Notes · MathematicsUK · A-Levels

Differentiation

Differentiation measures instantaneous rate of change. The A-Level develops the derivative from first principles, the standard results and the product, quotient and chain rules, the analysis of curves through stationary points and concavity, and implicit, parametric and connected rates of change. It is the calculus half that powers optimisation and kinematics.

4 sections~10 min reading time3 competenciesLevel Standard 1 · Advanced 3

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Exam profile

AO1 · Differentiate standard functions using the rules of differentiationAO2 · Justify the nature of stationary points and the shape of curvesAO3 · Solve optimisation and connected-rate problems in context

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basic level

AS-Level covers differentiation from first principles, powers of $x$ , tangents and normals, and stationary points.

higher level

The full A-Level adds the product, quotient and chain rules, the derivatives of $e^x$ , $\ln x$ and the trigonometric functions, implicit and parametric differentiation, and connected rates of change.

Depth

Text

Contents · 4 sections

Differentiation

The derivative and first principles#

StandardDfE GCE Mathematics G1AQA 7357 G

Chord tending to tangent

Fig. 1As the chord (secant) shrinks towards

x = 1

, its gradient tends to the tangent gradient

f'(1) = 2

The derivative measures the gradient of a curve at a point, defined as the limit of the gradient of a chord as the chord shrinks to nothing. Formally,

f'(x) = \lim_{h \to 0}\dfrac{f(x + h) - f(x)}{h}

. The fraction is the gradient of the chord joining

(x, f(x))

(x + h, f(x + h))

; as

h \to 0

the chord tends to the tangent, and its gradient tends to the derivative. This limit is what 'differentiation from first principles' means.

Carrying out the limit for

f(x) = x^2

shows the mechanism. The chord gradient is

\dfrac{(x + h)^2 - x^2}{h} = \dfrac{2xh + h^2}{h} = 2x + h

, and letting

h \to 0

gives

f'(x) = 2x

. The crucial step is that the

h

in the denominator cancels before you take the limit, avoiding the meaningless

\tfrac{0}{0}

; only after cancelling do you set

h = 0

First principles justify the general power rule

\dfrac{d}{dx}x^n = nx^{n-1}

, which holds for any rational

n

and is the workhorse for polynomials, roots and reciprocals once they are written as powers. The derivative of a sum is the sum of the derivatives, and a constant multiple factors out, so a whole polynomial is differentiated term by term.

The derivative is written interchangeably as

\dfrac{dy}{dx}

(Leibniz) or

f'(x)

(Lagrange), and its value at a point is the gradient of the tangent there. From it you get the equation of the tangent (gradient

f'(a)

through

(a, f(a))

) and the normal (perpendicular, gradient

-\dfrac{1}{f'(a)}

). Reading 'rate of change' as 'derivative' unlocks every application that follows.

f'(x) = \lim_{h \to 0}\frac{f(x + h) - f(x)}{h}

The derivative from first principles

The limiting gradient of the chord as its width $h$ tends to zero.

\frac{d}{dx}x^{n} = n x^{n-1}

The power rule

Holds for any rational $n$ ; differentiate polynomials term by term.

Worked example

Differentiation from first principles

Use first principles to differentiate $f(x) = 3x^2$ .

Form the chord gradient
$\dfrac{f(x + h) - f(x)}{h} = \dfrac{3(x + h)^2 - 3x^2}{h}$ .
Expand and simplify
$= \dfrac{3(x^2 + 2xh + h^2) - 3x^2}{h} = \dfrac{6xh + 3h^2}{h} = 6x + 3h$ .
$\frac{6xh + 3h^2}{h} = 6x + 3h$
Take the limit
As $h \to 0$ , $6x + 3h \to 6x$ .

Result: $f'(x) = 6x$ .

Typical mistakes

Setting $h = 0$ before cancelling it, producing $\tfrac{0}{0}$ .
Forgetting to convert roots and reciprocals to index form before applying the power rule.

Active revision

Differentiate $f(x) = x^2 - 3x$ from first principles, and hence find the gradient at $x = 4$ .

Active recall

Recall the key points — then reveal.

Sources: Mathematics: AS and A level content (GCE subject content) (Department for Education) · AQA A-level Mathematics 7357 specification (AQA)

Rules of differentiation#

AdvancedDfE GCE Mathematics G2-G3AQA 7357 G

A function and its gradient function

Fig. 2The derivative

f'(x) = 3x^2 - 3

is zero exactly where

f(x) = x^3 - 3x

has its turning points.

Beyond powers, you must know the standard derivatives:

\dfrac{d}{dx}e^{x} = e^{x}

\dfrac{d}{dx}\ln x = \dfrac{1}{x}

\dfrac{d}{dx}\sin x = \cos x

\dfrac{d}{dx}\cos x = -\sin x

and

\dfrac{d}{dx}\tan x = \sec^2 x

(with

x

in radians). These, combined with three rules for building functions, let you differentiate essentially anything on the specification.

The chain rule differentiates a composite function: if

y = f(g(x))

then

\dfrac{dy}{dx} = f'(g(x))\,g'(x)

, or in Leibniz form

\dfrac{dy}{dx} = \dfrac{dy}{du}\dfrac{du}{dx}

. In practice you 'differentiate the outside, keep the inside, times the derivative of the inside':

\dfrac{d}{dx}(3x + 1)^{5} = 5(3x + 1)^{4} \times 3

. It is the most frequently needed rule and the one most often forgotten.

The product rule differentiates a product: if

y = uv

then

\dfrac{dy}{dx} = u\dfrac{dv}{dx} + v\dfrac{du}{dx}

. The quotient rule differentiates a fraction: if

y = \dfrac{u}{v}

then

\dfrac{dy}{dx} = \dfrac{v\,u' - u\,v'}{v^2}

. The order of terms in the quotient rule matters because of the subtraction, so a reliable habit is to write

u

v

u'

and

v'

out separately before substituting.

Real problems often need two rules together — a product of a composite, or a quotient containing an exponential — so identifying the structure of the expression first is essential. Naming the pieces, differentiating each, and assembling with the correct rule, in that order, keeps the working clean and the marks secure.

\frac{d}{dx}f(g(x)) = f'(g(x))\,g'(x)

The chain rule

Differentiate the outer function, keep the inner, then multiply by the derivative of the inner.

\frac{d}{dx}(uv) = u v' + v u', \qquad \frac{d}{dx}\!\left(\frac{u}{v}\right) = \frac{v u' - u v'}{v^2}

Product and quotient rules

Mind the order of the subtraction in the quotient rule.

Worked example

Product and chain rules together

Differentiate $y = x^2 e^{2x}$ .

Identify the product
Let $u = x^2$ and $v = e^{2x}$ , so $u' = 2x$ and $v' = 2e^{2x}$ (chain rule on $e^{2x}$ ).
Apply the product rule
$\dfrac{dy}{dx} = u v' + v u' = x^2(2e^{2x}) + e^{2x}(2x)$ .
$\frac{dy}{dx} = 2x^2 e^{2x} + 2x e^{2x}$
Factorise
$= 2x e^{2x}(x + 1)$ .

Result: $\dfrac{dy}{dx} = 2x e^{2x}(x + 1)$ .

Typical mistakes

Omitting the inner derivative in the chain rule (e.g. writing $\tfrac{d}{dx}(3x+1)^5 = 5(3x+1)^4$ ).
Getting the subtraction the wrong way round in the quotient rule.

Active revision

Differentiate (a) $y = (2x^2 - 1)^4$ , (b) $y = x^2 e^{3x}$ , and (c) $y = \dfrac{\sin x}{x}$ .

Active recall

Recall the key points — then reveal.

Sources: Mathematics: AS and A level content (GCE subject content) (Department for Education) · AQA A-level Mathematics 7357 specification (AQA)

Stationary points and curve analysis#

AdvancedDfE GCE Mathematics G4-G5AQA 7357 G

Maximum and minimum of a cubic

Fig. 3

y = 2x^3 - 3x^2 - 12x + 5

has a maximum at

x = -1

and a minimum at

x = 2

, where

f'(x) = 0

A stationary point is where

\dfrac{dy}{dx} = 0

: the tangent is horizontal. Solving

f'(x) = 0

finds all such points, which may be local maxima, local minima or points of inflection. Stationary points are the key features that shape a curve and are the answer to every optimisation question, where the largest or smallest value of a quantity is required.

The nature of a stationary point is decided by the second derivative

\dfrac{d^2y}{dx^2}

, which measures how the gradient itself is changing. If

f''(x) > 0

at a stationary point the curve is concave up and the point is a minimum; if

f''(x) < 0

it is concave down and a maximum. If

f''(x) = 0

the test is inconclusive and you fall back on examining the sign of

f'(x)

on either side.

The sign of the first derivative describes where a function increases or decreases:

f'(x) > 0

means increasing,

f'(x) < 0

means decreasing. Combining intervals of increase and decrease with the locations of stationary points and the behaviour for large

|x|

produces a complete sketch. A point of inflection is where the concavity changes, i.e. where

f''(x)

changes sign.

Optimisation problems put this to work: express the quantity to be optimised as a function of a single variable (using any constraint to eliminate the others), differentiate, set the derivative to zero, and confirm the nature of the stationary point with the second derivative. Interpreting the result in context, and checking it against the physical range of the variable, completes the answer.

f'(x) = 0 \ \text{and}\ \begin{cases} f''(x) > 0 & \text{minimum} \\ f''(x) < 0 & \text{maximum} \end{cases}

The second-derivative test

The sign of the second derivative at a stationary point classifies it.

Worked example

Classifying stationary points

Find and classify the stationary points of $y = x^3 - 6x^2 + 9x + 1$ .

Differentiate and set to zero
$\dfrac{dy}{dx} = 3x^2 - 12x + 9 = 3(x - 1)(x - 3) = 0$ , so $x = 1$ or $x = 3$ .
Second derivative
$\dfrac{d^2y}{dx^2} = 6x - 12$ .
$\frac{d^2y}{dx^2} = 6x - 12$
Classify
At $x = 1$ : $6(1) - 12 = -6 < 0$ , a maximum ( $y = 5$ ). At $x = 3$ : $6(3) - 12 = 6 > 0$ , a minimum ( $y = 1$ ).

Result: Maximum at $(1, 5)$ and minimum at $(3, 1)$ .

Typical mistakes

Assuming $f''(x) > 0$ gives a maximum (it gives a minimum).
Not verifying the nature of the stationary point, or ignoring the valid range of the variable in optimisation.

Active revision

An open box is made from a square sheet of side 20 cm by cutting squares of side $x$ from the corners and folding up. Find the value of $x$ that maximises the volume, and confirm it is a maximum.

Active recall

Recall the key points — then reveal.

Sources: Mathematics: AS and A level content (GCE subject content) (Department for Education) · AQA A-level Mathematics 7357 specification (AQA)

Implicit, parametric and connected rates#

AdvancedDfE GCE Mathematics G6-G7AQA 7357 G

Implicit differentiation on a circle

Fig. 4On

x^2 + y^2 = 25

, implicit differentiation gives

\dfrac{dy}{dx} = -\dfrac{x}{y} = -\dfrac{3}{4}

(3, 4)

Not every curve is given as

y = f(x)

. Implicit differentiation handles equations relating

x

and

y

, such as a circle

x^2 + y^2 = 25

, by differentiating both sides with respect to

x

and treating

y

as a function of

x

(so

\dfrac{d}{dx}y^2 = 2y\dfrac{dy}{dx}

by the chain rule). Rearranging then gives

\dfrac{dy}{dx}

in terms of both

x

and

y

; for the circle,

\dfrac{dy}{dx} = -\dfrac{x}{y}

For a curve given parametrically by

x = f(t)

y = g(t)

, the gradient is found via the chain rule as

\dfrac{dy}{dx} = \dfrac{dy/dt}{dx/dt}

. You differentiate each coordinate with respect to the parameter and divide, then substitute the required value of

t

. This is how tangents to parametric curves — including the ellipse met in coordinate geometry — are obtained.

Connected rates of change link the rates of two quantities through a shared variable, usually time. If two quantities are related by an equation, differentiating with respect to

t

and using the chain rule connects their rates: for a sphere,

V = \tfrac{4}{3}\pi r^3

gives

\dfrac{dV}{dt} = 4\pi r^2\dfrac{dr}{dt}

, linking the rate of change of volume to that of radius. The chain rule

\dfrac{dV}{dt} = \dfrac{dV}{dr}\dfrac{dr}{dt}

is the engine.

These techniques appear in modelling contexts — an expanding balloon, a rising water level in a cone, a shadow lengthening — where you are given one rate and asked for another. The reliable method is to write down the equation relating the quantities, differentiate it with respect to time, substitute the known values at the instant in question, and solve for the required rate.

\frac{dy}{dx} = \frac{dy/dt}{dx/dt}

Parametric differentiation

Divide the two parameter-derivatives to get the gradient of the curve.

\frac{dV}{dt} = \frac{dV}{dr}\cdot\frac{dr}{dt}

Connected rates via the chain rule

Links the rate of one quantity to another through a shared variable.

Worked example

Connected rates of change

The radius of a circle increases at $0.2\ \text{cm s}^{-1}$ . Find the rate of increase of its area when $r = 10$ cm.

Relate the quantities
$A = \pi r^2$ , so $\dfrac{dA}{dr} = 2\pi r$ .
Chain rule in time
$\dfrac{dA}{dt} = \dfrac{dA}{dr}\cdot\dfrac{dr}{dt} = 2\pi r \times 0.2$ .
$\frac{dA}{dt} = 2\pi r \times 0.2$
Substitute r = 10
$\dfrac{dA}{dt} = 2\pi(10)(0.2) = 4\pi \approx 12.6\ \text{cm}^2\text{s}^{-1}$ .

Result: The area increases at $4\pi \approx 12.6\ \text{cm}^2\,\text{s}^{-1}$ .

Typical mistakes

Forgetting the $\dfrac{dy}{dx}$ factor when differentiating a $y$ -term implicitly.
Substituting numerical values before differentiating in a connected-rates problem.

Active revision

A spherical balloon is inflated so that its volume increases at $50\ \text{cm}^3\,\text{s}^{-1}$ . Find the rate at which the radius increases when $r = 5$ cm.

Active recall

Recall the key points — then reveal.

Sources: Mathematics: AS and A level content (GCE subject content) (Department for Education) · AQA A-level Mathematics 7357 specification (AQA)

References & sources

Sources

Department for Education

Mathematics: AS and A level content (GCE subject content)

AQA

AQA A-level Mathematics 7357 specification

Differentiation

Key points

Differentiation from first principles

Key points

Product and chain rules together

Key points

Classifying stationary points

Key points

Connected rates of change

Sources