dx} = \dfrac{dy}{du} \cdot \dfrac{du}{dx}.$

(GIVEN in the IAL formula booklet.)

Recipe. For $y = f(g(x))$ : $y' = f'(g(x)) \cdot g'(x)$ .

Worked example. Differentiate $y = (3 x^{2} - 1)^{5}$ .

Outer: $u^{5}$ , derivative $5 u^{4}$ . Inner: $u = 3 x^{2} - 1$ , derivative $6x$ .
Chain: $y' = 5(3 x^{2} - 1)^{4} \cdot 6x = 30 x (3 x^{2} - 1)^{4}$ .

Worked example. Differentiate $y = \sin(3x + 2)$ .

Outer: $\sin u$ , derivative $\cos u$ . Inner: $u = 3x + 2$ , derivative $3$ .
Chain: $y' = 3 \cos(3x + 2)$ .

Worked example. Differentiate $y = e^{x^{2}}$ .

Outer: $e^{u}$ , derivative $e^{u}$ . Inner: $u = x^{2}$ , derivative $2x$ .
Chain: $y' = 2x \, e^{x^{2}}$ .

Worked example. Differentiate $y = \ln(3 x^{2} + 1)$ .

Outer: $\ln u$ , derivative $1/u$ . Inner: $u = 3 x^{2} + 1$ , derivative $6x$ .
Chain: $y' = \dfrac{6x}{3 x^{2} + 1}$ .

Iterating the chain rule. For triple compositions $y = f(g(h(x)))$ : $y' = f'(g(h(x))) \cdot g'(h(x)) \cdot h'(x)$ . Just keep multiplying the chain.

Differentiate OUTER (leaving inner alone) MULTIPLIED BY derivative of inner.
Compositions: apply repeatedly.
Most-forgotten step: 'derivative of inner'.

Product rule

$(uv)' = u' v + u v'$ . Differentiate first times second plus first times derivative-of-second.

Formula.

$\dfrac{d}{dx}(u v) = u' v + u v' = u \dfrac{dv}{dx} + v \dfrac{du}{dx}.$

(GIVEN in the IAL formula booklet.)

Recipe. Identify $u$ and $v$ . Differentiate each. Apply the formula. Always factor the result if a common factor is obvious — Edexcel mark schemes prefer a factored answer.

Worked example. Differentiate $f(x) = x^{2} e^{x}$ .

$u = x^{2}$ , $v = e^{x}$ . $u' = 2x$ , $v' = e^{x}$ .
$f'(x) = 2x \cdot e^{x} + x^{2} \cdot e^{x} = x e^{x}(2 + x)$ .

Worked example. Differentiate $g(x) = x \sin x$ .

$u = x$ , $v = \sin x$ . $u' = 1$ , $v' = \cos x$ .
$g'(x) = \sin x + x \cos x$ .

When to use product vs simplifying first. If the product can be expanded into a polynomial or simpler form, expand first — it's quicker. Example: $y = x(x + 2) = x^{2} + 2x$ gives $y' = 2x + 2$ , no product rule needed. But $y = x e^{x}$ has no algebraic simplification — product rule is essential.

Higher-degree products. For $(uvw)' = u' v w + u v' w + u v w'$ — sum of three terms, each with one factor differentiated. P2 questions rarely go beyond two factors.

$(uv)' = u' v + u v'$ — PLUS sign in the middle.
Differentiate ONE at a time, keep the other.
Factor the result for cleanliness.

Quotient rule

$(u/v)' = (u' v - u v')/v^{2}$ . MINUS sign, square the bottom.

Formula.

$\dfrac{d}{dx}\!\left(\dfrac{u}{v}\right) = \dfrac{u' v - u v'}{v^{2}}.$

(GIVEN in the IAL formula booklet.)

Memory aid. "Low-D-high MINUS high-D-low, all over the SQUARE of below."

LOW = $v$ , HIGH = $u$ .
D-HIGH = $u'$ , D-LOW = $v'$ .
So: $v \cdot u' - u \cdot v'$ , all divided by $v^{2}$ .

Worked example. Differentiate $y = \dfrac{\sin x}{x^{2}}$ .

$u = \sin x$ , $v = x^{2}$ . $u' = \cos x$ , $v' = 2x$ .
$y' = \dfrac{\cos x \cdot x^{2} - \sin x \cdot 2x}{(x^{2})^{2}} = \dfrac{x \cos x - 2 \sin x}{x^{3}}$ (after cancelling one $x$ ).

Worked example. Differentiate $y = \dfrac{e^{x}}{1 + x}$ .

$u = e^{x}$ , $v = 1 + x$ . $u' = e^{x}$ , $v' = 1$ .
$y' = \dfrac{e^{x}(1 + x) - e^{x} \cdot 1}{(1 + x)^{2}} = \dfrac{e^{x}[(1 + x) - 1]}{(1 + x)^{2}} = \dfrac{x \, e^{x}}{(1 + x)^{2}}$ .

Alternative: negative power. Sometimes you can avoid the quotient rule by rewriting. $y = 1/x^{2} = x^{-2}$ , so $y' = -2 x^{-3} = -2/x^{3}$ . But for genuine quotients of two non-trivial functions, the quotient rule is unavoidable.

Sign trap. The MINUS sign in the numerator is the single biggest mark-loss in WMA12 calculus questions.

Numerator: $u' v - u v'$ (MINUS!).
Denominator: $v^{2}$ (SQUARED!).
Memory: 'low-D-high minus high-D-low'.
Rewrite as a negative power if possible.

Applications — tangents, normals, stationary points

Same techniques as P1, now with $\sin$ , $\cos$ , $e$ , $\ln$ in the function.

Tangent at $x = a$ . Slope $= f'(a)$ . Equation: $y - f(a) = f'(a)(x - a)$ .

Normal at $x = a$ . Slope $= -1/f'(a)$ . Equation: $y - f(a) = -\dfrac{1}{f'(a)} (x - a)$ .

Stationary points. Set $f'(x) = 0$ . Classify each using $f''$ or first-derivative sign changes.

$f''$ sign	Type
$f''(a) > 0$	Minimum
$f''(a) < 0$	Maximum
$f''(a) = 0$	Inconclusive — use first-derivative test

Worked example. Find and classify the stationary point of $f(x) = \ln x - x$ for $x > 0$ .

$f'(x) = 1/x - 1$ .
$f'(x) = 0$ : $1/x = 1$ , so $x = 1$ . Then $f(1) = 0 - 1 = -1$ .
$f''(x) = -1/x^{2}$ . $f''(1) = -1 < 0$ , so $(1, -1)$ is a MAXIMUM.

Worked example. Find the gradient of $y = x e^{x}$ at $x = 0$ .

$y' = e^{x} + x e^{x} = e^{x}(1 + x)$ (product rule, factored).
At $x = 0$ : $y'(0) = e^{0}(1 + 0) = 1$ .

Increasing / decreasing. $f$ is increasing where $f'(x) > 0$ and decreasing where $f'(x) < 0$ .

Tangent slope: $f'(a)$ . Normal slope: $-1/f'(a)$ .
Stationary points: $f' = 0$ , classify with $f''$ .
$f'' > 0$ at stationary point: minimum.
$f'' < 0$ at stationary point: maximum.

How it’s examined

Differentiation 2 appears in WMA12/01 P2 every sitting (Jan & Jun), typically as Q4-Q14 — worth $10$ - $18$ marks total. Typical questions: differentiate using chain rule (3-4 marks); product rule on $x^{n} e^{x}$ or $x \sin x$ (3-4 marks); quotient rule with simplification (4-5 marks); find stationary points and classify using $f''$ (6-8 marks); find equation of tangent or normal (5-7 marks). Mark schemes split into method (M) for rule structure, accuracy (A) for each term, and final-answer A1. A* students should aim for full marks — these questions reward clean, systematic working.

Sources: Pearson Edexcel International Advanced Level Mathematics specification (2018 — current for 2026 onwards); Edexcel IAL Mathematical Formulae and Statistical Tables (2018); WMA12/01 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 2

Step-by-step solutions to past-paper-style questions on differentiation 2, written exactly the way a tutor would explain them at the board.

1Chain rule on $(3x^{2} - 1)^{5}$ (3 marks, P2)
Core• Adapted from WMA12/01 January 2024 Q10• chain rule
Question
Differentiate $y = (3 x^{2} - 1)^{5}$ with respect to $x$ .
Step-by-step solution
1. Step 1
  Let $u = 3 x^{2} - 1$ , so $y = u^{5}$ . Then $\dfrac{dy}{du} = 5 u^{4}$ and $\dfrac{du}{dx} = 6x$ .
2. Step 2
  Chain rule: $\dfrac{dy}{dx} = \dfrac{dy}{du} \cdot \dfrac{du}{dx} = 5 u^{4} \cdot 6x = 30 x (3 x^{2} - 1)^{4}$ .
Answer
$\dfrac{dy}{dx} = 30 x (3 x^{2} - 1)^{4}$ .
Examiner tip
M1 for identifying the inner function. M1 for applying the chain rule. A1 for the fully correct derivative. The 'derivative of inside' factor ( $6x$ ) is the most-forgotten step — students write $5(3x^{2} - 1)^{4}$ and lose two marks.
2Product rule on $x^{2} e^{x}$ (3 marks, P2)
Core• Adapted from WMA12/01 June 2024 Q9• product rule
Question
Differentiate $f(x) = x^{2} e^{x}$ with respect to $x$ .
Step-by-step solution
1. Step 1
  Set $u = x^{2}$ , $v = e^{x}$ . Then $u' = 2x$ , $v' = e^{x}$ .
2. Step 2
  Product rule: $f'(x) = u' v + u v' = 2x \cdot e^{x} + x^{2} \cdot e^{x}$ .
3. Step 3
  Factor out $e^{x}$ and $x$ : $f'(x) = x e^{x}(2 + x)$ .
Answer
$f'(x) = e^{x}(x^{2} + 2x) = x e^{x}(x + 2)$ .
Examiner tip
M1 for the product rule structure. A1 for both terms correct. A1 for the factorised form (or any equivalent). Edexcel mark schemes accept both factored and unfactored answers; factoring shows cleaner thinking and often opens a follow-up part (stationary points) more easily.
3Quotient rule on $\dfrac{\sin x}{x^{2}}$ (4 marks, P2)
Extended• Adapted from WMA12/01 January 2023 Q10• quotient rule
Question
Find $\dfrac{dy}{dx}$ when $y = \dfrac{\sin x}{x^{2}}$ , $x \neq 0$ .
Step-by-step solution
1. Step 1
  Set $u = \sin x$ , $v = x^{2}$ . Then $u' = \cos x$ , $v' = 2x$ .
2. Step 2
  Quotient rule: $\dfrac{dy}{dx} = \dfrac{u' v - u v'}{v^{2}} = \dfrac{\cos x \cdot x^{2} - \sin x \cdot 2x}{x^{4}}$ .
3. Step 3
  Factor an $x$ from the numerator and cancel one from the denominator:
  $\dfrac{dy}{dx} = \dfrac{x \cos x - 2 \sin x}{x^{3}}$
Answer
$\dfrac{dy}{dx} = \dfrac{x \cos x - 2 \sin x}{x^{3}}$ .
Examiner tip
M1 for the quotient rule structure (correct ORDER of $u'v - uv'$ ). A1 for $\cos x \cdot x^{2}$ . A1 for $- \sin x \cdot 2x$ . A1 for the simplified final form. The minus sign in the middle is the biggest single mark-loser: students write $u'v + uv'$ in error.
4Stationary point of $f(x) = \ln x - x$ (5 marks, P2)
Extended• Adapted from WMA12/01 June 2023 Q11• ln, stationary point
Question
$f(x) = \ln x - x$ for $x > 0$ . Find the coordinates of the stationary point of $f$ , and determine its nature.
Step-by-step solution
1. Step 1
  Differentiate: $f'(x) = \dfrac{1}{x} - 1$ .
2. Step 2
  Set $f'(x) = 0$ : $\dfrac{1}{x} = 1$ , so $x = 1$ .
3. Step 3
  $y$ -coord: $f(1) = \ln 1 - 1 = 0 - 1 = -1$ . Stationary point at $(1, -1)$ .
4. Step 4
  Second derivative: $f''(x) = -\dfrac{1}{x^{2}}$ . At $x = 1$ , $f''(1) = -1 < 0$ , so the stationary point is a MAXIMUM.
Answer
Stationary point at $(1, -1)$ ; this is a maximum.
Examiner tip
M1 for $f'(x) = 1/x - 1$ . M1 for setting $= 0$ and solving. A1 for the point $(1, -1)$ . M1 for $f''(x)$ . A1 for the conclusion 'maximum because $f''(1) < 0$ '. The conclusion MUST reference the sign of $f''$ , not just be stated. Mark scheme: 'A1 dep on M for conclusion linked to the sign of $f''$ '.
5Chain rule on $\sin(3x + 2)$ and second derivative (5 marks, P2)
Extended• Adapted from WMA12/01 January 2024 Q13• chain rule, trig, second derivative
Question
$y = \sin(3x + 2)$ . (a) Find $\dfrac{dy}{dx}$ . (b) Find $\dfrac{d^{2} y}{dx^{2}}$ .
Step-by-step solution
1. Step 1
  (a) Let $u = 3x + 2$ , $y = \sin u$ . $\dfrac{dy}{du} = \cos u$ , $\dfrac{du}{dx} = 3$ . Chain rule:
  $\dfrac{dy}{dx} = 3 \cos(3x + 2)$
2. Step 2
  (b) Differentiate again. $\dfrac{d^{2} y}{dx^{2}} = 3 \cdot \dfrac{d}{dx}[\cos(3x + 2)]$ .
3. Step 3
  Chain rule again: $\dfrac{d}{dx}[\cos(3x + 2)] = -\sin(3x + 2) \cdot 3 = -3 \sin(3x + 2)$ .
4. Step 4
  So $\dfrac{d^{2} y}{dx^{2}} = 3 \cdot (-3 \sin(3x + 2)) = -9 \sin(3x + 2)$ .
Answer
(a) $\dfrac{dy}{dx} = 3 \cos(3x + 2)$ . (b) $\dfrac{d^{2} y}{dx^{2}} = -9 \sin(3x + 2)$ .
Examiner tip
(a) M1 for the chain rule structure. A1 for the answer. (b) M1 for applying the chain rule a second time. A1 for $-9 \sin(\ldots)$ . Note: $\dfrac{d^{2} y}{dx^{2}} = -9 y$ — i.e. $y$ satisfies $y'' + 9 y = 0$ , the simple-harmonic-motion ODE.

Key Formulae — Differentiation 2

The formulae you need to memorise for differentiation 2 on the Pearson Edexcel IAL XMA01 / YMA01 paper, with every variable defined in plain English and a note on when to use it.

Power rule
$\dfrac{d}{dx}(x^{n}) = n x^{n - 1}$
$n$
any real number
When to use
Differentiating polynomials, surds and reciprocal powers. NOT in the IAL formula booklet — memorise.
Example
$\dfrac{d}{dx}(x^{3}) = 3 x^{2}$ .
Derivatives of $\sin$ and $\cos$
$\dfrac{d}{dx}(\sin x) = \cos x, \qquad \dfrac{d}{dx}(\cos x) = -\sin x$
$x$
angle in radians
When to use
Differentiating trig functions — note these are valid ONLY when $x$ is in radians. NOT in the formula booklet.
Example
$\dfrac{d}{dx}(\sin 3x) = 3 \cos 3x$ (chain rule).
Derivative of $e^{x}$
$\dfrac{d}{dx}(e^{x}) = e^{x}$
When to use
$e^{x}$ is its own derivative. Combine with chain rule for $e^{f(x)}$ . NOT in the formula booklet.
Example
$\dfrac{d}{dx}(e^{2x}) = 2 e^{2x}$ .
Derivative of $\ln x$
$\dfrac{d}{dx}(\ln x) = \dfrac{1}{x} \quad (x > 0)$
$x$
positive real
When to use
Differentiating natural logs. NOT in the formula booklet.
Example
$\dfrac{d}{dx}[\ln(3x^{2})] = \dfrac{6x}{3 x^{2}} = \dfrac{2}{x}$ (chain rule).
Derivative of $a^{x}$
$\dfrac{d}{dx}(a^{x}) = a^{x} \ln a \quad (a > 0)$
$a$
positive base, $a \neq 1$
When to use
Differentiating non- $e$ exponentials. NOT in the formula booklet (derived from $a^{x} = e^{x \ln a}$ ).
Example
$\dfrac{d}{dx}(2^{x}) = 2^{x} \ln 2$ .
Chain rule
$\dfrac{dy}{dx} = \dfrac{dy}{du} \cdot \dfrac{du}{dx}$
$y$
outer function
$u$
inner function of $x$
When to use
Differentiating a composite function (function of a function). GIVEN in the IAL formula booklet (Pure Mathematics).
Example
$y = (3 x^{2} - 1)^{5}$ : set $u = 3 x^{2} - 1$ and apply.
Product rule
$\dfrac{d}{dx}(u v) = u' v + u v' = u \dfrac{dv}{dx} + v \dfrac{du}{dx}$
$u, v$
differentiable functions of $x$
When to use
Differentiating a product. GIVEN in the IAL formula booklet.
Example
$\dfrac{d}{dx}(x^{2} e^{x}) = 2x e^{x} + x^{2} e^{x} = x e^{x}(x + 2)$ .
Quotient rule
$\dfrac{d}{dx}\!\left(\dfrac{u}{v}\right) = \dfrac{u' v - u v'}{v^{2}}$
$u, v$
differentiable functions, $v \neq 0$
When to use
Differentiating a quotient when the denominator can't easily be re-expressed as a negative power. GIVEN in the IAL formula booklet.
Example
$\dfrac{d}{dx}\!\left(\dfrac{\sin x}{x^{2}}\right) = \dfrac{x \cos x - 2 \sin x}{x^{3}}$ .

Key Definitions and Keywords — Differentiation 2

Definitions to memorise and the exact keywords mark schemes credit for differentiation 2 answers — sharpened from recent examiner reports for the 2026 Pearson Edexcel IAL XMA01 / YMA01 sitting.

Chain rule
Examiner keyword
If $y$ is a function of $u$ , and $u$ is a function of $x$ , then $\dfrac{dy}{dx} = \dfrac{dy}{du} \cdot \dfrac{du}{dx}$ . Equivalently: differentiate the outside (leaving the inside alone), then multiply by the derivative of the inside.
Product rule
Examiner keyword
The derivative of $uv$ is $u' v + u v'$ . Memory aid: 'derivative of first times second, plus first times derivative of second.'
Quotient rule
Examiner keyword
The derivative of $u/v$ is $\dfrac{u' v - u v'}{v^{2}}$ . Note the MINUS in the numerator and the SQUARE in the denominator.
Stationary point
A point where $f'(x) = 0$ . Classified as a maximum, minimum, or point of inflection by looking at the sign of $f''(x)$ (or the sign changes of $f'$ around the point).
Second derivative
The derivative of the derivative: $\dfrac{d^{2} y}{dx^{2}} = \dfrac{d}{dx}\!\left(\dfrac{dy}{dx}\right)$ . Used for classifying stationary points and analysing concavity.

Common Mistakes and Misconceptions — Differentiation 2

The traps other students keep falling into on differentiation 2 questions — taken from recent Pearson Edexcel IAL XMA01 / YMA01 examiner reports and mark schemes — and how to avoid them.

✕Differentiating $(3 x^{2} - 1)^{5}$ as $5 (3 x^{2} - 1)^{4}$
WMA12/01 examiner reports — flagged annually
Why it happens
Forgetting to multiply by the derivative of the inside (chain rule's 'extra' factor).
How to avoid it
Always two factors: derivative-of-outside (evaluated at the inside) MULTIPLIED BY derivative-of-inside. Here: $5 (3 x^{2} - 1)^{4} \cdot 6 x = 30 x (3 x^{2} - 1)^{4}$ .
✕Writing $\dfrac{u' v + u v'}{v^{2}}$ in the quotient rule
Why it happens
Borrowing the plus sign from the product rule.
How to avoid it
Quotient rule has a MINUS in the numerator. Memory aid: 'low-D-high MINUS high-D-low, square the bottom and away we go.'
✕Writing $\dfrac{d}{dx}(\cos x) = \sin x$ (missing minus)
Why it happens
Mirror confusion with $\sin$ .
How to avoid it
$\dfrac{d}{dx}(\sin x) = \cos x$ (NO sign); $\dfrac{d}{dx}(\cos x) = -\sin x$ (MINUS). Memory rule: cosine 'falls'.
✕Differentiating $\sin x°$ as $\cos x°$
Why it happens
Not realising the derivative formulae for trig functions are valid ONLY in radians.
How to avoid it
Convert degrees to radians first: $\sin x° = \sin(\pi x / 180)$ . Then $\dfrac{d}{dx}[\sin(\pi x / 180)] = \dfrac{\pi}{180} \cos(\pi x / 180)$ .
✕Differentiating $\ln(-x)$ and reporting $\dfrac{1}{-x}$ (or worrying that the log is undefined)
Why it happens
Mixing up the domain restriction with the derivative formula.
How to avoid it
$\ln(-x)$ is defined for $x < 0$ , and by chain rule $\dfrac{d}{dx}[\ln(-x)] = \dfrac{-1}{-x} = \dfrac{1}{x}$ . The general rule for $|x|$ : $\dfrac{d}{dx}[\ln|x|] = \dfrac{1}{x}$ for all $x \neq 0$ .

Practice questions

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

Past paper style quiz

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Differentiation 2 — frequently asked questions

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

Differentiation 2

Short Study Notes in the form of Slides

Detailed Study Notes

What you’ll learn

How it’s examined

1Chain rule on (3x2−1)5(3x^{2} - 1)^{5}(3x2−1)5 (3 marks, P2)

2Product rule on x2exx^{2} e^{x}x2ex (3 marks, P2)

3Quotient rule on sin⁡xx2\dfrac{\sin x}{x^{2}}x2sinx​ (4 marks, P2)

4Stationary point of f(x)=ln⁡x−xf(x) = \ln x - xf(x)=lnx−x (5 marks, P2)

5Chain rule on sin⁡(3x+2)\sin(3x + 2)sin(3x+2) and second derivative (5 marks, P2)

Power rule

Derivatives of sin⁡\sinsin and cos⁡\coscos

Derivative of exe^{x}ex

Derivative of ln⁡x\ln xlnx

Derivative of axa^{x}ax

Chain rule

Product rule

Quotient rule

Chain rule

Product rule

Quotient rule

Stationary point

Second derivative

✕Differentiating (3x2−1)5(3 x^{2} - 1)^{5}(3x2−1)5 as 5(3x2−1)45 (3 x^{2} - 1)^{4}5(3x2−1)4

✕Writing u′v+uv′v2\dfrac{u' v + u v'}{v^{2}}v2u′v+uv′​ in the quotient rule

✕Writing ddx(cos⁡x)=sin⁡x\dfrac{d}{dx}(\cos x) = \sin xdxd​(cosx)=sinx (missing minus)

✕Differentiating sin⁡x°\sin x°sinx° as cos⁡x°\cos x°cosx°

✕Differentiating ln⁡(−x)\ln(-x)ln(−x) and reporting 1−x\dfrac{1}{-x}−x1​ (or worrying that the log is undefined)

Practice questions

Past paper style quiz

Assess your understanding

Differentiation 2 — frequently asked questions

1Chain rule on $(3x^{2} - 1)^{5}$ (3 marks, P2)

2Product rule on $x^{2} e^{x}$ (3 marks, P2)

3Quotient rule on $\dfrac{\sin x}{x^{2}}$ (4 marks, P2)

4Stationary point of $f(x) = \ln x - x$ (5 marks, P2)

5Chain rule on $\sin(3x + 2)$ and second derivative (5 marks, P2)

Derivatives of $\sin$ and $\cos$

Derivative of $e^{x}$

Derivative of $\ln x$

Derivative of $a^{x}$

✕Differentiating $(3 x^{2} - 1)^{5}$ as $5 (3 x^{2} - 1)^{4}$

✕Writing $\dfrac{u' v + u v'}{v^{2}}$ in the quotient rule

✕Writing $\dfrac{d}{dx}(\cos x) = \sin x$ (missing minus)

✕Differentiating $\sin x°$ as $\cos x°$

✕Differentiating $\ln(-x)$ and reporting $\dfrac{1}{-x}$ (or worrying that the log is undefined)