Differentiation

This lesson covers Differentiation as required by the A-Level Mathematics Pure 1 specification. Differentiation is a fundamental tool of calculus that allows us to find the rate of change of a function. It has applications in finding gradients, tangents, normals, stationary points, and solving optimisation problems.

Spec Mapping — AQA 7357 Section G Differentiation. This lesson covers first principles, the power rule, tangents and normals, stationary points and the second-derivative test in Section G. Refer to the official AQA specification document for exact wording.

---

The Gradient of a Curve

The gradient of a curve at any point is the gradient of the tangent to the curve at that point. Differentiation gives us a formula for this gradient.

The derivative of y = f(x) is written as:

$$\frac{dy}{dx} \quad \text{or} \quad f'(x)$$dxdy​orf′(x)

It represents the rate of change of y with respect to x.

---

Differentiating from First Principles

The derivative of f(x) from first principles is defined as:

$$f'(x) = \lim_{h \to 0} \frac{f(x + h) - f(x)}{h}$$f′(x)=h→0lim​hf(x+h)−f(x)​

Example: Differentiate f(x) = x² from first principles.

$$\begin{aligned} f(x + h) &= (x + h)^{2} = x^{2} + 2xh + h^{2} \\[4pt] f'(x) &= \lim_{h \to 0} \frac{(x^{2} + 2xh + h^{2}) - x^{2}}{h} \\ &= \lim_{h \to 0} \frac{2xh + h^{2}}{h} \\ &= \lim_{h \to 0} (2x + h) \\ &= 2x \end{aligned}$$f(x+h)f′(x)​=(x+h)2=x2+2xh+h2=h→0lim​h(x2+2xh+h2)−x2​=h→0lim​h2xh+h2​=h→0lim​(2x+h)=2x​

Exam Tip: First principles questions are commonly examined. Show every step clearly, and remember to take the limit as h → 0 at the end.

---

Standard Rules for Differentiation

Power Rule

If y = xⁿ, then dy/dx = nxⁿ⁻¹. This works for all rational values of n.

Function	Derivative
$y = x^n$y=xn	$\dfrac{dy}{dx} = nx^{n-1}$dxdy​=nxn−1
$y = ax^n$y=axn	$\dfrac{dy}{dx} = anx^{n-1}$dxdy​=anxn−1
$y = c$y=c (constant)	$\dfrac{dy}{dx} = 0$dxdy​=0
$y = x^3$y=x3	$\dfrac{dy}{dx} = 3x^2$dxdy​=3x2
$y = x^{-1} = \dfrac{1}{x}$y=x−1=x1​	$\dfrac{dy}{dx} = -x^{-2} = -\dfrac{1}{x^2}$dxdy​=−x−2=−x21​
$y = x^{1/2} = \sqrt{x}$y=x1/2=x​	$\dfrac{dy}{dx} = \tfrac{1}{2}x^{-1/2} = \dfrac{1}{2\sqrt{x}}$dxdy​=21​x−1/2=2x​1​

Sum/Difference Rule

If $y = f(x) \pm g(x)$y=f(x)±g(x), then $\dfrac{dy}{dx} = f'(x) \pm g'(x)$dxdy​=f′(x)±g′(x).

Example: Differentiate $y = 4x^3 - 7x^2 + 5x - 9$y=4x3−7x2+5x−9.

$$\frac{dy}{dx} = 12x^{2} - 14x + 5$$dxdy​=12x2−14x+5

---

Preparing Expressions for Differentiation

Before differentiating, you may need to expand brackets or rewrite fractions as powers of $x$x.

Example: Differentiate $y = \dfrac{3x^2 + 1}{x}$y=x3x2+1​.

Rewrite: $y = 3x + x^{-1}$y=3x+x−1.

$$dy/dx = 3 - x^{-2} = 3 - 1/x^{2}$$dy/dx=3−x−2=3−1/x2

Example: Differentiate $y = 3x^{-2} + 2\sqrt{x}$y=3x−2+2x​.

Write as $y = 3x^{-2} + 2x^{1/2}$y=3x−2+2x1/2.

$$\frac{dy}{dx} = -6x^{-3} + x^{-1/2} = -\frac{6}{x^{3}} + \frac{1}{\sqrt{x}}$$dxdy​=−6x−3+x−1/2=−x36​+x​1​

---

Tangents and Normals

Tangent

The tangent to a curve at a point has the same gradient as the curve at that point.

To find the equation of the tangent at x = a:

1. Find y = f(a) — the y-coordinate.
2. Find f'(a) — the gradient.
3. Use y − f(a) = f'(a)(x − a).

Normal

The normal is perpendicular to the tangent. Its gradient is −1/f'(a).

Example: Find the tangent and normal to y = x³ − 2x at x = 1.

At $x = 1$x=1: $y(1) = 1 - 2 = -1$y(1)=1−2=−1, so the point is $(1, -1)$(1,−1). The derivative is $\dfrac{dy}{dx} = 3x^{2} - 2$dxdy​=3x2−2, so the gradient at $x = 1$x=1 is $1$1.

Tangent: y − (−1) = 1(x − 1) → y = x − 2.

Normal: gradient = −1. y − (−1) = −1(x − 1) → y = −x.

---

Stationary Points

A stationary point occurs where dy/dx = 0. There are three types:

Type	Second Derivative	Description
Maximum	f''(x) < 0	Curve changes from increasing to decreasing
Minimum	f''(x) > 0	Curve changes from decreasing to increasing
Point of inflection	f''(x) = 0	Inconclusive — check gradient either side

The Second Derivative Test

The second derivative d²y/dx² = f''(x) is found by differentiating dy/dx.

Example: Find and classify the stationary points of y = x³ − 6x² + 9x + 1.

$$\begin{aligned} \frac{dy}{dx} &= 3x^{2} - 12x + 9 = 3(x^{2} - 4x + 3) = 3(x - 1)(x - 3) = 0 \\ x &= 1 \ \text{or} \ x = 3 \\[4pt] \frac{d^{2}y}{dx^{2}} &= 6x - 12 \\[4pt] \text{At } x = 1:\ & f''(1) = -6 < 0 \ \Rightarrow \ \text{local maximum at } (1, 5) \\ \text{At } x = 3:\ & f''(3) = 6 > 0 \ \Rightarrow \ \text{local minimum at } (3, 1) \end{aligned}$$dxdy​xdx2d2y​At x=1: At x=3: ​=3x2−12x+9=3(x2−4x+3)=3(x−1)(x−3)=0=1 or x=3=6x−12f′′(1)=−6<0 ⇒ local maximum at (1,5)f′′(3)=6>0 ⇒ local minimum at (3,1)​

---

Increasing and Decreasing Functions

• A function is increasing on an interval if f'(x) > 0 for all x in that interval.
• A function is decreasing on an interval if f'(x) < 0 for all x in that interval.

Example: Show that f(x) = x³ + 3x is an increasing function for all x.

$$f'(x) = 3x^{2} + 3 = 3(x^{2} + 1)$$f′(x)=3x2+3=3(x2+1)

Since x² ≥ 0 for all x, we have x² + 1 ≥ 1 > 0, so f'(x) > 0 for all x.

Therefore f(x) is increasing for all x. ∎

---

Synoptic Links

This topic connects to:

• Chain, product and quotient rules (aqa-alevel-maths-pure-2 / differentiation-techniques) — extends the power rule to composite, product and quotient functions.
• Variable acceleration (aqa-alevel-maths-mechanics / variable-acceleration) — kinematics with non-constant acceleration is calculus applied to s(t), v(t), a(t).
• Optimisation problems (aqa-alevel-maths-calculus-applications / optimisation-problems) — stationary-point methods drive real-world maxima/minima questions.

---

Summary

• The derivative f'(x) gives the gradient of a curve at any point.
• First principles: f'(x) = lim (h → 0) [f(x + h) − f(x)] / h.
• Power rule: if y = xⁿ, then dy/dx = nxⁿ⁻¹.
• Rewrite roots and fractions as powers before differentiating.
• Find tangent equations using the gradient; normals are perpendicular.
• Stationary points occur where dy/dx = 0; classify using the second derivative.
• A function is increasing where f'(x) > 0 and decreasing where f'(x) < 0.

Exam Tip: Always simplify expressions before differentiating — expand brackets and rewrite fractions as negative powers. When finding stationary points, clearly show that you have set dy/dx = 0, found the x-values, calculated the y-values, and used the second derivative to classify each point. If f''(x) = 0, you must check the gradient either side of the point.

A-Level Deep Dive: Differentiation

Spec mapping

AQA 7357 specification, Paper 1 — Pure Mathematics, Section G: "Understand and use the derivative of $f(x)$f(x) as the gradient of the tangent to the graph of $y = f(x)$y=f(x) at a general point $(x, y)$(x,y); the gradient of the tangent as a limit; interpretation as a rate of change; second-order derivatives; differentiation from first principles for small positive integer powers of $x$x and for $\sin x$sinx and $\cos x$cosx; differentiation of $x^n$xn for rational $n$n, sums and constant multiples; applications to gradients, tangents and normals, stationary points, increasing and decreasing functions." Synoptic reach is wide: chain, product and quotient rules (Pure 2, Section G continued); integration as the inverse of differentiation (Section H); coordinate geometry tangents and normals (Section E); and mechanics rates of change — velocity $= \dfrac{ds}{dt}$=dtds​, acceleration $= \dfrac{dv}{dt}$=dtdv​ — across Paper 2 Section R. The AQA formula booklet provides standard derivatives for $\sin x$sinx, $\cos x$cosx, $\tan x$tanx, $e^{kx}$ekx and $\ln x$lnx, but the power rule $\dfrac{d}{dx}x^n = nx^{n-1}$dxd​xn=nxn−1 must be memorised.

---

Specimen Question (Modelled on the AQA Paper Format)

Note: this question is constructed to model AQA Paper 1/2/3 style; it is not a reproduction of any published past paper.

Worked example with full mark scheme

Question (8 marks): The curve $C$C has equation $y = x^3 - 3x^2 - 9x + 5$y=x3−3x2−9x+5.

(a) Find $\dfrac{dy}{dx}$dxdy​ and hence determine the coordinates of the stationary points of $C$C. (5)

(b) Use the second derivative test to classify each stationary point as a local maximum or local minimum. (3)

Solution with mark scheme:

(a) Step 1 — differentiate.

$\dfrac{dy}{dx} = 3x^2 - 6x - 9$dxdy​=3x2−6x−9

M1 — applying the power rule term-by-term. Common slip: dropping the constant $+5$+5 to give $3x^2 - 6x - 9 + 0$3x2−6x−9+0 correctly, but writing $+5$+5 on the derivative line.

A1 — fully correct first derivative.

Step 2 — set the derivative equal to zero.

$3x^2 - 6x - 9 = 0 \implies x^2 - 2x - 3 = 0 \implies (x - 3)(x + 1) = 0$3x2−6x−9=0⟹x2−2x−3=0⟹(x−3)(x+1)=0

M1 — equating $\dfrac{dy}{dx}$dxdy​ to zero and reducing to a soluble quadratic. Dividing by 3 first is rewarded as a clean technique.

So $x = 3$x=3 or $x = -1$x=−1.

Step 3 — find the corresponding $y$y-values.

At $x = 3$x=3: $y = 27 - 27 - 27 + 5 = -22$y=27−27−27+5=−22. At $x = -1$x=−1: $y = -1 - 3 + 9 + 5 = 10$y=−1−3+9+5=10.

A1 — both stationary points correctly identified as $(3, -22)$(3,−22) and $(-1, 10)$(−1,10). A single arithmetic slip in either $y$y-value loses this mark; the M-marks before it stand.

A1 — answers presented as coordinate pairs, not just $x$x-values. Examiners reward the question's phrasing exactly.

(b) Step 1 — second derivative.

$\dfrac{d^2 y}{dx^2} = 6x - 6$dx2d2y​=6x−6

M1 — differentiating $\dfrac{dy}{dx}$dxdy​ correctly.

Step 2 — evaluate at each stationary point.

At $x = 3$x=3: $\dfrac{d^2 y}{dx^2} = 18 - 6 = 12 > 0$dx2d2y​=18−6=12>0, so $(3, -22)$(3,−22) is a local minimum.

At $x = -1$x=−1: $\dfrac{d^2 y}{dx^2} = -6 - 6 = -12 < 0$dx2d2y​=−6−6=−12<0, so $(-1, 10)$(−1,10) is a local maximum.

A1 — correct sign analysis and conclusion at both points.

A1 — full classification stated with the correct labels (local max/local min). Writing only "max" and "min" without specifying which point is which loses this mark.

Total: 8 marks (M3 A5).

---

Specimen question modelled on the AQA 7357 Paper 1 format

Question (6 marks): The curve $C$C has equation $y = 2x^3 - 9x^2 + 12x$y=2x3−9x2+12x.

(a) Find an equation of the tangent to $C$C at the point $P$P where $x = 2$x=2, giving your answer in the form $y = mx + c$y=mx+c. (4)

(b) Find an equation of the normal to $C$C at $P$P, giving your answer in the form $ax + by + c = 0$ax+by+c=0 where $a$a, $b$b, $c$c are integers. (2)

Mark scheme decomposition by AO:

(a)

• M1 (AO1.1a) — differentiating: $\dfrac{dy}{dx} = 6x^2 - 18x + 12$dxdy​=6x2−18x+12.
• M1 (AO1.1b) — substituting $x = 2$x=2: gradient $= 24 - 36 + 12 = 0$=24−36+12=0, and $y$y-coordinate at $P$P: $y = 16 - 36 + 24 = 4$y=16−36+24=4, so $P = (2, 4)$P=(2,4).
• A1 (AO1.1b) — recognising that the tangent has gradient $0$0, hence is horizontal.
• A1 (AO2.5) — equation of tangent: $y = 4$y=4, written in the form $y = mx + c$y=mx+c with $m = 0$m=0, $c = 4$c=4.

(b)

• M1 (AO1.1b) — recognising that a horizontal tangent has a vertical normal, so the normal is $x = 2$x=2.
• A1 (AO2.5) — rearranged to integer-coefficient form: $x - 2 = 0$x−2=0, i.e. $1x + 0y - 2 = 0$1x+0y−2=0.

Total: 6 marks split AO1 = 5, AO2 = 1. A subtle examiner trap: students rush to compute $-1/m$−1/m for the normal and divide by zero. Recognising that $m = 0$m=0 implies a vertical normal is the AO2 reasoning step.

---

Synoptic links

Connects to:

1. Pure 2, Section G — Chain, product and quotient rules: the power rule generalises to differentiating $(\text{inner})^n$(inner)n via the chain rule: $\dfrac{d}{dx}(2x + 1)^5 = 5(2x + 1)^4 \cdot 2$dxd​(2x+1)5=5(2x+1)4⋅2. Without confidence in the power rule, the chain rule has nothing to apply to.

2. Section H — Integration: integration is defined as the inverse of differentiation. $\int x^n\,dx = \dfrac{x^{n+1}}{n + 1} + C$∫xndx=n+1xn+1​+C for $n \neq -1$n=−1 is literally the power rule run backwards. Every fluency error in differentiation propagates into integration.

3. Section E — Coordinate geometry: the equation of a tangent at $(x_0, y_0)$(x0​,y0​) is $y - y_0 = f'(x_0)(x - x_0)$y−y0​=f′(x0​)(x−x0​), and the normal uses gradient $-1/f'(x_0)$−1/f′(x0​). These connect the calculus content of Section G with the line-equation content of Section E.

4. Paper 2, Section R — Kinematics: if $s(t)$s(t) is displacement, then $v(t) = \dfrac{ds}{dt}$v(t)=dtds​ and $a(t) = \dfrac{dv}{dt} = \dfrac{d^2 s}{dt^2}$a(t)=dtdv​=dt2d2s​. Mechanics rates-of-change questions are differentiation questions in disguise — same calculus, physical context.

5. Section J — Numerical methods (Pure 2): the Newton-Raphson iteration $x_{n+1} = x_n - \dfrac{f(x_n)}{f'(x_n)}$xn+1​=xn​−f′(xn​)f(xn​)​ uses the derivative directly. Failure of the iteration when $f'(x_n) = 0$f′(xn​)=0 is exactly the stationary-point condition.

---

Mark-scheme literacy

Differentiation questions on 7357 are a near-even split between AO1 fluency and AO2 reasoning, with AO3 modelling reserved for kinematics or optimisation contexts: