When do I use the product rule vs. the chain rule?

Use the product rule when you are differentiating a function that is one function MULTIPLIED by another, like $f(x) = x^2 e^x$. Use the chain rule when you have a function INSIDE another function (a composite function), like $f(x) = e^{(x^2)}$. Sometimes you need both, for example in $f(x) = x^2 e^{3x}$ you use the product rule overall, but need the chain rule to differentiate the $e^{3x}$ part.

Is there an easier way than the quotient rule?

Sometimes, yes. You can often rewrite a quotient as a product and use the product rule. For example, $y = \frac{\ln x}{x^2}$ can be written as $y = (\ln x)(x^{-2})$. Differentiating this with the product rule can sometimes feel simpler and lead to fewer sign errors. However, you must be comfortable with both methods.

What's the point of the second derivative test if it can be inconclusive?

The second derivative test is usually much faster than testing points on either side of the stationary point. It's only inconclusive in specific cases (like $y=x^3$ or $y=x^4$ at $x=0$). For most exam questions, it works perfectly. You just need to remember the backup method (testing the sign of $f'(x)$) for when $f''(x)=0$.

In connected rates of change, how do I know which variable to eliminate?

Look at the rate you are given and the rate you need to find. For example, if you are given $\frac{dV}{dt}$ and need to find $\frac{dh}{dt}$, your chain rule will be $\frac{dh}{dt} = \frac{dh}{dV} \times \frac{dV}{dt}$. This tells you that you need a relationship between $h$ and $V$. Therefore, you should write your volume formula $V$ in terms of $h$ only, eliminating other variables like the radius $r$.

Where can I learn Differentiation for Cambridge 9709?

This free lesson on MarkScheme covers syllabus point 2.4 (Differentiation) for Cambridge A-Level Mathematics. It includes visual revision, key points, and links to real past-paper questions with mark-scheme marking at markscheme.app/courses/9709/2-4-differentiation.

Is Differentiation (2.4) on 9709 free to study?

Yes. The Mathematics 9709 course on MarkScheme is 100% free — including premium visual lessons, exam tips, and past-paper marking. No paywall for Differentiation.

Differentiation 9709 — Free A-Level Maths

Differentiation

Welcome to the next level of differentiation. In Pure Mathematics 1, you mastered the basics of finding gradients and locating stationary points for polynomials. Now, in Pure Mathematics 2, we equip you with more powerful tools to differentiate a wider range of functions, including products, quotients, and combinations involving exponential and logarithmic functions. These skills are essential for solving more complex and realistic problems.

BY THE END, YOU CAN…

1Apply the product rule and quotient rule to differentiate complex functions.
2Locate stationary points and determine their nature using the second derivative test.
3Solve problems involving connected rates of change using the chain rule.
4Formulate and solve optimisation problems by finding and verifying maximum or minimum values.

In simple terms

A friendly intro before the formal notes — no formulas yet.

The Calculus of Change

Differentiation finds the instantaneous rate of change of a function. In P2, we learn powerful new rules to handle more complex functions and apply them to real-world problems like optimisation and related rates.

THINK OF IT LIKE…

Imagine you're filling a strangely shaped container with water. Differentiation tells you how quickly the water level is rising at any specific moment, even if the container's walls are curved. Connected rates of change is like knowing how fast you're pouring water in (volume per second) and using that to calculate how fast the water level (height per second) is rising.

1
P2: product and quotient rules; chain rule for composites.
2
Stationary points: f′(x) = 0 — max, min, or point of inflection.
3
Connected rates of change — link dy/dt and dx/dt.
4
Optimisation problems — form equation, differentiate, verify max/min.

Explore the concept

Use the live diagram and synced steps — play it or tap a step card to walk through.

Key formulas

Tap any symbol to reveal exactly what it means and its units.

y = uv

Tap a symbol — great for exam definitions

$\frac{dy}{dt} = \frac{dy}{dx} \times \frac{dx}{dt}$

Tap a symbol — great for exam definitions

Full topic notes

Formal explanation with the rigour you need for the exam.

The Product and Quotient Rules

When a function is formed by multiplying two other functions together (e.g., $y = x^2 \sin x$ ), we cannot simply differentiate each part and multiply the results. We must use the Product Rule. Similarly, when a function is a fraction with functions in the numerator and denominator (e.g., $y = \frac{e^x}{x+1}$ ), we need the Quotient Rule.

Product Rule: If $y = uv$ , then $\frac{dy}{dx} = u\frac{dv}{dx} + v\frac{du}{dx}$

Quotient Rule: If $y = \frac{u}{v}$ , then $\frac{dy}{dx} = \frac{v\frac{du}{dx} - u\frac{dv}{dx}}{v^2}$

Stationary Points and Their Nature

As before, stationary points occur where the gradient of the curve is zero, i.e., where $\frac{dy}{dx} = 0$ . These points can be local maxima, local minima, or points of inflection. To determine the nature of a stationary point, we use the second derivative test. This involves finding the second derivative, $\frac{d^2y}{dx^2}$ or $f''(x)$ , and evaluating its sign at the stationary point.

Find stationary points by solving $\frac{dy}{dx} = 0$ .
Calculate the second derivative, $\frac{d^2y}{dx^2}$ .
At a stationary point $x=a$ : If $\frac{d^2y}{dx^2} > 0$ , it is a local minimum.
At a stationary point $x=a$ : If $\frac{d^2y}{dx^2} < 0$ , it is a local maximum.
If $\frac{d^2y}{dx^2} = 0$ , the test is inconclusive. You must test the sign of $\frac{dy}{dx}$ on either side of the point.

Connected Rates of Change

In many real-world scenarios, several quantities change with time and are related to each other. For example, the volume, radius, and surface area of an inflating spherical balloon all change with time. Connected rates of change problems use the chain rule to find the rate of change of one quantity when the rate of change of another is known.

The Chain Rule for rates: $\frac{dy}{dt} = \frac{dy}{dx} \times \frac{dx}{dt}$

EXAM TIP

In connected rates of change problems, always start by writing down what you know (e.g., $\frac{dV}{dt} = 5$ ) and what you want to find (e.g., $\frac{dr}{dt}$ ). Then, write down the chain rule relationship that connects these rates, such as $\frac{dr}{dt} = \frac{dr}{dV} \times \frac{dV}{dt}$ . This helps structure your solution.

Worked examples

See the formulas applied — reveal one step at a time, like the exam.

EXAMPLE 1

A curve has equation $y = \frac{\ln x}{x^2}$ . Find the exact coordinates of its stationary point and determine its nature.

1
The function is a quotient, so we use the quotient rule with $u = \ln x$ and $v = x^2$ . $u' = \frac{du}{dx} = \frac{1}{x}$ and $v' = \frac{dv}{dx} = 2x$ .

EXAMPLE 2

A water tank is in the shape of an inverted cone with base radius 2 m and height 6 m. Water is being pumped into the tank at a constant rate of $0.1 \text{ m}^3\text{min}^{-1}$ . Find the rate at which the water level is rising when the depth of the water is 3 m. (The volume of a cone is $V = \frac{1}{3}\pi r^2 h$ ).

1
Let $h$ be the depth of the water and $r$ be the radius of the water surface at time $t$ . We are given $\frac{dV}{dt} = 0.1$ . We want to find $\frac{dh}{dt}$ when $h=3$ .

How it all connects

The big idea sits in the middle — tap a linked idea to explore the link.

MAIN IDEADifferentiation

Tap a linked idea to see how it connects back to the main topic — that connection is what examiners reward.

Glossary

Try to recall each definition before you reveal it.

Quick check

Answer in your head first — then tap to check. No pressure.

Revision flashcards

Flip the card. Test yourself before the exam.

QUESTION · 1 / 10

What is the Product Rule for differentiation?

TAP OR SPACE TO FLIP

ANSWER

If $y = uv$ , where $u$ and $v$ are functions of $x$ , then $\frac{dy}{dx} = u\frac{dv}{dx} + v\frac{du}{dx}$ . In words: 'first times derivative of the second, plus second times derivative of the first'.

TAP OR SPACE TO FLIP BACK

1 / 10 · ← → space

Key takeaways

Review these before you close the topic — retrieval beats re-reading.

✓
Find stationary points by solving $\frac{dy}{dx} = 0$ .
✓
Calculate the second derivative, $\frac{d^2y}{dx^2}$ .
✓
At a stationary point $x=a$ : If $\frac{d^2y}{dx^2} > 0$ , it is a local minimum.
✓
At a stationary point $x=a$ : If $\frac{d^2y}{dx^2} < 0$ , it is a local maximum.
✓
If $\frac{d^2y}{dx^2} = 0$ , the test is inconclusive. You must test the sign of $\frac{dy}{dx}$ on either side of the point.

Practice — then mark it

The whole point: a real Cambridge question, marked mark-by-mark.

Real past paper[0]

Practice Differentiation Problems

Do it on paper → mark itMARKED MARK-BY-MARK · B1 / M1 / A1 · OFFICIAL SCHEME

Extra simulations & links

PhET, GeoGebra and other curated tools — open in a new tab.

Frequently asked

✓

Checkpoint

One marked question is worth ten re-reads — close the loop before you move on.

Checkpoint

Reading it isn’t knowing it — prove it.

Before you move on: do Practice Differentiation Problems on paper, snap a photo, and get examiner-style feedback on exactly where you win and lose marks.

Attempt & get marked →Takes about a minute · you’ll land back on this lesson

← PREVIOUS2.3 Trigonometry NEXT →2.5 Integration

Differentiation

BY THE END, YOU CAN…

1Apply the product rule and quotient rule to differentiate complex functions.
2Locate stationary points and determine their nature using the second derivative test.
3Solve problems involving connected rates of change using the chain rule.
4Formulate and solve optimisation problems by finding and verifying maximum or minimum values.

In simple terms

A friendly intro before the formal notes — no formulas yet.

The Calculus of Change

THINK OF IT LIKE…

1
P2: product and quotient rules; chain rule for composites.
2
Stationary points: f′(x) = 0 — max, min, or point of inflection.
3
Connected rates of change — link dy/dt and dx/dt.
4
Optimisation problems — form equation, differentiate, verify max/min.

Explore the concept

Use the live diagram and synced steps — play it or tap a step card to walk through.

Key formulas

Tap any symbol to reveal exactly what it means and its units.

y = uv

Tap a symbol — great for exam definitions

$\frac{dy}{dt} = \frac{dy}{dx} \times \frac{dx}{dt}$

Tap a symbol — great for exam definitions

Full topic notes

Formal explanation with the rigour you need for the exam.

The Product and Quotient Rules

Product Rule: If $y = uv$ , then $\frac{dy}{dx} = u\frac{dv}{dx} + v\frac{du}{dx}$

Quotient Rule: If $y = \frac{u}{v}$ , then $\frac{dy}{dx} = \frac{v\frac{du}{dx} - u\frac{dv}{dx}}{v^2}$

Stationary Points and Their Nature

Find stationary points by solving $\frac{dy}{dx} = 0$ .
Calculate the second derivative, $\frac{d^2y}{dx^2}$ .
At a stationary point $x=a$ : If $\frac{d^2y}{dx^2} > 0$ , it is a local minimum.
At a stationary point $x=a$ : If $\frac{d^2y}{dx^2} < 0$ , it is a local maximum.
If $\frac{d^2y}{dx^2} = 0$ , the test is inconclusive. You must test the sign of $\frac{dy}{dx}$ on either side of the point.

Connected Rates of Change

The Chain Rule for rates: $\frac{dy}{dt} = \frac{dy}{dx} \times \frac{dx}{dt}$

EXAM TIP

Worked examples

See the formulas applied — reveal one step at a time, like the exam.

EXAMPLE 1

A curve has equation $y = \frac{\ln x}{x^2}$ . Find the exact coordinates of its stationary point and determine its nature.

1
The function is a quotient, so we use the quotient rule with $u = \ln x$ and $v = x^2$ . $u' = \frac{du}{dx} = \frac{1}{x}$ and $v' = \frac{dv}{dx} = 2x$ .

EXAMPLE 2

1
Let $h$ be the depth of the water and $r$ be the radius of the water surface at time $t$ . We are given $\frac{dV}{dt} = 0.1$ . We want to find $\frac{dh}{dt}$ when $h=3$ .

How it all connects

The big idea sits in the middle — tap a linked idea to explore the link.

MAIN IDEADifferentiation

Tap a linked idea to see how it connects back to the main topic — that connection is what examiners reward.

Glossary

Try to recall each definition before you reveal it.

Quick check

Answer in your head first — then tap to check. No pressure.

Revision flashcards

Flip the card. Test yourself before the exam.

QUESTION · 1 / 10

What is the Product Rule for differentiation?

TAP OR SPACE TO FLIP

ANSWER

TAP OR SPACE TO FLIP BACK

1 / 10 · ← → space

Key takeaways

Review these before you close the topic — retrieval beats re-reading.

✓
Find stationary points by solving $\frac{dy}{dx} = 0$ .
✓
Calculate the second derivative, $\frac{d^2y}{dx^2}$ .
✓
At a stationary point $x=a$ : If $\frac{d^2y}{dx^2} > 0$ , it is a local minimum.
✓
At a stationary point $x=a$ : If $\frac{d^2y}{dx^2} < 0$ , it is a local maximum.
✓
If $\frac{d^2y}{dx^2} = 0$ , the test is inconclusive. You must test the sign of $\frac{dy}{dx}$ on either side of the point.

Practice — then mark it

The whole point: a real Cambridge question, marked mark-by-mark.

Real past paper[0]

Practice Differentiation Problems

Do it on paper → mark itMARKED MARK-BY-MARK · B1 / M1 / A1 · OFFICIAL SCHEME

Extra simulations & links

PhET, GeoGebra and other curated tools — open in a new tab.

Frequently asked

✓

Checkpoint

One marked question is worth ten re-reads — close the loop before you move on.

Checkpoint

Reading it isn’t knowing it — prove it.

Before you move on: do Practice Differentiation Problems on paper, snap a photo, and get examiner-style feedback on exactly where you win and lose marks.

Attempt & get marked →Takes about a minute · you’ll land back on this lesson

← PREVIOUS2.3 Trigonometry NEXT →2.5 Integration

Revise Differentiation for Cambridge 9709 (2.4)

In simple terms

Explore the concept

Key formulas

Full topic notes

The Product and Quotient Rules

Stationary Points and Their Nature

Connected Rates of Change

Worked examples

How it all connects

Glossary

Quick check

Revision flashcards

Key takeaways

Practice — then mark it

Practice Differentiation Problems

Extra simulations & links

Frequently asked

Checkpoint

Reading it isn’t knowing it — prove it.

In simple terms

Explore the concept

Key formulas

Full topic notes

The Product and Quotient Rules

Stationary Points and Their Nature

Connected Rates of Change

Worked examples

How it all connects

Glossary

Quick check

Revision flashcards

Key takeaways

Practice — then mark it

Practice Differentiation Problems

Extra simulations & links

Frequently asked

Checkpoint

Reading it isn’t knowing it — prove it.

Discuss Differentiation