Library › Differentiation

Additional Mathematics

Differentiation

Revise Differentiation for Additional Mathematics — revision notes and instant AI marking. Free to start.

📖 Revision notes · preview

Cambridge (CIE) IGCSE — Additional Maths

Differentiation

The Big Idea: Differentiation is just a machine that turns "how a curve is shaped" into "how steep it is at any single point" — and once you know the steepness everywhere, you can find maximums, minimums, tangents, and rates of change for almost anything.

Summary — What This Chapter Covers

The gradient of a curve at a point = the gradient of the tangent at that point.
The derivative (written dy/dx) is a formula that gives you the gradient at any x-value.
Power rule: bring the power down, subtract one from it.
Special derivatives for sin, cos, tan, eˣ, and ln x — memorise these cold.
Chain rule for "function of a function" (composite functions).
Product rule for two functions multiplied together.
Quotient rule for one function divided by another.
Using derivatives to find tangents, normals, and whether a function is increasing/decreasing.
Second derivatives to classify stationary points as maximum, minimum, or point of inflection.
Modelling & optimisation — using calculus to solve real "maximise/minimise" problems.
Connected rates of change — chaining multiple derivatives together with the chain rule.

1. What Is Differentiation, Really?

The Gradient of a Curve

For a straight line, "gradient" is easy — it's the same everywhere, so you just do rise-over-run between any two points. But a curve is a problem: its steepness is constantly changing. The gradient at the bottom of a U-shape is different from the gradient near the top.

So mathematicians defined the gradient of a curve at a single point as the gradient of the tangent — the straight line that just grazes the curve at that exact point, touching it but not slicing through it.

Think of it like this

Imagine you're riding a bike along the curve. At any exact instant, you're heading in a specific direction — that direction, extended into a straight line, is the tangent. Differentiation is the tool that tells you exactly which direction you're facing at any point on the ride.

Here's the key insight that makes calculus work: a tangent might touch the curve at one point but could still cross the curve somewhere completely different further along. That's fine — we only care about what's happening locally, right at our point of interest.

The Derivative — a "Gradient Machine"

The derivative of a function is itself a new function — one that tells you the gradient for any x-value you plug in. It's often called the gradient function.

For example, the derivative of y = x² is dy/dx = 2x. That single formula tells you:

At x = 1, the gradient is 2(1) = 2
At x = 5, the gradient is 2(5) = 10

One formula, infinite answers. That's the power of differentiation — you calculate it once and can then find the gradient anywhere on the curve instantly.

Vocabulary check

"Rate of change of..." and "gradient of..." mean exactly the same thing in these questions. If a question asks for a rate of change, it wants you to differentiate.

2. Differentiating Powers of x

This is the foundation rule everything else builds on. Powers of x follow a beautifully simple pattern: bring the power down in front, then subtract 1 from the power.

Power Rule If y = axⁿ then dy/dx = anxⁿ⁻¹ In words: multiply by the old power, then reduce the power by 1

Two special cases are easy to forget but come up constantly:

If y = ax then dy/dx = a a straight line has constant gradient

If y = a then dy/dx = 0 a constant never changes — zero slope

You can differentiate term by term for any sum or difference of powers of x. But products and quotients cannot be split up like this — you must expand or simplify first before you can differentiate term by term.

Common Mistake

Students often try to differentiate y = (2x − 3)(x² − 4) by differentiating each bracket separately and multiplying. That's wrong! You must expand first to get a sum of powers of x, THEN differentiate term by term. (Unless you use the product rule — see Section 4.)

Handling fractions

If x is in the denominator, like y = 4/x, rewrite it as a negative power first: y = 4x⁻¹. Only then apply the power rule.

Worked Example

Find dy/dx for y = 5x³ + 2x + 3/x² + 8

Step 1 — rewrite the fraction as a negative power:
y = 5x³ + 2x + 3x⁻² + 8

Step 2 — apply the power rule to each term:
dy/dx = 15x² + 2 − 6x⁻³

Step 3 — rewrite negative power as a fraction if needed:
dy/dx = 15x² + 2 − 6/x³

Practice Question 1

Find dy/dx for y = (2x + 3)²

Practice Question 2

Find dy/dx for y = (8x⁶ − x³) / 2x⁴

3. Differentiating Special Functions

Trig Functions

These three results must simply be memorised — there's no shortcut. Crucially, they only work when angles are measured in radians. If your calculator is in degree mode during an exam, you will get every trig-calculus question wrong.

y =	dy/dx =
sin x	cos x
cos x	−sin x
tan x	sec²x

For the linear function (ax + b) inside the trig function, or any general function f(x), you get an extra factor from the chain rule (more on this in Section 4):

General Trig Chain Rule Results y = sin(f(x)) → dy/dx = f′(x)cos(f(x)) | y = cos(f(x)) → dy/dx = −f′(x)sin(f(x)) | y = tan(f(x)) → dy/dx = f′(x)sec²(f(x))

Pattern to notice

Every time you differentiate a "co-" function (cosine), a minus sign appears. It's incredibly easy to forget this minus sign under exam pressure — flag it every single time.

Exponentials and Logarithms

Core Results y = eˣ → dy/dx = eˣ | y = ln x → dy/dx = 1/x eˣ is the only function that differentiates to itself!

For linear inputs and general functions:

y =	dy/dx =
e^(ax+b)	ae^(ax+b)
ln(ax+b)	a / (ax+b)
e^f(x)	f′(x)e^f(x)
ln(f(x))	f′(x) / f(x)

The #1 Most Common Mistake in this Whole Chapter

The derivative of ln(kx) with respect to x is 1/x, NOT k/x! The constant k cancels out. Similarly, the derivative of e^(kx) is ke^(kx), NOT kxe^(kx−1) — you never "bring the power down" for exponentials, because x is in the power, not being raised to one.

Worked Example

A curve has equation y = e⁻³ˣ + 2ln x. Find the gradient at x = 2, giving your answer in the form a + be^c.

Step 1 — differentiate each term separately:
dy/dx = −3e⁻³ˣ + 2(1/x)

Step 2 — substitute x = 2:
dy/dx = −3e⁻⁶ + 2(1/2) = −3e⁻⁶ + 1

Step 3 — rearrange to required form (do NOT evaluate on calculator!):
dy/dx = 1 − 3e⁻⁶

Practice Question

Differentiate y = 3sin 4x − cos(2x − 3) with respect to x.

4. The Chain Rule

The chain rule is for differentiating composite functions — a "function of a function." The giveaway is that the variable x doesn't "appear alone"; something is being done to x before the outer function acts on it.

Spotting the difference

sin x — x appears alone, this is NOT composite.
sin(3x + 2) — x is tripled and shifted BEFORE sine acts on it. This IS composite, and needs the chain rule.

Chain Rule If y = f(u) and u = g(x), then dy/dx = dy/du × du/dx Differentiate the "outside", multiply by the derivative of the "inside"

The 3-step method:

Step 1 — Identify the two functions: rewrite y as a function of u, and u as a function of x.
Step 2 — Differentiate each part separately: find dy/du and du/dx.
Step 3 — Multiply them together, then substitute u back in terms of x.

The mental shortcut

Once confident, say this in your head: "differentiate the outer function, ignoring the inner one — then multiply by the derivative of the inner function." This lets you skip formal substitution for most exam questions.

Worked Example

Find the derivative of y = (x² − 5x + 7)⁷

Step 1 — identify functions:
y = u⁷, where u = x² − 5x + 7

Step 2 — differentiate separately:
dy/du = 7u⁶ du/dx = 2x − 5

Step 3 — multiply and substitute back:
dy/dx = 7u⁶(2x−5) = 7(x²−5x+7)⁶(2x−5)

dy/dx = 7(2x − 5)(x² − 5x + 7)⁶

Two extremely common patterns worth knowing directly:

y = (ax+b)ⁿ
dy/dx = an(ax+b)ⁿ⁻¹

y = √(ax+b)
dy/dx = a / (2√(ax+b))

Practice Question

Find the derivative of y = sin(e²ˣ)

5. The Product Rule

Use this when two different functions of x are multiplied together — not composite (function of a function), but a genuine product like x²eˣ or sin x cos x.

Product Rule If y = uv, then dy/dx = u(dv/dx) + v(du/dx) Shorthand: y′ = uv′ + vu′

Don't confuse this with the chain rule!

sin(cos x) is composite (cosine goes INSIDE sine) → chain rule.
sin x · cos x is a product (two separate functions multiplied) → product rule.

Method: Identify u and v, differentiate each separately to get u′ and v′, then plug into the formula. Arranging u, v, u′, v′ in a square with opposite diagonals matching can help you keep track.

Worked Example

Differentiate (5sinx − 7cosx)e^3x+2

u = 5sinx − 7cosx u′ = 5cosx + 7sinx (don't forget the minus sign flips when differentiating cos!)

v = e³ˣ⁺² v′ = 3e³ˣ⁺² (chain rule needed here)

dy/dx = (5sinx−7cosx)(3e³ˣ⁺²) + (e³ˣ⁺²)(5cosx+7sinx)

dy/dx = (22sinx − 16cosx)e³ˣ⁺² after simplifying

Practice Question

Differentiate y = 5x ln(5x) with respect to x.

6. The Quotient Rule

Use this when one function is divided by another and both the top and bottom depend on x. (If only the numerator is a constant, it's often faster to rewrite using negative powers instead.)

Quotient Rule If y = u/v, then dy/dx = [v(du/dx) − u(dv/dx)] / v² Shorthand: y′ = (vu′ − uv′) / v² — order matters because of the minus sign!

Watch the order!

Unlike the product rule, the quotient rule has a minus sign, so getting u and v the wrong way round completely changes your answer. Always write "v times u-dash MINUS u times v-dash" — top function's derivative comes first.

Forgot the formula in the exam?

Rewrite y = u/v as y = u × v⁻¹ and use the product rule instead — it gives the same result. The quotient rule is really just a shortcut for this.

Worked Example

Differentiate f(x) = cos2x / (x² + 2x + 1)

g(x) = cos2x, g′(x) = −2sin2x h(x) = x²+2x+1, h′(x) = 2x+2

f′(x) = [(x²+2x+1)(−2sin2x) − (cos2x)(2x+2)] / (x²+2x+1)²

Since x²+2x+1 = (x+1)², this simplifies nicely to:

f′(x) = −2[cos2x + (x+1)sin2x] / (x+1)³

Practice Question

Differentiate f(x) = sinx / cosx using the quotient rule, and show it simplifies to sec²x.

7. Applications of Differentiation

Finding Gradients & Approximate Changes

To find the gradient of a curve at a specific point: differentiate to get dy/dx, then substitute the x-coordinate in.

For small changes, gradient ≈ (change in y) / (change in x), so change in y ≈ gradient × change in x. This is a handy shortcut for estimating how y shifts when x nudges slightly.

Three phrases, one meaning

"Find the gradient of the curve at x=2", "find the gradient of the tangent at x=2", and "find the rate of change of y with respect to x at x=2" are ALL asking for exactly the same calculation.

Increasing & Decreasing Functions

The Sign of the Gradient Tells You Everything dy/dx > 0 → function is increasing | dy/dx < 0 → function is decreasing

To find the range of x-values where a function is increasing or decreasing: differentiate, then solve the inequality f′(x) > 0 or f′(x) < 0. Sketching the gradient function's graph (usually a parabola) helps you see which interval satisfies the inequality.

Worked Example

For what values of x is y = 2x³ − 3x² + 5 a decreasing function?

dy/dx = 6x² − 6x. Solve 6x² − 6x < 0 → 6x(x−1) < 0

Roots at x=0 and x=1. Since this is a positive (u-shaped) quadratic, it's negative BETWEEN the roots.

The function is decreasing for 0 < x < 1

Tangents & Normals

A tangent touches the curve without crossing at that point; a normal is the straight line perpendicular to the tangent at that same point. To find the equation of either, you need a point and a gradient — then plug into y − y₁ = m(x − x₁).

Tangent & Normal Gradients Tangent gradient at (x₁,y₁) = f′(x₁) | Normal gradient = −1 / f′(x₁) Perpendicular lines have gradients that are negative reciprocals: m₁ × m₂ = −1

Worked Example

f(x) = 2x⁴ + 3x⁻². Find the tangent at x=1 in the form y=mx+c.

f′(x) = 8x³ − 6x⁻³. At x=1: gradient = 8−6 = 2

f(1) = 2+3 = 5, so the point is (1,5)

y − 5 = 2(x − 1) → y = 2x + 3

For the normal: gradient = −1/2, so y − 5 = −½(x−1) → x + 2y − 11 = 0

Practice Question

A curve has equation y = x³ − 2x² − x + 3. Find the equation of the tangent at the point where x = 2.

Second Derivatives

Differentiate the derivative again and you get the second derivative, written d²y/dx². Note carefully where the powers of 2 sit — differentiating twice with respect to x twice.

If the first derivative tells you the gradient (rate of change of y), the second derivative tells you the rate of change of the gradient itself — in other words, whether the curve is bending upward (concave up, like a U) or downward (concave down, like an n).

Exam habit

Even if you're confident, ALWAYS write down the first derivative as a separate line before finding the second. If you slip up on the second derivative, you still bank marks for the correct first one.

Stationary Points & Their Nature

A stationary point is anywhere the gradient equals zero — the curve is momentarily flat. There are three types: local maximum (peak), local minimum (valley), and point of inflection (a flat "shoulder" that isn't a peak or valley).

To find stationary points:

Step 1 — Find dy/dx
Step 2 — Solve dy/dx = 0 to find the x-coordinates
Step 3 — Substitute back into the original equation to get y-coordinates

To classify each one, use the (usually faster) second derivative test:

Second Derivative Test d²y/dx² > 0 → local minimum | d²y/dx² < 0 → local maximum | d²y/dx² = 0 → inconclusive, use Method A instead

If the second derivative gives zero

You must fall back on "Method A" — check the sign of the first derivative just to the left and just to the right of the point. Negative-then-positive means a minimum (u-shape); positive-then-negative means a maximum (n-shape); same sign both sides means a point of inflection.

Worked Example

Find and classify the stationary points of y = x²(2x² − 4)

Expand: y = 2x⁴ − 4x². dy/dx = 8x³ − 8x = 8x(x−1)(x+1)

Stationary points at x = 0, 1, −1 → points (0,0), (1,−2), (−1,−2)

d²y/dx² = 24x² − 8. At x=0: −8 (negative → max). At x=±1: 16 (positive → min)

(0,0) is a local maximum; (1,−2) and (−1,−2) are both local minimums

Modelling & Optimisation

Optimisation problems ask you to find the biggest or smallest possible value of something — maximum volume, minimum cost, minimum material used. The method is always the same recipe: build a formula, differentiate it, set the derivative to zero, solve, and use the second derivative to confirm you've found a max or min (not just any stationary point).

Typical exam structure

These questions are usually two parts: first "Show that V = ..." (derive the formula yourself), then "Find the maximum/minimum ..." (use calculus). Even if you can't do part (a), you can still use the given formula to attempt part (b) — don't give up on the whole question!

Practice Question

A cuboid has volume V = 12x − 20x². Find the value of x that maximises V, and prove it is a maximum.

Connected Rates of Change

When a problem involves more than two variables (like volume, radius, and time all at once), you use the chain rule to "connect" rates you know to a rate you want to find.

Connected Rates Formula dV/dt = dV/ds × ds/dt Chain together what you know with what you want to find

Watch for the keyword "rate" — it always signals a derivative with respect to time is involved. "Increasing at a rate of 4 cm/s" means dh/dt = 4; "decreasing" means the rate is negative.

Worked Example

The volume of a cube increases at a constant rate of 3 cm³/s. Find the rate of change of the side length s when s = 5cm.

We know dV/dt = 3. We want ds/dt.

By the chain rule: dV/dt = dV/ds × ds/dt

For a cube, V = s³, so dV/ds = 3s². Rearranging: ds/dt = (dV/dt) / (dV/ds) = 3/(3s²) = 1/s²

At s=5: ds/dt = 1/25

The side length is increasing at 1/25 cm/s

Easy to lose track

These problems throw a lot of letters at you at once. Before writing anything, list clearly: which letters are the variables, which are constants, and which rate you actually need to find. Confusing a constant for a variable (or vice versa) is the single biggest reason these questions go wrong.

What to Memorise

Function y =	Derivative dy/dx =
axⁿ	anxⁿ⁻¹
ax	a
a (constant)	0
sin x	cos x
cos x	−sin x
tan x	sec²x
eˣ	eˣ
ln x	1/x
sin(ax+b)	a cos(ax+b)
cos(ax+b)	−a sin(ax+b)
tan(ax+b)	a sec²(ax+b)
e^ax+b	ae^ax+b
ln(ax+b)	a/(ax+b)

The three big rules:

Chain Rule: dy/dx = dy/du × du/dx

Product Rule: y′ = uv′ + vu′

Quotient Rule: y′ = (vu′ − uv′) / v²

Tangent/normal equations:

Tangent: y − y₁ = f′(x₁)(x − x₁) | Normal: y − y₁ = [−1/f′(x₁)](x − x₁)

Stationary point classification:

d²y/dx² > 0 → minimum | d²y/dx² < 0 → maximum | = 0 → check first derivative either side

Concepts Checklist

I understand that the gradient of a curve at a point = gradient of the tangent there
I can differentiate any power of x, including negative and fractional powers
I know I must rewrite fractions and roots as powers of x before differentiating
I have memorised the derivatives of sin x, cos x, and tan x
I have memorised the derivatives of eˣ and ln x
I can identify when a function is composite and needs the chain rule
I can identify when to use the product rule vs the chain rule
I can apply the quotient rule correctly, keeping the order of terms right
I can find the equation of a tangent and a normal to a curve
I can determine intervals where a function is increasing or decreasing
I can find and classify stationary points using the second derivative test
I can set up and solve optimisation (modelling) problems
I can solve connected rates of change problems using the chain rule

Exam Tips & Common Traps

Radians, always radians

Trig calculus results only work in radians. If your calculator is in degree mode during a trig differentiation question, every numerical answer will be wrong.

ln(kx) derivative

The derivative of ln(kx) is 1/x — the constant k always cancels out. Do NOT write k/x.

Products and quotients can't be split

You cannot differentiate (2x−3)(x²−4) by treating each bracket separately unless you use the product rule properly. Either expand first, or apply uv′+vu′.

Quotient rule order

Because of the minus sign in vu′ − uv′, getting the order backwards flips your answer's sign. Always start with "v" (the denominator).

Show your working, always

Write down the first derivative on its own line even in second-derivative or stationary-point questions — you bank method marks even if a later step goes wrong.

Match the requested form

If a question asks for the answer "in the form y = mx + c" or "ax + by + d = 0 where a, b, d are integers", always rearrange fully — marks are often reserved specifically for the correct final form.

Don't evaluate when asked to leave exact

If a question asks for an answer "in the form a + be^c", do not punch it into your calculator for a decimal — leave it in exact form as requested.

Sketch to check

For increasing/decreasing and stationary point questions, a quick rough sketch of the gradient function (or the original curve) is a fast way to sanity-check your inequality direction or point classification.

Also in the full note

Exam Tips & Common Traps
Finding Gradients & Approximate Changes
Increasing & Decreasing Functions
Tangents & Normals
Stationary Points & Their Nature
Modelling & Optimisation

What's inside

📖 Revision notes ✦ AI flashcards ✓ Instant AI marking

Read the full Differentiation notes free

That's the preview — create a free account to read the rest, plus flashcards and practice questions with instant AI marking. No credit card.

Unlock the full notes free →