What is differentiation in the context of Edexcel International A Levels Pure Mathematics 4?

Differentiation is a fundamental concept in calculus that involves finding the derivative of a function. In the context of Edexcel International A Levels Pure Mathematics 4, it is used to determine the rate at which a function is changing at any given point. This is crucial for solving problems related to rates of change, optimization, and curve sketching.

How do you differentiate a composite function in Pure Mathematics 4?

To differentiate a composite function, you use the chain rule. The chain rule states that if you have a composite function y = f(g(x)), the derivative is found by differentiating the outer function f with respect to the inner function g, and then multiplying by the derivative of the inner function g with respect to x. This is expressed as dy/dx = f'(g(x)) * g'(x).

What are some common differentiation techniques covered in Pure Mathematics 4?

In Pure Mathematics 4, common differentiation techniques include the power rule, product rule, quotient rule, and chain rule. These rules help in differentiating various types of functions, such as polynomials, trigonometric functions, exponential functions, and logarithmic functions. Mastery of these techniques is essential for solving complex calculus problems.

How is differentiation used to find the maximum and minimum points of a function?

Differentiation is used to find the maximum and minimum points of a function by determining where its derivative is zero or undefined, which indicates potential stationary points. By analyzing the second derivative, you can determine the nature of these stationary points: if the second derivative is positive, the point is a local minimum; if negative, a local maximum. This process is known as the second derivative test.

What role does differentiation play in solving real-world problems in Pure Mathematics 4?

Differentiation plays a crucial role in solving real-world problems by allowing us to model and analyze dynamic systems. It helps in optimizing functions to find maximum efficiency or minimum cost, analyzing motion through velocity and acceleration, and understanding the behavior of physical systems. In Pure Mathematics 4, these applications are explored through various problem-solving exercises that demonstrate the practical utility of differentiation.

Why is "$\dfrac{d}{dx}(y^{2}) = 2y$ — forgetting the $\dfrac{dy}{dx}$" a common mistake in Differentiation?

Why it happens: Students treat y as a constant variable and miss the chain rule. How to avoid it: Every time a y-term is differentiated with respect to x, multiply by dy/dx. Say in your head: 'y is a function of x, so I need the chain rule.' Source: WMA14/01 examiner reports — flagged annually

Why is "Forgetting the product rule on $xy$ terms" a common mistake in Differentiation?

Why it happens: Students see xy and just write y + xdy/dx correctly OR write a single term, missing one half. How to avoid it: Always apply product rule: d/dx(xy) = y + xdy/dx. The standard product rule is unchanged — just one factor brings a dy/dx.

Why is "Writing $\dfrac{d}{dx}(3^{x}) = x \cdot 3^{x-1}$ (treating it like a power function)" a common mistake in Differentiation?

Why it happens: Confusion between x^n (power function, base variable, exponent fixed) and a^x (exponential function, base fixed, exponent variable). How to avoid it: Power function: d/dx(x^n) = nx^n-1. Exponential: d/dx(a^x) = a^x a. Different functions, different rules. Source: WMA14/01 examiner reports — flagged annually

Why is "Substituting $x, y$ values before differentiating" a common mistake in Differentiation?

Why it happens: Students see specific point coordinates and substitute too eagerly. How to avoid it: Differentiate the GENERAL equation first; substitute the point coordinates only AFTER you have the implicit derivative. Otherwise you lose the dependence on the variables you're differentiating.

Why is "Omitting units in connected-rates answers" a common mistake in Differentiation?

Why it happens: Students focus on the numerical computation and forget the physical context. How to avoid it: Volume rates: cm^3/s. Area rates: cm^2/s. Length rates: cm/s. State units in EVERY final answer — examiners deduct A1 for omission.

Pure Mathematics 4Edexcel International A Levels Mathematics (XMA01-YMA01)

Differentiation

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Differentiation (P4) Study Notes — Edexcel IAL Mathematics P4 (WMA14/01, 2018 spec — 2026 onwards)

Implicit differentiation (every $y$ -term picks up a $\dfrac{dy}{dx}$ ), differentiation of $a^{x}$ , inverse-trig derivatives, and connected rates of change. P4 builds on the P3 standard derivatives and extends to relations where $y$ is not isolated. Tangents, normals, and rates problems are heavy with implicit working.

At a glance

Implicit differentiation: $\dfrac{d}{dx}f(y) = f'(y) \dfrac{dy}{dx}$ — chain rule.
Product rule for mixed terms: $\dfrac{d}{dx}(xy) = y + x\dfrac{dy}{dx}$ .
$a^{x}$ derivative: $\dfrac{d}{dx}(a^{x}) = a^{x} \ln a$ .
Inverse trig derivatives (in formula booklet): $\dfrac{d}{dx}\arcsin x = \dfrac{1}{\sqrt{1 - x^{2}}}$ , $\dfrac{d}{dx}\arctan x = \dfrac{1}{1 + x^{2}}$ .
Connected rates: $\dfrac{dA}{dt} = \dfrac{dA}{dr} \cdot \dfrac{dr}{dt}$ .
Tangent at $(a, b)$ : substitute $x = a, y = b$ into the implicit derivative AFTER differentiating.
Units matter in connected-rates final answers.

What you’ll learn

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

P4 4.1 — Differentiate functions defined implicitly.
P4 4.2 — Differentiate exponential functions of the form $a^{x}$ for positive $a \neq 1$ .
P4 4.3 — Use the chain rule to solve connected rates-of-change problems, including those involving implicit relationships.

Implicit differentiation

Differentiate both sides; every $y$ -term picks up a $\dfrac{dy}{dx}$ .

Idea. A curve $F(x, y) = 0$ defines $y$ as a (possibly multi-valued) function of $x$ . To find $\dfrac{dy}{dx}$ without solving for $y$ explicitly, differentiate both sides with respect to $x$ and apply the chain rule whenever a $y$ -term appears.

Chain rule consequence.

$\dfrac{d}{dx}f(y) = f'(y) \cdot \dfrac{dy}{dx}.$

Term	Derivative w.r.t. $x$
$y^{2}$	$2y \dfrac{dy}{dx}$
$y^{3}$	$3y^{2} \dfrac{dy}{dx}$
$\sin y$	$\cos y \cdot \dfrac{dy}{dx}$
$e^{y}$	$e^{y} \dfrac{dy}{dx}$
$\ln y$	$\dfrac{1}{y} \dfrac{dy}{dx}$

Mixed (product) terms need the product rule, with one factor still bringing $\dfrac{dy}{dx}$ .

Term	Derivative w.r.t. $x$
$xy$	$y + x\dfrac{dy}{dx}$
$x^{2}y$	$2xy + x^{2}\dfrac{dy}{dx}$
$xy^{2}$	$y^{2} + 2xy\dfrac{dy}{dx}$
$x^{2}y^{3}$	$2xy^{3} + 3x^{2}y^{2}\dfrac{dy}{dx}$

Worked example 1. Find $\dfrac{dy}{dx}$ for $x^{2} + y^{2} = 25$ .

Differentiate: $2x + 2y \dfrac{dy}{dx} = 0$ .
Solve: $\dfrac{dy}{dx} = -\dfrac{x}{y}$ .

Worked example 2. Find $\dfrac{dy}{dx}$ for $x^{2}y + xy^{2} = 6$ .

Differentiate: $(2xy + x^{2}\dfrac{dy}{dx}) + (y^{2} + 2xy\dfrac{dy}{dx}) = 0$ .
Collect: $(x^{2} + 2xy)\dfrac{dy}{dx} = -(2xy + y^{2})$ .
Solve: $\dfrac{dy}{dx} = -\dfrac{2xy + y^{2}}{x^{2} + 2xy} = -\dfrac{y(2x + y)}{x(x + 2y)}$ .

Worked example 3 — gradient at a point. Find the gradient of $x^{3} + y^{3} = 6xy$ at $(3, 3)$ .

Differentiate: $3x^{2} + 3y^{2} \dfrac{dy}{dx} = 6y + 6x \dfrac{dy}{dx}$ .
Collect: $(3y^{2} - 6x) \dfrac{dy}{dx} = 6y - 3x^{2}$ , so $\dfrac{dy}{dx} = \dfrac{6y - 3x^{2}}{3y^{2} - 6x} = \dfrac{2y - x^{2}}{y^{2} - 2x}$ .
Substitute $(3, 3)$ : $\dfrac{6 - 9}{9 - 6} = \dfrac{-3}{3} = -1$ .

Differentiate term by term; treat $y$ as a function of $x$ .
Every $y^{n}$ becomes $n y^{n-1} \dfrac{dy}{dx}$ .
Mixed $x^{m} y^{n}$ terms need product rule.
Collect $\dfrac{dy}{dx}$ on one side, solve.

Tangent and normal lines on implicit curves

Find the gradient via implicit differentiation, then use point-gradient form.

Strategy.

Implicit-differentiate to get $\dfrac{dy}{dx}$ .
Substitute the given point $(a, b)$ to evaluate the gradient $m$ at that point.
Use $y - b = m(x - a)$ for the tangent.
For the normal, use gradient $-1/m$ .

Worked example. Find the tangent and normal to $x^{2} + y^{2} = 25$ at $(3, 4)$ .

$\dfrac{dy}{dx} = -\dfrac{x}{y}$ (from earlier).
At $(3, 4)$ : $m = -3/4$ .
Tangent: $y - 4 = -\dfrac{3}{4}(x - 3)$ , i.e. $3x + 4y = 25$ .
Normal gradient: $4/3$ . Normal: $y - 4 = \dfrac{4}{3}(x - 3)$ , i.e. $4x - 3y = 0$ .

Vertical tangents. If at $(a, b)$ the gradient $\dfrac{dy}{dx}$ is undefined (e.g. $-x/y$ with $y = 0$ ), the curve has a vertical tangent. Equation $x = a$ .

Horizontal tangents. If $\dfrac{dy}{dx} = 0$ at $(a, b)$ , the tangent is horizontal: $y = b$ .

Stationary point hunting on implicit curves. Set $\dfrac{dy}{dx} = 0$ — this gives an equation in $x, y$ which, combined with the curve equation, locates the stationary points. The curve $x^{2} + y^{2} = 25$ has horizontal tangents where $-x/y = 0$ , i.e. $x = 0$ — points $(0, 5)$ and $(0, -5)$ .

Differentiate, substitute the point, get the gradient.
Tangent: $y - b = m(x - a)$ .
Normal: gradient $-1/m$ .
Vertical tangent when $\dfrac{dy}{dx}$ undefined.

Differentiation of $a^{x}$

Rewrite as $e^{x \ln a}$ and chain-rule.

Derivation. $a^{x} = e^{\ln(a^{x})} = e^{x \ln a}$ . Differentiating:

$\dfrac{d}{dx}(a^{x}) = \dfrac{d}{dx}(e^{x \ln a}) = e^{x \ln a} \cdot \ln a = a^{x} \ln a.$

Standard results.

Function	Derivative
$a^{x}$	$a^{x} \ln a$
$a^{kx}$	$k a^{kx} \ln a$
$a^{f(x)}$	$a^{f(x)} \ln a \cdot f'(x)$

Worked example 1. Differentiate $y = 2^{x}$ . Answer: $\dfrac{dy}{dx} = 2^{x} \ln 2$ .

Worked example 2. Differentiate $y = 5^{3x + 1}$ .

Use the chain rule with $u = 3x + 1$ : $\dfrac{dy}{dx} = 5^{u} \ln 5 \cdot \dfrac{du}{dx} = 5^{3x + 1} \ln 5 \cdot 3 = 3 \cdot 5^{3x + 1} \ln 5$ .

Worked example 3. Find $\dfrac{dy}{dx}$ where $y = x \cdot 2^{x}$ .

Product rule: $\dfrac{dy}{dx} = 2^{x} + x \cdot 2^{x} \ln 2 = 2^{x}(1 + x \ln 2)$ .

Special case $a = e$ . Then $\ln e = 1$ , recovering $\dfrac{d}{dx}(e^{x}) = e^{x}$ .

$\dfrac{d}{dx}(a^{x}) = a^{x} \ln a$ .
Chain rule for $a^{f(x)}$ : multiply by $f'(x)$ .
$\ln a$ is just a constant — keep it as $\ln a$ , don't try to compute a decimal in exact-form answers.

Inverse trigonometric derivatives

Three standard derivatives provided in the formula booklet.

Formula booklet results.

Function	Derivative	Domain
$\arcsin x$	$\dfrac{1}{\sqrt{1 - x^{2}}}$	$
$\arccos x$	$-\dfrac{1}{\sqrt{1 - x^{2}}}$	$
$\arctan x$	$\dfrac{1}{1 + x^{2}}$	all real $x$

Derivation of $\arctan x$ (a typical P4 'prove that' exercise).

Let $y = \arctan x$ . Then $\tan y = x$ with $y \in (-\pi/2, \pi/2)$ .

Differentiate implicitly: $\sec^{2}y \dfrac{dy}{dx} = 1$ .
Use $\sec^{2}y = 1 + \tan^{2}y = 1 + x^{2}$ .
So $\dfrac{dy}{dx} = \dfrac{1}{1 + x^{2}}$ .

Chain rule with inverse trig.

$\dfrac{d}{dx}\arctan(3x) = \dfrac{1}{1 + (3x)^{2}} \cdot 3 = \dfrac{3}{1 + 9x^{2}}$ .
$\dfrac{d}{dx}\arcsin(x^{2}) = \dfrac{1}{\sqrt{1 - x^{4}}} \cdot 2x = \dfrac{2x}{\sqrt{1 - x^{4}}}$ .

Worked example. Differentiate $y = x \arcsin x + \sqrt{1 - x^{2}}$ .

Product rule on $x \arcsin x$ : $\arcsin x + \dfrac{x}{\sqrt{1 - x^{2}}}$ .
Chain rule on $\sqrt{1 - x^{2}} = (1 - x^{2})^{1/2}$ : $\dfrac{1}{2}(1 - x^{2})^{-1/2} \cdot (-2x) = -\dfrac{x}{\sqrt{1 - x^{2}}}$ .
Sum: $\arcsin x + \dfrac{x}{\sqrt{1 - x^{2}}} - \dfrac{x}{\sqrt{1 - x^{2}}} = \arcsin x$ .

$\dfrac{d}{dx}\arcsin x = \dfrac{1}{\sqrt{1 - x^{2}}}$ .
$\dfrac{d}{dx}\arccos x = -\dfrac{1}{\sqrt{1 - x^{2}}}$ .
$\dfrac{d}{dx}\arctan x = \dfrac{1}{1 + x^{2}}$ .
Chain rule: multiply by $f'(x)$ if the argument is $f(x)$ .

Connected rates of change

Use the chain rule to relate time-derivatives of different quantities.

Chain rule. If $A$ depends on $r$ , and $r$ depends on $t$ , then

$\dfrac{dA}{dt} = \dfrac{dA}{dr} \cdot \dfrac{dr}{dt}.$

Spherical balloon archetype. Volume $V = \tfrac{4}{3}\pi r^{3}$ , surface area $S = 4\pi r^{2}$ . Given $\dfrac{dV}{dt}$ , find $\dfrac{dS}{dt}$ .

$\dfrac{dV}{dr} = 4\pi r^{2}$ . Use $\dfrac{dV}{dt} = \dfrac{dV}{dr} \cdot \dfrac{dr}{dt}$ to find $\dfrac{dr}{dt}$ .
$\dfrac{dS}{dr} = 8\pi r$ . Use $\dfrac{dS}{dt} = \dfrac{dS}{dr} \cdot \dfrac{dr}{dt}$ .

Worked example. Spherical balloon, $\dfrac{dV}{dt} = 30 \text{ cm}^{3}/\text{s}$ , $r = 5$ cm.

$\dfrac{dV}{dr} = 4\pi(25) = 100\pi$ .
$\dfrac{dr}{dt} = \dfrac{30}{100\pi} = \dfrac{3}{10\pi}$ cm/s.
$\dfrac{dS}{dr} = 8\pi(5) = 40\pi$ .
$\dfrac{dS}{dt} = 40\pi \cdot \dfrac{3}{10\pi} = 12$ cm $^{2}$ /s.

Conical-tank archetype. Water in a cone, given $\dfrac{dV}{dt}$ , find $\dfrac{dh}{dt}$ .

Cone volume $V = \tfrac{1}{3}\pi r^{2} h$ .
Similar triangles: $r/h = r_{\text{rim}}/h_{\text{rim}}$ (constants), so $r = k h$ for some $k$ .
$V = \tfrac{1}{3}\pi k^{2} h^{3}$ , and $\dfrac{dV}{dh} = \pi k^{2} h^{2}$ .
Apply chain rule.

Implicit rates. When the relationship between variables is implicit (e.g. $x^{2} + xy + y^{2} = 7$ ), differentiate the relation IMPLICITLY with respect to time, treating both $x$ and $y$ as functions of $t$ :

$2x \dfrac{dx}{dt} + \dfrac{dx}{dt} y + x \dfrac{dy}{dt} + 2y \dfrac{dy}{dt} = 0$ .

Substitute known values to find the unknown rate.

$\dfrac{dA}{dt} = \dfrac{dA}{dr} \cdot \dfrac{dr}{dt}$ — chain rule.
Compute geometric derivatives first ( $\dfrac{dV}{dr}, \dfrac{dS}{dr}$ , etc.).
Use given rate to find the bridging rate $\dfrac{dr}{dt}$ .
Always state UNITS in the final answer.

How it’s examined

Differentiation (P4) appears in every WMA14/01 paper — typically Q6 or Q7, $7$ - $10$ marks. Patterns: 'A curve has equation $F(x, y) = 0$ ; find $\dfrac{dy}{dx}$ at $P$ ' (6 marks); 'Find the equation of the tangent at $P$ ' (further 3 marks); 'Show that the derivative of $\arctan x$ is ...' (4-5 marks, implicit method); 'A spherical / conical / cubic body...; find $\dfrac{dS}{dt}$ ' (8 marks, connected rates). Mark schemes are strict on UNITS for rates and on the $\dfrac{dy}{dx}$ multiplier in implicit work.

Sources: Pearson Edexcel International Advanced Level Mathematics specification (2018 — current for 2026 onwards); Edexcel IAL Mathematical Formulae and Statistical Tables (2018) — inverse trig derivatives given; WMA14/01 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

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

1Implicit differentiation — circle (4 marks, P4)
Core• Adapted from WMA14/01 January 2024 Q8• implicit differentiation, circle
Question
A curve has equation $x^{2} + y^{2} = 25$ . Find $\dfrac{dy}{dx}$ in terms of $x$ and $y$ .
Step-by-step solution
1. Step 1
  Differentiate both sides with respect to $x$ . Treat $y$ as a function of $x$ — so $\dfrac{d}{dx}(y^{2}) = 2y \dfrac{dy}{dx}$ by the chain rule.
  $\dfrac{d}{dx}(x^{2}) + \dfrac{d}{dx}(y^{2}) = \dfrac{d}{dx}(25)$
2. Step 2
  Evaluate each piece.
  $2x + 2y \dfrac{dy}{dx} = 0$
3. Step 3
  Solve for $\dfrac{dy}{dx}$ :
  $\dfrac{dy}{dx} = -\dfrac{x}{y}$
Answer
$\dfrac{dy}{dx} = -\dfrac{x}{y}$ .
Examiner tip
M1 for differentiating both sides with respect to $x$ . A1 for the $2y \dfrac{dy}{dx}$ from the chain rule. M1 for isolating $\dfrac{dy}{dx}$ . A1 for the simplified expression. The chain-rule application is the focus: students who write $\dfrac{d}{dx}(y^{2}) = 2y$ (forgetting $\dfrac{dy}{dx}$ ) lose the A1.
2Implicit differentiation with a product term (6 marks, P4)
Extended• Adapted from WMA14/01 June 2024 Q7• implicit differentiation, product rule
Question
Find $\dfrac{dy}{dx}$ for the curve $x^{2}y + 3xy^{2} = 10$ .
Step-by-step solution
1. Step 1
  Differentiate term by term. For $x^{2}y$ , use the product rule: $\dfrac{d}{dx}(x^{2}y) = 2xy + x^{2}\dfrac{dy}{dx}$ .
2. Step 2
  For $3xy^{2}$ , also product rule: $\dfrac{d}{dx}(3xy^{2}) = 3y^{2} + 3x \cdot 2y \dfrac{dy}{dx} = 3y^{2} + 6xy \dfrac{dy}{dx}$ .
3. Step 3
  RHS: $\dfrac{d}{dx}(10) = 0$ . Combine.
  $2xy + x^{2}\dfrac{dy}{dx} + 3y^{2} + 6xy \dfrac{dy}{dx} = 0$
4. Step 4
  Collect $\dfrac{dy}{dx}$ terms on one side:
  $(x^{2} + 6xy)\dfrac{dy}{dx} = -(2xy + 3y^{2})$
5. Step 5
  Solve.
  $\dfrac{dy}{dx} = -\dfrac{2xy + 3y^{2}}{x^{2} + 6xy} = -\dfrac{y(2x + 3y)}{x(x + 6y)}$
Answer
$\dfrac{dy}{dx} = -\dfrac{y(2x + 3y)}{x(x + 6y)}$ (provided $x(x + 6y) \neq 0$ ).
Examiner tip
M1 for using product rule on $x^{2}y$ . A1 for $2xy + x^{2} \dfrac{dy}{dx}$ . M1 for product rule on $3xy^{2}$ . A1 for $3y^{2} + 6xy \dfrac{dy}{dx}$ . M1 for collecting $\dfrac{dy}{dx}$ terms. A1 for the final expression. The product rule must be applied to every mixed term — students often miss one of the two contributions.
3Tangent to an implicit curve (6 marks, P4)
Extended• Adapted from WMA14/01 January 2023 Q7• implicit differentiation, tangent line
Question
The curve $C$ has equation $x^{3} + 3xy^{2} - y^{3} = 5$ . The point $P(1, 2)$ lies on $C$ . Find the equation of the tangent to $C$ at $P$ .
Step-by-step solution
1. Step 1
  Differentiate implicitly. $\dfrac{d}{dx}(x^{3}) = 3x^{2}$ ; $\dfrac{d}{dx}(3xy^{2}) = 3y^{2} + 6xy\dfrac{dy}{dx}$ ; $\dfrac{d}{dx}(y^{3}) = 3y^{2}\dfrac{dy}{dx}$ .
  $3x^{2} + 3y^{2} + 6xy\dfrac{dy}{dx} - 3y^{2}\dfrac{dy}{dx} = 0$
2. Step 2
  Substitute $P(1, 2)$ :
  $3(1)^{2} + 3(4) + 6(1)(2)\dfrac{dy}{dx} - 3(4)\dfrac{dy}{dx} = 0$
3. Step 3
  Simplify: $3 + 12 + (12 - 12)\dfrac{dy}{dx} = 0$ , so $15 + 0 = 0$ .
4. Step 4
  This gives $15 = 0$ , which is impossible — meaning the $\dfrac{dy}{dx}$ coefficient is zero AND there's a remainder. This indicates a vertical tangent (or $P$ does not lie on $C$ ). Re-check: $1 + 3(1)(4) - 8 = 1 + 12 - 8 = 5$ ✓. So $P$ is on the curve and the tangent IS vertical.
5. Step 5
  Vertical tangent at $P(1, 2)$ : equation $x = 1$ .
6. Step 6
  Alternative check: switch variables — differentiate implicitly with respect to $y$ instead, find $\dfrac{dx}{dy} = 0$ at $P$ , confirming the vertical tangent.
Answer
Tangent: $x = 1$ (vertical tangent at $P$ ).
Examiner tip
M1 for implicit differentiation. A1 for each correctly handled implicit term. M1 for substituting $P$ . A1 for recognising the vertical-tangent case. A1 for $x = 1$ . Note: the more typical exam variant gives a non-zero gradient; the vertical case is rarer but tests the deeper concept. Standard variant: gradient $= -5/2$ , tangent $y - 2 = -5/2(x - 1)$ , i.e. $5x + 2y = 9$ . Always show full working.
4Differentiating $a^{x}$ (4 marks, P4)
Core• Adapted from WMA14/01 June 2023 Q6• exponential, differentiation
Question
Find $\dfrac{dy}{dx}$ for $y = 3^{x}$ .
Step-by-step solution
1. Step 1
  Rewrite $3^{x}$ using $e$ : $3^{x} = e^{x \ln 3}$ (since $3 = e^{\ln 3}$ ).
2. Step 2
  Differentiate using the chain rule (with $e^{u}$ , $u = x \ln 3$ ):
  $\dfrac{dy}{dx} = e^{x \ln 3} \cdot \ln 3 = 3^{x} \ln 3$
3. Step 3
  Result.
  $\dfrac{d}{dx}(3^{x}) = 3^{x} \ln 3$
Answer
$\dfrac{dy}{dx} = 3^{x} \ln 3$ .
Examiner tip
M1 for rewriting $a^{x} = e^{x \ln a}$ (or quoting the general result $\dfrac{d}{dx}(a^{x}) = a^{x} \ln a$ ). A1 for the chain-rule factor $\ln 3$ . A1 for $3^{x} \ln 3$ . B1 for clear working. Booklet: this is NOT given — memorise $\dfrac{d}{dx}(a^{x}) = a^{x} \ln a$ .
5Connected rates of change — implicit (6 marks, P4)
Challenge• Adapted from WMA14/01 January 2024 Q9• connected rates, implicit
Question
A spherical balloon is being inflated. Its volume $V$ and surface area $S$ satisfy $V = \dfrac{4}{3}\pi r^{3}$ and $S = 4\pi r^{2}$ . Given that the volume is increasing at $30 \, \text{cm}^{3}/\text{s}$ at the instant when $r = 5 \, \text{cm}$ , find the rate of increase of the surface area at that instant.
Step-by-step solution
1. Step 1
  Use the chain rule: $\dfrac{dS}{dt} = \dfrac{dS}{dr} \cdot \dfrac{dr}{dt}$ .
2. Step 2
  Find $\dfrac{dr}{dt}$ from $V$ : $\dfrac{dV}{dt} = \dfrac{dV}{dr} \cdot \dfrac{dr}{dt}$ . $\dfrac{dV}{dr} = 4\pi r^{2}$ . At $r = 5$ : $\dfrac{dV}{dr} = 100\pi$ . Hence $30 = 100\pi \cdot \dfrac{dr}{dt}$ , so $\dfrac{dr}{dt} = \dfrac{30}{100\pi} = \dfrac{3}{10\pi}$ .
3. Step 3
  Now $\dfrac{dS}{dr} = 8\pi r$ . At $r = 5$ : $\dfrac{dS}{dr} = 40\pi$ .
4. Step 4
  Combine:
  $\dfrac{dS}{dt} = 40\pi \cdot \dfrac{3}{10\pi} = 12 \;\; \text{cm}^{2}/\text{s}$
Answer
$\dfrac{dS}{dt} = 12 \, \text{cm}^{2}/\text{s}$ .
Examiner tip
M1 for the chain rule setup. A1 for $\dfrac{dV}{dr} = 4\pi r^{2}$ . M1 for finding $\dfrac{dr}{dt}$ . A1 for $\dfrac{dr}{dt} = 3/(10\pi)$ . M1 for $\dfrac{dS}{dr} = 8\pi r$ . A1 for $\dfrac{dS}{dt} = 12$ . Units matter — examiners deduct if cm $^{2}$ /s is omitted.
6Derivative of $\arctan x$ (4 marks, P4)
Extended• Adapted from WMA14/01 June 2024 Q6• inverse trig, implicit
Question
Use implicit differentiation to show that $\dfrac{d}{dx}(\arctan x) = \dfrac{1}{1 + x^{2}}$ .
Step-by-step solution
1. Step 1
  Let $y = \arctan x$ , so $\tan y = x$ , with $-\pi/2 < y < \pi/2$ .
2. Step 2
  Differentiate implicitly: $\dfrac{d}{dx}(\tan y) = \dfrac{d}{dx}(x)$ .
  $\sec^{2}y \cdot \dfrac{dy}{dx} = 1$
3. Step 3
  So $\dfrac{dy}{dx} = \dfrac{1}{\sec^{2}y}$ .
4. Step 4
  Use $\sec^{2}y = 1 + \tan^{2}y = 1 + x^{2}$ .
  $\dfrac{dy}{dx} = \dfrac{1}{1 + x^{2}}$
Answer
$\dfrac{d}{dx}(\arctan x) = \dfrac{1}{1 + x^{2}}$ . $\square$
Examiner tip
B1 for the substitution $\tan y = x$ . M1 for implicit differentiation. A1 for $\sec^{2}y \dfrac{dy}{dx} = 1$ . M1 for using $\sec^{2}y = 1 + \tan^{2}y$ . A1 for the final form $\dfrac{1}{1 + x^{2}}$ . Booklet: inverse-trig derivatives ARE given — students can quote $\dfrac{d}{dx}(\arctan x) = \dfrac{1}{1 + x^{2}}$ directly, but exam often asks 'prove' to test implicit-differentiation technique.

Key Formulae — Differentiation

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

Implicit differentiation rule
$\dfrac{d}{dx}\left[f(y)\right] = f'(y) \cdot \dfrac{dy}{dx}$
$y$
implicit function of $x$
$f'(y)$
derivative of $f$ with respect to $y$
When to use
Every time a $y$ -term is differentiated with respect to $x$ — multiply by $\dfrac{dy}{dx}$ . NOT in the formula booklet — this is the chain rule.
Example
$\dfrac{d}{dx}(y^{2}) = 2y \dfrac{dy}{dx}$ ; $\dfrac{d}{dx}(\sin y) = \cos y \cdot \dfrac{dy}{dx}$ .
Derivative of $a^{x}$
$\dfrac{d}{dx}(a^{x}) = a^{x} \ln a$
$a$
positive constant base, $a > 0$ , $a \neq 1$
When to use
Whenever you differentiate an exponential function whose base is NOT $e$ . IS in the IAL formula booklet but easy to forget — memorise.
Example
$\dfrac{d}{dx}(3^{x}) = 3^{x} \ln 3$ ; $\dfrac{d}{dx}(2^{x}) = 2^{x} \ln 2$ .
Derivative of $\arctan x$
$\dfrac{d}{dx}(\arctan x) = \dfrac{1}{1 + x^{2}}$
$\arctan x$
inverse tangent function
When to use
For inverse-trig derivatives. IS in the IAL formula booklet (table of inverse trig derivatives provided).
Example
$\dfrac{d}{dx}\arctan(2x) = \dfrac{1}{1 + (2x)^{2}} \cdot 2 = \dfrac{2}{1 + 4x^{2}}$ by chain rule.
Derivatives of $\arcsin x$ and $\arccos x$
$\dfrac{d}{dx}(\arcsin x) = \dfrac{1}{\sqrt{1 - x^{2}}}; \quad \dfrac{d}{dx}(\arccos x) = -\dfrac{1}{\sqrt{1 - x^{2}}}$
$x$
input with $|x| < 1$
When to use
For inverse sine and cosine derivatives, valid for $|x| < 1$ . Provided in the IAL formula booklet.
Example
$\dfrac{d}{dx}\arcsin(3x) = \dfrac{3}{\sqrt{1 - 9x^{2}}}$ via the chain rule.
Connected rates of change
$\dfrac{dA}{dt} = \dfrac{dA}{dr} \cdot \dfrac{dr}{dt}$
$A$
quantity of interest (volume, area, ...)
$r$
intermediate variable (radius, length, ...)
$t$
time
When to use
For 'rate of change' problems where one quantity changes with time via an intermediate variable. Chain rule. NOT specifically in the formula booklet — generic chain-rule application.
Example
Spherical balloon: $\dfrac{dV}{dt} = 4\pi r^{2} \cdot \dfrac{dr}{dt}$ .

Key Definitions and Keywords — Differentiation

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

Implicit function
Examiner keyword
A relation $F(x, y) = 0$ that defines $y$ as a function of $x$ without giving a closed form $y = f(x)$ . Example: $x^{2} + y^{2} = 25$ defines $y$ as $\pm\sqrt{25 - x^{2}}$ — implicit because we don't see $y = \ldots$ written explicitly.
Implicit differentiation
Examiner keyword
Differentiating both sides of an equation $F(x, y) = 0$ with respect to $x$ , treating $y$ as a function of $x$ (so the chain rule applies: $\dfrac{d}{dx}f(y) = f'(y) \dfrac{dy}{dx}$ ). The resulting expression involves both $x$ , $y$ , and $\dfrac{dy}{dx}$ .
Connected rates of change
Examiner keyword
Problems where two or more quantities depend on time and are related to each other. The chain rule relates their time-derivatives.
Example
Volume and surface area of a sphere are connected via the radius — knowing $dV/dt$ lets you find $dS/dt$ .
Vertical tangent
A tangent line at a point where $\dfrac{dy}{dx}$ is undefined (denominator vanishes, numerator does not). The tangent is vertical: equation $x = a$ for some constant $a$ .
Example
On $x^{2} + y^{2} = 25$ , at $(5, 0)$ : $\dfrac{dy}{dx} = -x/y$ is undefined (division by zero) — tangent is $x = 5$ .

Common Mistakes and Misconceptions — Differentiation

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

✕ $\dfrac{d}{dx}(y^{2}) = 2y$ — forgetting the $\dfrac{dy}{dx}$
WMA14/01 examiner reports — flagged annually
Why it happens
Students treat $y$ as a constant variable and miss the chain rule.
How to avoid it
Every time a $y$ -term is differentiated with respect to $x$ , multiply by $\dfrac{dy}{dx}$ . Say in your head: ' $y$ is a function of $x$ , so I need the chain rule.'
✕Forgetting the product rule on $xy$ terms
Why it happens
Students see $xy$ and just write $y + x\dfrac{dy}{dx}$ correctly OR write a single term, missing one half.
How to avoid it
Always apply product rule: $\dfrac{d}{dx}(xy) = y + x\dfrac{dy}{dx}$ . The standard product rule is unchanged — just one factor brings a $\dfrac{dy}{dx}$ .
✕Writing $\dfrac{d}{dx}(3^{x}) = x \cdot 3^{x-1}$ (treating it like a power function)
WMA14/01 examiner reports — flagged annually
Why it happens
Confusion between $x^{n}$ (power function, base variable, exponent fixed) and $a^{x}$ (exponential function, base fixed, exponent variable).
How to avoid it
Power function: $\dfrac{d}{dx}(x^{n}) = nx^{n-1}$ . Exponential: $\dfrac{d}{dx}(a^{x}) = a^{x} \ln a$ . Different functions, different rules.
✕Substituting $x, y$ values before differentiating
Why it happens
Students see specific point coordinates and substitute too eagerly.
How to avoid it
Differentiate the GENERAL equation first; substitute the point coordinates only AFTER you have the implicit derivative. Otherwise you lose the dependence on the variables you're differentiating.
✕Omitting units in connected-rates answers
Why it happens
Students focus on the numerical computation and forget the physical context.
How to avoid it
Volume rates: cm $^{3}$ /s. Area rates: cm $^{2}$ /s. Length rates: cm/s. State units in EVERY final answer — examiners deduct A1 for omission.

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 — frequently asked questions

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

Differentiation

Short Study Notes in the form of Slides

Detailed Study Notes

What you’ll learn

How it’s examined

1Implicit differentiation — circle (4 marks, P4)

2Implicit differentiation with a product term (6 marks, P4)

3Tangent to an implicit curve (6 marks, P4)

4Differentiating axa^{x}ax (4 marks, P4)

5Connected rates of change — implicit (6 marks, P4)

6Derivative of arctan⁡x\arctan xarctanx (4 marks, P4)

Implicit differentiation rule

Derivative of axa^{x}ax

Derivative of arctan⁡x\arctan xarctanx

Derivatives of arcsin⁡x\arcsin xarcsinx and arccos⁡x\arccos xarccosx

Connected rates of change

Implicit function

Implicit differentiation

Connected rates of change

Vertical tangent

✕ddx(y2)=2y\dfrac{d}{dx}(y^{2}) = 2ydxd​(y2)=2y — forgetting the dydx\dfrac{dy}{dx}dxdy​

✕Forgetting the product rule on xyxyxy terms

✕Writing ddx(3x)=x⋅3x−1\dfrac{d}{dx}(3^{x}) = x \cdot 3^{x-1}dxd​(3x)=x⋅3x−1 (treating it like a power function)

✕Substituting x,yx, yx,y values before differentiating

✕Omitting units in connected-rates answers

Practice questions

Past paper style quiz

Video lesson

Differentiation — frequently asked questions

4Differentiating $a^{x}$ (4 marks, P4)

6Derivative of $\arctan x$ (4 marks, P4)

Derivative of $a^{x}$

Derivative of $\arctan x$

Derivatives of $\arcsin x$ and $\arccos x$

✕ $\dfrac{d}{dx}(y^{2}) = 2y$ — forgetting the $\dfrac{dy}{dx}$

✕Forgetting the product rule on $xy$ terms

✕Writing $\dfrac{d}{dx}(3^{x}) = x \cdot 3^{x-1}$ (treating it like a power function)

✕Substituting $x, y$ values before differentiating