What is the difference between $\frac{dy}{dx}$ and $\frac{\delta y}{\delta x}$?

$\frac{\delta y}{\delta x}$ (using the Greek letter delta) represents a finite change in y divided by a finite change in x, like the gradient of a chord between two points. $\frac{dy}{dx}$ represents the limit of this ratio as the change in x becomes infinitesimally small. It is the true gradient of the curve at a single point, not an approximation between two points.

Can I use the Product Rule or Quotient Rule in Paper 1?

No. The Product Rule, Quotient Rule, and Chain Rule are part of the P3 (Pure Mathematics 3) syllabus. For P1, you are only expected to differentiate sums of terms of the form $ax^n$. If you see a product or quotient, you must first expand or divide it to get a sum of terms before differentiating.

What does it mean if the gradient $\frac{dy}{dx}$ is negative?

A negative gradient means that the function is decreasing at that point. As you move from left to right along the x-axis, the y-value of the curve is going down. This corresponds to a tangent line that slopes downwards.

How do I find where a curve has a specific gradient, for example, a gradient of 5?

First, find the gradient function by differentiating to get $\frac{dy}{dx}$. Then, set your gradient function equal to the desired gradient. For example, solve the equation $\frac{dy}{dx} = 5$. The solution(s) for $x$ will give you the x-coordinate(s) where the curve has that gradient.

Where can I learn Differentiation for Cambridge 9709?

This free lesson on MarkScheme covers syllabus point 1.7 (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/9709/1-7-differentiation.

Is Differentiation (1.7) 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 differentiation. While you're familiar with the gradient of a straight line being constant, most functions in mathematics are curves with a gradient that changes continuously. Differentiation is the calculus tool that allows us to find the exact gradient of a curve at any given point. This has profound applications, from optimising business profits to modelling the motion of planets.

BY THE END, YOU CAN…

1Understand the concept of the derivative, denoted as dy/dx or f'(x), as the gradient of the tangent to a curve.
2Apply the power rule for differentiation to functions of the form xⁿ, including cases where n is a negative or fractional integer.
3Use differentiation to find the gradient of a curve at a specific point, and hence find the equations of the tangent and the normal at that point.
4Find the coordinates of stationary points on a curve by setting the derivative equal to zero.

In simple terms

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

The Gradient Machine

Differentiation is a tool that gives you a new function, called the derivative, which tells you the steepness (gradient) of your original function at any given point. It's like having a 'gradient-o-meter' for any curve.

THINK OF IT LIKE…

Imagine you're on a rollercoaster. The track is a curve, y = f(x). At any moment, your steepness is changing. Differentiation is like building a device that, given your position 'x' along the track, instantly tells you how steep the track is right at that point. A steep climb has a large positive gradient, a sharp drop has a large negative gradient, and the very top of a hill, just for an instant, is perfectly flat – its gradient is zero.

1
A function y = f(x) has a gradient that varies along the curve.
2
The tangent line at a point shows the instantaneous gradient dy/dx.
3
Where the curve is steep, |dy/dx| is large; at a turning point, dy/dx = 0.
4
Power rule: if y = xⁿ then dy/dx = nxⁿ⁻¹.

Explore the concept

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

INTERACTIVESTEP 1 / 3 · Tangent

Drag the point: the tangent’s steepness is the gradient f′(x).

Key formulas

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

$y = ax^n$

Tap a symbol — great for exam definitions

$m_{\text{normal}} = -\frac{1}{m_{\text{tangent}}}$

Tap a symbol — great for exam definitions

Full topic notes

Formal explanation with the rigour you need for the exam.

The Gradient Function

For a curve given by an equation $y = f(x)$ , its steepness changes as $x$ changes. We can define a new function, called the derivative or the gradient function, which gives us the gradient for any value of $x$ . This function is denoted by $\frac{dy}{dx}$ or $f'(x)$ . For example, if we find that the derivative of a curve is $f'(x) = 2x$ , this means that at $x=1$ , the gradient is $2(1)=2$ , and at $x=-3$ , the gradient is $2(-3)=-6$ . The process of finding this gradient function is called differentiation.

The Power Rule: Your Main Tool for P1

The most fundamental rule for differentiation in this course is the power rule. It provides a simple algorithm for differentiating any term that can be expressed as a power of $x$ . You must be fluent in applying this rule to polynomials, as well as terms involving roots and reciprocals, which must first be converted to index form.

If $y = ax^n$ , then its derivative is $\frac{dy}{dx} = anx^{n-1}$

Multiply by the power: Bring the old power down and multiply it by the coefficient.
Reduce the power by one: Subtract 1 from the old power to get the new power.
Constants: A constant term (e.g., +5) differentiates to 0.
Sums/Differences: Differentiate the function term by term.
Preparation is key: Always rewrite terms like $\sqrt[b]{x^a}$ as $x^{a/b}$ and $\frac{a}{x^b}$ as $ax^{-b}$ before you differentiate.

Finding Equations of Tangents and Normals

A common exam question is to find the equation of a tangent or a normal to a curve at a specific point. The tangent is a line that touches the curve at that point and has the same gradient. The normal is a line perpendicular to the tangent at that same point. The process involves finding the gradient via differentiation and then using the standard formula for the equation of a straight line, $y - y_1 = m(x - x_1)$ .

Gradient of Normal: $m_{\text{normal}} = -\frac{1}{m_{\text{tangent}}}$

EXAM TIP

A very common mistake is to use the tangent gradient when asked for the normal's equation. After you find $\frac{dy}{dx}$ at the point, immediately write down $m_{\text{tangent}} = ...$ and $m_{\text{normal}} = ...$ on your script. This simple step prevents you from mixing them up under pressure.

Worked examples

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

EXAMPLE 1

Given the curve $f(x) = 2x^3 - 5x^2 + x + 9$ . (a) Find $f'(x)$ . (b) Find the gradient of the curve at the point where $x=2$ .

1
(a) Find the derivative $f'(x)$ We differentiate the function term by term using the power rule. $f(x) = 2x^3 - 5x^2 + 1x^1 + 9x^0$ $f'(x) = (3 \times 2)x^{3-1} - (2 \times 5)x^{2-1} + (1 \times 1)x^{1-1} + 0$ $f'(x) = 6x^2 - 10x + 1x^0$ Since $x^0 = 1$ , we have: $f'(x) = 6x^2 - 10x + 1$

EXAMPLE 2

Find the equation of the normal to the curve $y = x^2 + \frac{8}{x}$ at the point where $x=2$ . Give your answer in the form $ax+by+c=0$ , where $a, b$ and $c$ are integers.

1
The solution requires several steps:

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 / 11

What is differentiation?

TAP OR SPACE TO FLIP

ANSWER

It is the process of finding the derivative, or gradient function, of a function. This new function tells you the instantaneous rate of change or the gradient of the original function at any point.

TAP OR SPACE TO FLIP BACK

1 / 11 · ← → space

Key takeaways

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

✓
Multiply by the power: Bring the old power down and multiply it by the coefficient.
✓
Reduce the power by one: Subtract 1 from the old power to get the new power.
✓
Constants: A constant term (e.g., +5) differentiates to 0.
✓
Sums/Differences: Differentiate the function term by term.
✓
Preparation is key: Always rewrite terms like $\sqrt[b]{x^a}$ as $x^{a/b}$ and $\frac{a}{x^b}$ as $ax^{-b}$ before you differentiate.

Practice — then mark it

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

Real past paper[7]

9709/12 · Q5

A curve has equation y = x³ − 6x² + 9x + 2. Find the coordinates of the stationary points and determine their nature.

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 9709/12 · Q5 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

← PREVIOUS1.6 Series NEXT →1.8 Integration

Differentiation

BY THE END, YOU CAN…

1Understand the concept of the derivative, denoted as dy/dx or f'(x), as the gradient of the tangent to a curve.
2Apply the power rule for differentiation to functions of the form xⁿ, including cases where n is a negative or fractional integer.
3Use differentiation to find the gradient of a curve at a specific point, and hence find the equations of the tangent and the normal at that point.
4Find the coordinates of stationary points on a curve by setting the derivative equal to zero.

In simple terms

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

The Gradient Machine

THINK OF IT LIKE…

1
A function y = f(x) has a gradient that varies along the curve.
2
The tangent line at a point shows the instantaneous gradient dy/dx.
3
Where the curve is steep, |dy/dx| is large; at a turning point, dy/dx = 0.
4
Power rule: if y = xⁿ then dy/dx = nxⁿ⁻¹.

Explore the concept

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

INTERACTIVESTEP 1 / 3 · Tangent

Drag the point: the tangent’s steepness is the gradient f′(x).

Key formulas

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

$y = ax^n$

Tap a symbol — great for exam definitions

$m_{\text{normal}} = -\frac{1}{m_{\text{tangent}}}$

Tap a symbol — great for exam definitions

Full topic notes

Formal explanation with the rigour you need for the exam.

The Gradient Function

The Power Rule: Your Main Tool for P1

If $y = ax^n$ , then its derivative is $\frac{dy}{dx} = anx^{n-1}$

Multiply by the power: Bring the old power down and multiply it by the coefficient.
Reduce the power by one: Subtract 1 from the old power to get the new power.
Constants: A constant term (e.g., +5) differentiates to 0.
Sums/Differences: Differentiate the function term by term.
Preparation is key: Always rewrite terms like $\sqrt[b]{x^a}$ as $x^{a/b}$ and $\frac{a}{x^b}$ as $ax^{-b}$ before you differentiate.

Finding Equations of Tangents and Normals

Gradient of Normal: $m_{\text{normal}} = -\frac{1}{m_{\text{tangent}}}$

EXAM TIP

Worked examples

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

EXAMPLE 1

Given the curve $f(x) = 2x^3 - 5x^2 + x + 9$ . (a) Find $f'(x)$ . (b) Find the gradient of the curve at the point where $x=2$ .

1
(a) Find the derivative $f'(x)$ We differentiate the function term by term using the power rule. $f(x) = 2x^3 - 5x^2 + 1x^1 + 9x^0$ $f'(x) = (3 \times 2)x^{3-1} - (2 \times 5)x^{2-1} + (1 \times 1)x^{1-1} + 0$ $f'(x) = 6x^2 - 10x + 1x^0$ Since $x^0 = 1$ , we have: $f'(x) = 6x^2 - 10x + 1$

EXAMPLE 2

Find the equation of the normal to the curve $y = x^2 + \frac{8}{x}$ at the point where $x=2$ . Give your answer in the form $ax+by+c=0$ , where $a, b$ and $c$ are integers.

1
The solution requires several steps:

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 / 11

What is differentiation?

TAP OR SPACE TO FLIP

ANSWER

It is the process of finding the derivative, or gradient function, of a function. This new function tells you the instantaneous rate of change or the gradient of the original function at any point.

TAP OR SPACE TO FLIP BACK

1 / 11 · ← → space

Key takeaways

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

✓
Multiply by the power: Bring the old power down and multiply it by the coefficient.
✓
Reduce the power by one: Subtract 1 from the old power to get the new power.
✓
Constants: A constant term (e.g., +5) differentiates to 0.
✓
Sums/Differences: Differentiate the function term by term.
✓
Preparation is key: Always rewrite terms like $\sqrt[b]{x^a}$ as $x^{a/b}$ and $\frac{a}{x^b}$ as $ax^{-b}$ before you differentiate.

Practice — then mark it

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

Real past paper[7]

9709/12 · Q5

A curve has equation y = x³ − 6x² + 9x + 2. Find the coordinates of the stationary points and determine their nature.

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 9709/12 · Q5 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

← PREVIOUS1.6 Series NEXT →1.8 Integration

Revise Differentiation for Cambridge 9709 (1.7)

In simple terms

Explore the concept

Key formulas

Full topic notes

The Gradient Function

The Power Rule: Your Main Tool for P1

Finding Equations of Tangents and Normals

Worked examples

How it all connects

Glossary

Quick check

Revision flashcards

Key takeaways

Practice — then mark it

9709/12 · Q5

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 Gradient Function

The Power Rule: Your Main Tool for P1

Finding Equations of Tangents and Normals

Worked examples

How it all connects

Glossary

Quick check

Revision flashcards

Key takeaways

Practice — then mark it

9709/12 · Q5

Extra simulations & links

Frequently asked

Checkpoint

Reading it isn’t knowing it — prove it.

Discuss Differentiation