What are the key concepts covered in Differentiation 3 for Edexcel International A Levels Pure Mathematics 3?

Differentiation 3 in Edexcel International A Levels Pure Mathematics 3 focuses on advanced differentiation techniques. Key concepts include implicit differentiation, differentiation of inverse trigonometric functions, and the application of differentiation in solving real-world problems such as optimization and related rates.

How do you perform implicit differentiation in Pure Mathematics 3?

Implicit differentiation is used when a function is not given explicitly as y = f(x). To differentiate implicitly, differentiate both sides of the equation with respect to x, applying the chain rule where necessary, and then solve for dy/dx. This technique is particularly useful for equations involving products or powers of x and y.

What is the importance of differentiating inverse trigonometric functions in Differentiation 3?

Differentiating inverse trigonometric functions is crucial in Differentiation 3 as it extends the range of functions students can analyze. Understanding these derivatives allows students to solve complex calculus problems involving arcsin, arccos, and arctan, which are common in advanced mathematics and physics applications.

Can you explain how differentiation is used in optimization problems in Pure Mathematics 3?

In Pure Mathematics 3, differentiation is used in optimization to find maximum or minimum values of functions. By finding the derivative and setting it to zero, students can identify critical points. Further analysis using the second derivative test or analyzing the sign changes of the first derivative helps determine whether these points are maxima, minima, or points of inflection.

What are related rates problems and how are they solved using differentiation?

Related rates problems involve finding the rate at which one quantity changes with respect to another. In Differentiation 3, these problems are solved by identifying the relationship between the variables, differentiating this relationship with respect to time, and then substituting known values to find the unknown rate. This technique is widely used in physics and engineering to model dynamic systems.

Why is "Differentiating $\sin(2x)$ as $\cos(2x)$ (without the factor of $2$)" a common mistake in Differentiation 3?

Why it happens: Treating (2x) as if it were x. How to avoid it: ALWAYS apply the chain rule when the argument is not just x. d/dx(2x) = 2(2x). The inner derivative du/dx = 2 MUST appear. Source: WMA13/01 examiner reports — flagged annually

Why is "Using degrees when differentiating trig" a common mistake in Differentiation 3?

Why it happens: Not converting back to radians for calculus. How to avoid it: The derivatives d/dx x = x etc. hold ONLY in radians. If a question gives angles in degrees, convert first.

Why is "$\dfrac{d}{dx}\ln(f(x)) = \dfrac{1}{f(x)}$ (forgetting the chain)" a common mistake in Differentiation 3?

Why it happens: Memorising the rule without the chain. How to avoid it: d/dx(f(x)) = f'(x)/f(x). The derivative of the inner is in the numerator. For f(x) = x, f'(x) = 1 and you recover 1/x.

Why is "Writing $\dfrac{dy}{dx} = \dfrac{dx/dt}{dy/dt}$ (upside down)" a common mistake in Differentiation 3?

Why it happens: Misremembering the order. How to avoid it: dy/dx = dy/dt/dx/dt — 'y on top, x on bottom' matches the LHS direction. Mnemonic: cancel the dt.

Why is "Using the product rule when chain rule is required for connected rates" a common mistake in Differentiation 3?

Why it happens: Confusing two independent rates with one chained relationship. How to avoid it: Connected rates: ONE chain dA/dt = dA/dB · dB/dt. Write it down BEFORE substituting. If you see a product of two time-dependent factors, that's the product rule, a different setting.

Pure Mathematics 3Edexcel International A Levels Mathematics (XMA01-YMA01)

Differentiation 3

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Differentiation 3 Study Notes — Edexcel IAL Mathematics P3 (WMA13/01, 2018 spec — 2026 onwards)

Differentiate trig functions (all six), $e^{f(x)}$ and $\ln f(x)$ via the chain rule, parametric curves $\bigl(x(t), y(t)\bigr)$ , and use the second derivative for concavity and connected rates of change. The IAL formula booklet lists derivatives of $\sin$ , $\cos$ , $\tan$ , $\sec$ , $\csc$ , $\cot$ , $e^{x}$ , $\ln x$ — chained versions you derive each time.

At a glance

$\dfrac{d}{dx}\sin x = \cos x$ , $\dfrac{d}{dx}\cos x = -\sin x$ , $\dfrac{d}{dx}\tan x = \sec^{2}x$ .
$\dfrac{d}{dx}\sec x = \sec x\tan x$ ; $\dfrac{d}{dx}\csc x = -\csc x\cot x$ ; $\dfrac{d}{dx}\cot x = -\csc^{2}x$ .
$\dfrac{d}{dx}e^{f(x)} = f'(x) e^{f(x)}$ .
$\dfrac{d}{dx}\ln f(x) = \dfrac{f'(x)}{f(x)}$ .
Parametric: $\dfrac{dy}{dx} = \dfrac{dy/dt}{dx/dt}$ .
$f''(x) > 0$ at stationary → min; $f''(x) < 0$ → max; $= 0$ inconclusive.
Connected rates: $\dfrac{dA}{dt} = \dfrac{dA}{dB} \cdot \dfrac{dB}{dt}$ .

What you’ll learn

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

P3 6.1 — Differentiate trigonometric functions including reciprocal trig.
P3 6.2 — Differentiate $e^{f(x)}$ and $\ln f(x)$ using the chain rule.
P3 6.3 — Use parametric differentiation for curves given as $\bigl(x(t), y(t)\bigr)$ .
P3 6.4 — Use higher derivatives to classify stationary points and identify concavity.
P3 6.5 — Solve connected-rates-of-change problems.

Derivatives of trig functions

Six trig derivatives — memorise patterns and 'co'-anti-symmetry.

Six core derivatives.

Function	Derivative
$\sin x$	$\cos x$
$\cos x$	$-\sin x$
$\tan x$	$\sec^{2}x$
$\sec x$	$\sec x \tan x$
$\csc x$	$-\csc x \cot x$
$\cot x$	$-\csc^{2}x$

Pattern: 'co'-derivatives have minus signs. $\cos$ , $\csc$ , $\cot$ — all give negative derivatives.

Always in radians. These identities hold ONLY when $x$ is in radians. Convert degrees to radians before differentiating.

Chained version. With chain rule: $\dfrac{d}{dx}\sin(f(x)) = f'(x) \cos(f(x)), \quad \dfrac{d}{dx}\tan(f(x)) = f'(x) \sec^{2}(f(x)),$ and similarly for the others.

Worked example. $\dfrac{d}{dx}\sin(2x^{2} + 3) = (4x) \cos(2x^{2} + 3)$ — applying the chain.

Deriving sec / csc / cot derivatives. Each is a chain rule application on $(\cos x)^{-1}$ , $(\sin x)^{-1}$ , $(\tan x)^{-1}$ . P3 'show that' questions sometimes ask for the derivation:

$\dfrac{d}{dx}\sec x = \dfrac{d}{dx}(\cos x)^{-1} = -(\cos x)^{-2} \cdot (-\sin x) = \dfrac{\sin x}{\cos^{2}x} = \sec x \tan x$ .

Six core derivatives — three with minus signs ('co'-derivatives).
Radians only.
Chain rule essential for compound arguments.
Reciprocal derivatives derivable from $u^{-1}$ chain.

Derivatives of $e^{f(x)}$ and $\ln f(x)$

Chain rule applied to $e$ and $\ln$ .

Standard rules. $\dfrac{d}{dx} e^{x} = e^{x}, \qquad \dfrac{d}{dx} \ln x = \dfrac{1}{x}.$

Chained versions. $\dfrac{d}{dx} e^{f(x)} = f'(x) e^{f(x)}, \qquad \dfrac{d}{dx} \ln f(x) = \dfrac{f'(x)}{f(x)}.$

Worked examples.

$\dfrac{d}{dx} e^{3x^{2} - 1} = (6x) e^{3x^{2} - 1}$ .
$\dfrac{d}{dx} \ln(\sin x) = \dfrac{\cos x}{\sin x} = \cot x$ .
$\dfrac{d}{dx} \ln(x^{2} + 1) = \dfrac{2x}{x^{2} + 1}$ .
$\dfrac{d}{dx} e^{\sin x} = \cos x \cdot e^{\sin x}$ .

Log laws BEFORE differentiation. Often easier to simplify first: $\dfrac{d}{dx} \ln(x^{2}(x + 1)) = \dfrac{d}{dx}\bigl(2\ln x + \ln(x + 1)\bigr) = \dfrac{2}{x} + \dfrac{1}{x + 1}.$

Doing log-laws first sidesteps the quotient/product rule.

Logarithmic differentiation. For complicated products / quotients / powers, take $\ln$ first: $y = \dfrac{x^{2}(x + 1)^{3}}{(x - 1)} \Rightarrow \ln y = 2\ln x + 3\ln(x + 1) - \ln(x - 1).$ Differentiate both sides implicitly with respect to $x$ , then multiply by $y$ .

$\dfrac{d}{dx} e^{f(x)} = f'(x) e^{f(x)}$ .
$\dfrac{d}{dx} \ln f(x) = \dfrac{f'(x)}{f(x)}$ .
Simplify with log laws BEFORE differentiating.
Logarithmic differentiation for products / quotients.

Parametric differentiation

$\dfrac{dy}{dx} = \dfrac{dy/dt}{dx/dt}$ .

Setup. A curve given parametrically as $x = x(t)$ , $y = y(t)$ for some parameter $t$ (often time, or angle).

Gradient formula. $\dfrac{dy}{dx} = \dfrac{dy/dt}{dx/dt}.$

Derivation: chain rule. $\dfrac{dy}{dt} = \dfrac{dy}{dx} \cdot \dfrac{dx}{dt}$ , so $\dfrac{dy}{dx} = \dfrac{dy/dt}{dx/dt}$ , provided $\dfrac{dx}{dt} \neq 0$ .

Worked example. $x = 3\cos t$ , $y = \sin 2t$ , $0 \leq t \leq \pi$ . Find $\dfrac{dy}{dx}$ at $t = \pi/3$ .

$\dfrac{dx}{dt} = -3\sin t$ .
$\dfrac{dy}{dt} = 2\cos 2t$ (chain on $\sin 2t$ ).
$\dfrac{dy}{dx} = \dfrac{2\cos 2t}{-3\sin t}$ .

At $t = \pi/3$ : $\cos(2\pi/3) = -1/2$ , $\sin(\pi/3) = \sqrt{3}/2$ . $\dfrac{dy}{dx} = \dfrac{2 \cdot (-1/2)}{-3 \cdot \sqrt{3}/2} = \dfrac{-1}{-3\sqrt{3}/2} = \dfrac{2}{3\sqrt{3}} = \dfrac{2\sqrt{3}}{9}.$

Stationary points. $\dfrac{dy}{dx} = 0$ when $\dfrac{dy}{dt} = 0$ (provided $\dfrac{dx}{dt} \neq 0$ ).

Tangent and normal. Same as Cartesian curves: gradient is $\dfrac{dy}{dx}$ at the relevant parameter value. Tangent: $y - y_{1} = m(x - x_{1})$ where $(x_{1}, y_{1}) = (x(t_{0}), y(t_{0}))$ and $m = \dfrac{dy}{dx}|_{t_{0}}$ .

Cartesian equation. Sometimes you can eliminate $t$ to get a Cartesian equation. E.g. $x = 3\cos t$ , $y = 3\sin t$ gives $x^{2} + y^{2} = 9$ .

$\dfrac{dy}{dx} = \dfrac{dy/dt}{dx/dt}$ (provided $dx/dt \neq 0$ ).
Compute $dy/dt$ and $dx/dt$ separately.
Stationary when $dy/dt = 0$ (and $dx/dt \neq 0$ ).
Eliminate parameter for Cartesian if useful.

Higher derivatives and concavity

$f''(x)$ tells you concavity. Use to classify stationary points.

Second derivative. $f''(x) = \dfrac{d}{dx}f'(x) = \dfrac{d^{2}y}{dx^{2}}$ . The derivative of the derivative.

Classifying stationary points. At $x_{0}$ where $f'(x_{0}) = 0$ :

$f''(x_{0}) > 0$ → local MINIMUM (concave up).
$f''(x_{0}) < 0$ → local MAXIMUM (concave down).
$f''(x_{0}) = 0$ → INCONCLUSIVE; use a sign-change test on $f'$ instead.

Concavity. A curve is:

Concave up (convex) where $f''(x) > 0$ — looks like a cup.
Concave down where $f''(x) < 0$ — looks like a cap.
A point of inflection is where $f''$ changes sign.

Worked example. Classify the stationary point of $f(x) = x e^{-x}$ .

$f'(x) = e^{-x} - x e^{-x} = e^{-x}(1 - x)$ . $f'(x) = 0 \Rightarrow x = 1$ .
$f''(x) = -e^{-x}(1 - x) + e^{-x}(-1) = e^{-x}(-1 + x - 1) = e^{-x}(x - 2)$ .
At $x = 1$ : $f''(1) = e^{-1}(1 - 2) = -e^{-1} < 0$ . So $(1, e^{-1})$ is a local MAX.

Inflection points. Where $f''$ changes sign:

From $f''(x) = e^{-x}(x - 2)$ , $f''$ changes sign at $x = 2$ .
Above example has an inflection at $(2, 2 e^{-2})$ .

Notation reminder. $\dfrac{d^{2}y}{dx^{2}}$ — the '2' is on TOP of the $d$ in the numerator (an operator squared) and on the BOTTOM with the $x$ (because we're differentiating with respect to $x$ twice).

$f''(x_{0}) > 0$ → min; $< 0$ → max.
$f''(x_{0}) = 0$ → use sign-change test.
Concave up ⇔ $f''(x) > 0$ .
Inflection ⇔ $f''$ changes sign.

Connected rates of change

Chain rule for related variables.

Setup. Two related quantities $A$ , $B$ both change with time $t$ , linked by a static formula $A = f(B)$ (e.g. volume = $\tfrac{4}{3}\pi r^{3}$ links volume and radius for a sphere).

Chain rule. $\dfrac{dA}{dt} = \dfrac{dA}{dB} \cdot \dfrac{dB}{dt}.$

Strategy.

Identify both rates. Mark which is GIVEN and which is REQUIRED.
Write the static formula relating $A$ and $B$ .
Differentiate to get $\dfrac{dA}{dB}$ .
Apply the chain rule and solve for the unknown rate.

Worked example. A spherical balloon is inflated at $25 \text{ cm}^{3}\text{s}^{-1}$ . Find the rate of increase of radius when $r = 5$ cm.

Given: $\dfrac{dV}{dt} = 25$ . Want: $\dfrac{dr}{dt}$ when $r = 5$ .
Formula: $V = \tfrac{4}{3}\pi r^{3}$ .
$\dfrac{dV}{dr} = 4\pi r^{2}$ . At $r = 5$ : $\dfrac{dV}{dr} = 100\pi$ .
Chain: $\dfrac{dV}{dt} = \dfrac{dV}{dr} \cdot \dfrac{dr}{dt}$ , so $25 = 100\pi \cdot \dfrac{dr}{dt}$ .
$\dfrac{dr}{dt} = \dfrac{25}{100\pi} = \dfrac{1}{4\pi} \approx 0.0796 \text{ cm/s}$ .

Worked example 2. A conical container has fixed half-angle. Water flows in at $V'(t) = q$ (constant). Find $\dfrac{dh}{dt}$ as a function of $h$ .

Cone: $V = \tfrac{1}{3}\pi r^{2} h$ . Geometry: $r = h \tan\alpha$ (where $\alpha$ = half-angle).
So $V = \tfrac{1}{3}\pi h^{3} \tan^{2}\alpha$ . Differentiate: $\dfrac{dV}{dh} = \pi h^{2} \tan^{2}\alpha$ .
$\dfrac{dV}{dt} = q = \dfrac{dV}{dh} \cdot \dfrac{dh}{dt} \Rightarrow \dfrac{dh}{dt} = \dfrac{q}{\pi h^{2}\tan^{2}\alpha}$ .

Units. Always include the correct units in the final answer.

$\dfrac{dA}{dt} = \dfrac{dA}{dB} \cdot \dfrac{dB}{dt}$ (chain rule).
Static formula → differentiate → chain.
Substitute SPECIFIC values only after the chain step.
Include units (cm/s, $\mathrm{m}^{3}/$ s, etc.).

Memorise this

Verbatim phrases and definitions Cambridge mark schemes credit.

$\dfrac{d}{dx}\sin x = \cos x$ , $\dfrac{d}{dx}\cos x = -\sin x$ , $\dfrac{d}{dx}\tan x = \sec^{2}x$ .
$\dfrac{d}{dx}\sec x = \sec x\tan x$ .
$\dfrac{d}{dx}\csc x = -\csc x\cot x$ .
$\dfrac{d}{dx}\cot x = -\csc^{2}x$ .
$\dfrac{d}{dx} e^{f(x)} = f'(x) e^{f(x)}$ .
$\dfrac{d}{dx} \ln f(x) = \dfrac{f'(x)}{f(x)}$ .
Parametric: $\dfrac{dy}{dx} = \dfrac{dy/dt}{dx/dt}$ .
Connected rates: $\dfrac{dA}{dt} = \dfrac{dA}{dB} \cdot \dfrac{dB}{dt}$ .

How it’s examined

Differentiation 3 is a backbone of WMA13/01 — appears as a $10$ - $15$ -mark question every sitting. Sub-parts: differentiate a chained trig / exp / log function ( $3$ - $4$ marks each); find a stationary point ( $4$ - $5$ marks); classify it with the second derivative ( $2$ marks); for parametric problems: $\dfrac{dy}{dx}$ at a point ( $4$ - $5$ marks), tangent / normal equation ( $3$ marks). Connected rates is a $6$ - $8$ mark stand-alone question. Common errors: forgetting the chain factor, mixing degrees with radians, substituting values before differentiating in connected-rates problems.

Sources: Pearson Edexcel International Advanced Level Mathematics specification (2018 — current for 2026 onwards); Edexcel IAL Mathematical Formulae and Statistical Tables (2018); WMA13/01 Examiner Reports — January 2023, June 2023, January 2024, June 2024. Last reviewed 2026-05-12.

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Step-by-step worked examples — Differentiation 3

Step-by-step solutions to past-paper-style questions on differentiation 3, written exactly the way a tutor would explain them at the board.

1Differentiate $y = \sin(2x^{2} + 3)$ (4 marks, P3)
Core• Adapted from WMA13/01 January 2024 Q2• chain rule, trig derivative
Question
Find $\dfrac{dy}{dx}$ for $y = \sin(2x^{2} + 3)$ .
Step-by-step solution
1. Step 1
  Chain rule: let $u = 2x^{2} + 3$ , so $y = \sin u$ . Then $\dfrac{dy}{du} = \cos u$ and $\dfrac{du}{dx} = 4x$ .
2. Step 2
  Combine:
  $\dfrac{dy}{dx} = \cos u \cdot 4x = 4x \cos(2x^{2} + 3)$
Answer
$\dfrac{dy}{dx} = 4x \cos(2x^{2} + 3)$ .
Examiner tip
M1 identify chain rule. A1 derivative of outer ( $\cos u$ ). A1 derivative of inner ( $4x$ ). A1 final combined form. Always state 'using the chain rule' or write the $u$ substitution explicitly — it earns the M mark cleanly.
2Differentiate $y = e^{3x^{2} - 1}$ (3 marks, P3)
Core• Adapted from WMA13/01 June 2024 Q1• chain rule, exponential derivative
Question
Find $\dfrac{dy}{dx}$ for $y = e^{3x^{2} - 1}$ .
Step-by-step solution
1. Step 1
  Chain rule. Outer: $e^{u}$ differentiates to $e^{u}$ . Inner: $u = 3x^{2} - 1$ , so $\dfrac{du}{dx} = 6x$ .
2. Step 2
  Combine:
  $\dfrac{dy}{dx} = e^{3x^{2} - 1} \cdot 6x = 6x \, e^{3x^{2} - 1}$
Answer
$\dfrac{dy}{dx} = 6x \, e^{3x^{2} - 1}$ .
Examiner tip
M1 chain rule. A1 derivative of inner $= 6x$ . A1 final form. Crucial: $\dfrac{d}{dx} e^{f(x)} = f'(x) e^{f(x)}$ . Many students forget to multiply by $f'(x)$ .
3Differentiate $y = \ln(\sin x)$ (3 marks, P3)
Core• Adapted from WMA13/01 January 2023 Q3• chain rule, natural log derivative
Question
Find $\dfrac{dy}{dx}$ for $y = \ln(\sin x)$ .
Step-by-step solution
1. Step 1
  $\dfrac{d}{dx} \ln f(x) = \dfrac{f'(x)}{f(x)}$ . Here $f(x) = \sin x$ , $f'(x) = \cos x$ .
2. Step 2
  Result:
  $\dfrac{dy}{dx} = \dfrac{\cos x}{\sin x} = \cot x$
Answer
$\dfrac{dy}{dx} = \cot x$ .
Examiner tip
M1 quote $\dfrac{f'}{f}$ form. A1 substitution. A1 simplification to $\cot x$ . Recognising $\dfrac{\cos x}{\sin x}$ as $\cot x$ is required for the final A — leaving as a fraction is fine but the simplified form scores cleaner.
4Prove $\dfrac{d}{dx}(\sec x) = \sec x \tan x$ (4 marks, P3)
Extended• Adapted from WMA13/01 June 2024 Q4• proof, secant derivative
Question
Starting from $\sec x = (\cos x)^{-1}$ , prove that $\dfrac{d}{dx}(\sec x) = \sec x \tan x$ .
Step-by-step solution
1. Step 1
  $\sec x = (\cos x)^{-1}$ . Chain rule with outer $u^{-1}$ :
  $\dfrac{d}{dx}(\cos x)^{-1} = -1 \cdot (\cos x)^{-2} \cdot \dfrac{d}{dx}(\cos x)$
2. Step 2
  $\dfrac{d}{dx}(\cos x) = -\sin x$ :
  $= -(\cos x)^{-2} \cdot (-\sin x) = \dfrac{\sin x}{\cos^{2}x}$
3. Step 3
  Rewrite using $\sec$ and $\tan$ :
  $= \dfrac{1}{\cos x} \cdot \dfrac{\sin x}{\cos x} = \sec x \tan x. \;\;\blacksquare$
Answer
$\dfrac{d}{dx}(\sec x) = \sec x \tan x$ .
Examiner tip
M1 quote $\sec x = (\cos x)^{-1}$ and apply chain. A1 derivative of $\cos x$ . M1 simplify into $\sin x / \cos^{2} x$ . A1 recognise as $\sec x \tan x$ . 'Show / prove' demands every step. Even though $\dfrac{d}{dx} \sec x$ is in the formula booklet, the proof is a standard P3 exam request.
5Parametric curve — find $\dfrac{dy}{dx}$ at a parameter value (5 marks, P3)
Extended• Adapted from WMA13/01 January 2024 Q9• parametric, chain rule
Question
A curve has parametric equations $x = 3\cos t$ , $y = \sin 2t$ , for $0 \leq t \leq \pi$ . Find $\dfrac{dy}{dx}$ at $t = \dfrac{\pi}{3}$ .
Step-by-step solution
1. Step 1
  $\dfrac{dx}{dt} = -3 \sin t$ . $\dfrac{dy}{dt} = 2 \cos 2t$ (chain rule for $\sin 2t$ ).
2. Step 2
  Parametric derivative:
  $\dfrac{dy}{dx} = \dfrac{dy/dt}{dx/dt} = \dfrac{2\cos 2t}{-3\sin t}$
3. Step 3
  At $t = \pi/3$ : $\cos(2\pi/3) = -1/2$ , $\sin(\pi/3) = \sqrt{3}/2$ .
  $\dfrac{dy}{dx} = \dfrac{2 \cdot (-1/2)}{-3 \cdot \sqrt{3}/2} = \dfrac{-1}{-3\sqrt{3}/2} = \dfrac{2}{3\sqrt{3}} = \dfrac{2\sqrt{3}}{9}$
Answer
$\dfrac{dy}{dx} = \dfrac{2\sqrt{3}}{9}$ at $t = \pi/3$ .
Examiner tip
M1 each for $dx/dt$ and $dy/dt$ . A1 each for correct expressions. M1 $dy/dx = (dy/dt)/(dx/dt)$ . A1 substituted value. A1 rationalised final form. Common errors: forgetting the chain in $\dfrac{d}{dt}\sin 2t = 2\cos 2t$ (not $\cos 2t$ ).
6Connected rate of change — expanding sphere (6 marks, P3)
Extended• Adapted from WMA13/01 June 2023 Q9• connected rates, chain rule
Question
Air is pumped into a spherical balloon at a constant rate of $25 \text{ cm}^{3}\text{s}^{-1}$ . Find the rate at which the radius is increasing when the radius is $5 \text{ cm}$ . (Volume of a sphere: $V = \tfrac{4}{3}\pi r^{3}$ .)
Step-by-step solution
1. Step 1
  Identify: $\dfrac{dV}{dt} = 25$ (given). Want: $\dfrac{dr}{dt}$ when $r = 5$ .
2. Step 2
  Chain rule: $\dfrac{dV}{dt} = \dfrac{dV}{dr} \cdot \dfrac{dr}{dt}$ . Differentiate $V$ :
  $\dfrac{dV}{dr} = 4\pi r^{2}$
3. Step 3
  At $r = 5$ : $\dfrac{dV}{dr} = 4\pi (25) = 100\pi$ .
4. Step 4
  Solve for $\dfrac{dr}{dt}$ :
  $25 = 100\pi \cdot \dfrac{dr}{dt} \;\Rightarrow\; \dfrac{dr}{dt} = \dfrac{25}{100\pi} = \dfrac{1}{4\pi}$
Answer
$\dfrac{dr}{dt} = \dfrac{1}{4\pi} \approx 0.0796 \text{ cm/s}$ .
Examiner tip
B1 state $dV/dt = 25$ . M1 chain rule linking the three rates. A1 expression for $dV/dr$ . M1 substitute $r = 5$ . A1 isolate $dr/dt$ . A1 final form ( $\dfrac{1}{4\pi}$ exact, or $\approx 0.080$ to 2 s.f.). Units MATTER in modelling questions — always include cm/s.

Key Formulae — Differentiation 3

The formulae you need to memorise for differentiation 3 on the Pearson Edexcel IAL XMA01 / YMA01 paper, with every variable defined in plain English and a note on when to use it.

Derivatives of trig functions
$\dfrac{d}{dx}(\sin x) = \cos x, \;\; \dfrac{d}{dx}(\cos x) = -\sin x, \;\; \dfrac{d}{dx}(\tan x) = \sec^{2}x$
$x$
in radians (essential — the derivatives don't hold in degrees)
When to use
Whenever differentiating a trig function. IN the IAL formula booklet.
Example
$\dfrac{d}{dx}(\tan(3x)) = 3\sec^{2}(3x)$ (chain rule).
Derivatives of reciprocal trig
$\dfrac{d}{dx}(\sec x) = \sec x \tan x, \;\; \dfrac{d}{dx}(\csc x) = -\csc x \cot x, \;\; \dfrac{d}{dx}(\cot x) = -\csc^{2}x$
$x$
in radians, excluding asymptote points
When to use
Whenever differentiating $\sec$ , $\csc$ , $\cot$ . IN the IAL formula booklet. Mnemonic: 'co' derivatives all have a minus sign.
Example
$\dfrac{d}{dx}(\sec(2x)) = 2\sec(2x)\tan(2x)$ .
Derivatives of $e^{f(x)}$ and $\ln f(x)$
$\dfrac{d}{dx}(e^{f(x)}) = f'(x) e^{f(x)}, \qquad \dfrac{d}{dx}(\ln f(x)) = \dfrac{f'(x)}{f(x)}$
$f(x)$
any differentiable function
$f'(x)$
derivative of $f(x)$
When to use
Whenever an exponent or log has anything more complex than $x$ . The chained forms are NOT explicitly in the formula booklet — derive via chain rule each time.
Example
$\dfrac{d}{dx}(\ln(x^{2} + 1)) = \dfrac{2x}{x^{2} + 1}$ .
Parametric differentiation
$\dfrac{dy}{dx} = \dfrac{dy/dt}{dx/dt} \;\; (\text{provided } dx/dt \neq 0)$
$x = x(t), y = y(t)$
parametric equations of the curve
$t$
parameter (often interpreted as time)
When to use
Whenever a curve is given as $\bigl(x(t), y(t)\bigr)$ and you need the gradient. NOT in the formula booklet — derivation via chain rule $\dfrac{dy}{dt} = \dfrac{dy}{dx} \cdot \dfrac{dx}{dt}$ .
Example
$x = 3\cos t$ , $y = \sin 2t$ gives $\dfrac{dy}{dx} = \dfrac{2\cos 2t}{-3\sin t}$ .
Connected rates of change
$\dfrac{dA}{dt} = \dfrac{dA}{dB} \cdot \dfrac{dB}{dt}$
$A, B$
two related quantities
$t$
time (or other independent variable)
When to use
Modelling problems where one rate is known and another is asked for, linked via a static (geometric) formula. NOT in the formula booklet — write down the chain explicitly.
Example
Sphere: $\dfrac{dV}{dt} = 4\pi r^{2} \cdot \dfrac{dr}{dt}$ .

Key Definitions and Keywords — Differentiation 3

Definitions to memorise and the exact keywords mark schemes credit for differentiation 3 answers — sharpened from recent examiner reports for the 2026 Pearson Edexcel IAL XMA01 / YMA01 sitting.

Chain rule
Examiner keyword
$\dfrac{dy}{dx} = \dfrac{dy}{du} \cdot \dfrac{du}{dx}$ — differentiate the outer function, evaluated at the inner, then multiply by the derivative of the inner. The essential tool for composite functions.
Second derivative ( $f''(x)$ or $\dfrac{d^{2}y}{dx^{2}}$ )
Examiner keyword
The derivative of the derivative. Sign at a stationary point: $f''(x_{0}) > 0 \Rightarrow$ local minimum; $f''(x_{0}) < 0 \Rightarrow$ local maximum; $f''(x_{0}) = 0 \Rightarrow$ inconclusive (need a sign-change test).
Concavity
A curve is concave up (or convex) where $f''(x) > 0$ — bends upward like a cup; concave down where $f''(x) < 0$ — bends downward. Points where the concavity changes are points of inflection.
Parametric equations
Examiner keyword
A curve described by $\big(x(t), y(t)\big)$ — coordinates expressed as functions of a parameter $t$ . Differentiation uses $\dfrac{dy}{dx} = \dfrac{dy/dt}{dx/dt}$ .
Connected rates of change
Examiner keyword
Two quantities $A$ , $B$ that change with time, linked by a static relation $A = f(B)$ . The chain rule connects $\dfrac{dA}{dt}$ to $\dfrac{dB}{dt}$ via $\dfrac{dA}{dB}$ .

Common Mistakes and Misconceptions — Differentiation 3

The traps other students keep falling into on differentiation 3 questions — taken from recent Pearson Edexcel IAL XMA01 / YMA01 examiner reports and mark schemes — and how to avoid them.

✕Differentiating $\sin(2x)$ as $\cos(2x)$ (without the factor of $2$ )
WMA13/01 examiner reports — flagged annually
Why it happens
Treating $\sin(2x)$ as if it were $\sin x$ .
How to avoid it
ALWAYS apply the chain rule when the argument is not just $x$ . $\dfrac{d}{dx}\sin(2x) = 2\cos(2x)$ . The inner derivative $\dfrac{du}{dx} = 2$ MUST appear.
✕Using degrees when differentiating trig
Why it happens
Not converting back to radians for calculus.
How to avoid it
The derivatives $\dfrac{d}{dx}\sin x = \cos x$ etc. hold ONLY in radians. If a question gives angles in degrees, convert first.
✕ $\dfrac{d}{dx}\ln(f(x)) = \dfrac{1}{f(x)}$ (forgetting the chain)
Why it happens
Memorising the rule without the chain.
How to avoid it
$\dfrac{d}{dx}\ln(f(x)) = \dfrac{f'(x)}{f(x)}$ . The derivative of the inner is in the numerator. For $f(x) = x$ , $f'(x) = 1$ and you recover $\dfrac{1}{x}$ .
✕Writing $\dfrac{dy}{dx} = \dfrac{dx/dt}{dy/dt}$ (upside down)
Why it happens
Misremembering the order.
How to avoid it
$\dfrac{dy}{dx} = \dfrac{dy/dt}{dx/dt}$ — ' $y$ on top, $x$ on bottom' matches the LHS direction. Mnemonic: cancel the $dt$ .
✕Using the product rule when chain rule is required for connected rates
Why it happens
Confusing two independent rates with one chained relationship.
How to avoid it
Connected rates: ONE chain $\dfrac{dA}{dt} = \dfrac{dA}{dB} \cdot \dfrac{dB}{dt}$ . Write it down BEFORE substituting. If you see a product of two time-dependent factors, that's the product rule, a different setting.

Practice questions

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

Past paper style quiz

Get a report showing which sub-topics you've nailed and which ones still need work.

Video lesson

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

Differentiation 3 — frequently asked questions

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

Differentiation 3

Short Study Notes in the form of Slides

Detailed Study Notes

What you’ll learn

How it’s examined

1Differentiate y=sin⁡(2x2+3)y = \sin(2x^{2} + 3)y=sin(2x2+3) (4 marks, P3)

2Differentiate y=e3x2−1y = e^{3x^{2} - 1}y=e3x2−1 (3 marks, P3)

3Differentiate y=ln⁡(sin⁡x)y = \ln(\sin x)y=ln(sinx) (3 marks, P3)

4Prove ddx(sec⁡x)=sec⁡xtan⁡x\dfrac{d}{dx}(\sec x) = \sec x \tan xdxd​(secx)=secxtanx (4 marks, P3)

5Parametric curve — find dydx\dfrac{dy}{dx}dxdy​ at a parameter value (5 marks, P3)

6Connected rate of change — expanding sphere (6 marks, P3)

Derivatives of trig functions

Derivatives of reciprocal trig

Derivatives of ef(x)e^{f(x)}ef(x) and ln⁡f(x)\ln f(x)lnf(x)

Parametric differentiation

Connected rates of change

Chain rule

Second derivative (f′′(x)f''(x)f′′(x) or d2ydx2\dfrac{d^{2}y}{dx^{2}}dx2d2y​)

Concavity

Parametric equations

Connected rates of change

✕Differentiating sin⁡(2x)\sin(2x)sin(2x) as cos⁡(2x)\cos(2x)cos(2x) (without the factor of 222)

✕Using degrees when differentiating trig

✕ddxln⁡(f(x))=1f(x)\dfrac{d}{dx}\ln(f(x)) = \dfrac{1}{f(x)}dxd​ln(f(x))=f(x)1​ (forgetting the chain)

✕Writing dydx=dx/dtdy/dt\dfrac{dy}{dx} = \dfrac{dx/dt}{dy/dt}dxdy​=dy/dtdx/dt​ (upside down)

✕Using the product rule when chain rule is required for connected rates

Practice questions

Past paper style quiz

Video lesson

Differentiation 3 — frequently asked questions

1Differentiate $y = \sin(2x^{2} + 3)$ (4 marks, P3)

2Differentiate $y = e^{3x^{2} - 1}$ (3 marks, P3)

3Differentiate $y = \ln(\sin x)$ (3 marks, P3)

4Prove $\dfrac{d}{dx}(\sec x) = \sec x \tan x$ (4 marks, P3)

5Parametric curve — find $\dfrac{dy}{dx}$ at a parameter value (5 marks, P3)

Derivatives of $e^{f(x)}$ and $\ln f(x)$

Second derivative ( $f''(x)$ or $\dfrac{d^{2}y}{dx^{2}}$ )

✕Differentiating $\sin(2x)$ as $\cos(2x)$ (without the factor of $2$ )

✕ $\dfrac{d}{dx}\ln(f(x)) = \dfrac{1}{f(x)}$ (forgetting the chain)

✕Writing $\dfrac{dy}{dx} = \dfrac{dx/dt}{dy/dt}$ (upside down)