What is connected rates of change?

When two quantities both change with time, the chain rule connects their rates: dVdt = dVdrdrdt. This lets you find one rate from another, for example the rate a radius grows from the rate a volume is added.

Differentiate (d)/(dx)y^3 implicitly. [1 mark]

CCEA CCEA 2021-style practice: Differentiate y = x^2 e^3x, and find the x-coordinates of any stationary points.

Use the product rule with u = x^2 and v = e^3x, so u' = 2x and v' = 3e^3x: (dy)/(dx) = u'v + uv' = 2x\,e^3x + x^2(3e^3x) = e^3x(2x + 3x^2) = x e^3x(2 + 3x). Stationary points where (dy)/(dx) = 0: since e^3x 0, either x = 0 or 2 + 3x = 0, giving x = 0 or x = -(2)/(3). Markers reward the product rule, factoring out e^3x, and both stationary x-values.

CCEA CCEA 2019-style practice: A curve has equation x^2 + xy + y^2 = 7. Find (dy)/(dx) in terms of x and y by implicit differentiation.

Differentiate each term with respect to x, using the product rule on xy: 2x + (y + x(dy)/(dx)) + 2y(dy)/(dx) = 0. Collect the (dy)/(dx) terms: x(dy)/(dx) + 2y(dy)/(dx) = -2x - y, so (dy)/(dx)(x + 2y) = -(2x + y). Therefore (dy)/(dx) = -(2x + y)/(x + 2y). Markers reward differentiating y^2 as 2y(dy)/(dx), the product rule on xy, collecting terms, and the final expression.

Northern IrelandMathsSyllabus dot point

How do the chain, product and quotient rules let us differentiate any combination of functions, including implicitly?

Differentiating exponential, logarithmic and trigonometric functions, the chain, product and quotient rules, implicit differentiation, and connected rates of change.

A CCEA A-Level Mathematics answer on differentiating exponential, logarithmic and trigonometric functions, the chain, product and quotient rules, implicit differentiation, and using the chain rule for connected rates of change.

Generated by Claude Opus 4.813 min answerUpdated 2026-06-16

Reviewed by: AI editorial process; not yet individually human-reviewed

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What this dot point is asking

CCEA wants you to differentiate exponential, logarithmic and trigonometric functions, apply the chain, product and quotient rules, differentiate implicitly when $y$ is not isolated, and use the chain rule to connect rates of change. This is the heart of A2 calculus and supports every applied differentiation problem.

The answer

Differentiating standard functions

The chain, product and quotient rules

So $\frac{d}{dx}e^{3x} = 3e^{3x}$ (chain rule), and $\frac{d}{dx}(x\sin x) = \sin x + x\cos x$ (product rule). The skill is recognising which rule a function needs: a function of a function (such as $\sin(x^2)$ ) calls for the chain rule, a clear product (such as $x^2\ln x$ ) calls for the product rule, and a single fraction with a variable denominator (such as $\frac{x}{x+1}$ ) calls for the quotient rule. Many questions combine two rules, for example differentiating $x^2\sin 3x$ uses the product rule with the chain rule inside it, so identifying the structure first is half the work.

Implicit differentiation

Connected rates of change

When two quantities both change with time, the chain rule connects their rates: $\dfrac{dV}{dt} = \dfrac{dV}{dr}\dfrac{dr}{dt}$ . This lets you find one rate from another, for example the rate a radius grows from the rate a volume is added.

Worked example: a connected rate of change

Find how fast a radius grows

Air is pumped into a spherical balloon at $50\,\text{cm}^3\,\text{s}^{-1}$ . Find the rate of increase of the radius when $r = 5\,\text{cm}$ . The volume is $V = \tfrac{4}{3}\pi r^3$ .

step Differentiate the volume

\frac{dV}{dr} = 4\pi r^2.

step Apply the chain rule

\frac{dV}{dt} = \frac{dV}{dr}\frac{dr}{dt} \Rightarrow 50 = 4\pi r^2 \frac{dr}{dt}.

step Solve at r = 5

\frac{dr}{dt} = \frac{50}{4\pi (5)^2} = \frac{50}{100\pi} = \frac{1}{2\pi} \approx 0.159\,\text{cm s}^{-1}.

The chain rule links the known volume rate to the unknown radius rate.

Examples in context

Example 1. Cooling and growth models. A quantity decaying as $T = T_0 e^{-kt}$ has rate $\frac{dT}{dt} = -kT_0 e^{-kt} = -kT$ , proportional to itself. Differentiating exponentials shows why such models have a constant proportional rate.

Example 2. A sliding ladder. As the foot of a ladder slides out at a known rate, the top slides down at a rate found by differentiating $x^2 + y^2 = L^2$ implicitly with respect to time. Implicit differentiation and connected rates combine in this classic problem.

Try this

Q1. Differentiate $y = e^{5x}$ . [1 mark]

Cue. $\frac{dy}{dx} = 5e^{5x}$ .

Q2. Differentiate $y = x^2\ln x$ . [2 marks]

Cue. Product rule: $2x\ln x + x^2 \cdot \frac{1}{x} = 2x\ln x + x$ .

Q3. Differentiate $\frac{d}{dx}y^3$ implicitly. [1 mark]

Cue. $3y^2\frac{dy}{dx}$ .

Exam-style practice questions

Practice questions written in the style of CCEA exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

CCEA 20216 marksDifferentiate

y = x^2 e^{3x}

, and find the

x

-coordinates of any stationary points.

Show worked answer →

Use the product rule with $u = x^2$ and $v = e^{3x}$ , so $u' = 2x$ and $v' = 3e^{3x}$ :

$\frac{dy}{dx} = u'v + uv' = 2x\,e^{3x} + x^2(3e^{3x}) = e^{3x}(2x + 3x^2) = x e^{3x}(2 + 3x).$

Stationary points where $\frac{dy}{dx} = 0$ : since $e^{3x} \ne 0$ , either $x = 0$ or $2 + 3x = 0$ , giving $x = 0$ or $x = -\frac{2}{3}$ .

Markers reward the product rule, factoring out $e^{3x}$ , and both stationary $x$ -values.

CCEA 20196 marksA curve has equation

x^2 + xy + y^2 = 7

. Find

\frac{dy}{dx}

in terms of

x

and

y

by implicit differentiation.

Show worked answer →

Differentiate each term with respect to $x$ , using the product rule on $xy$ :

$2x + \left(y + x\frac{dy}{dx}\right) + 2y\frac{dy}{dx} = 0.$

Collect the $\frac{dy}{dx}$ terms:

$x\frac{dy}{dx} + 2y\frac{dy}{dx} = -2x - y,$ so $\frac{dy}{dx}(x + 2y) = -(2x + y).$

Therefore $\frac{dy}{dx} = -\frac{2x + y}{x + 2y}$ .

Markers reward differentiating $y^2$ as $2y\frac{dy}{dx}$ , the product rule on $xy$ , collecting terms, and the final expression.

Related dot points

Sources & how we know this

CCEA GCE Mathematics specification — CCEA (2018)