Differentiation Techniques

This lesson covers the chain rule, product rule, quotient rule, and implicit differentiation — extending your differentiation toolkit for more complex functions.

The Chain Rule

If y = f(g(x)), then:

dy/dx = f’(g(x)) × g’(x)

Or, if y = f(u) and u = g(x):

dy/dx = (dy/du) × (du/dx)

Worked Example 1

Differentiate y = (3x + 2)⁵.

Let u = 3x + 2, so y = u⁵.

dy/du = 5u⁴, du/dx = 3

dy/dx = 5(3x + 2)⁴ × 3

= 15(3x + 2)⁴

Worked Example 2

Differentiate y = sin(x²).

dy/dx = cos(x²) × 2x

= 2x cos(x²)

Worked Example 3

Differentiate y = e^(3x²+1).

dy/dx = e^(3x²+1) × 6x

= 6x e^(3x²+1)

Worked Example 4

Differentiate y = ln(2x + 5).

dy/dx = 1/(2x + 5) × 2

= 2/(2x + 5)

The Product Rule

If y = u × v, where u and v are both functions of x:

dy/dx = u (dv/dx) + v (du/dx)

Worked Example 5

Differentiate y = x² sin x.

u = x², v = sin x
du/dx = 2x, dv/dx = cos x

dy/dx = x² cos x + 2x sin x

= x² cos x + 2x sin x

Worked Example 6

Differentiate y = (2x + 1)eˣ.

u = 2x + 1, v = eˣ
du/dx = 2, dv/dx = eˣ

dy/dx = (2x + 1)eˣ + 2eˣ = eˣ(2x + 1 + 2)

= eˣ(2x + 3)

The Quotient Rule

If y = u/v, where u and v are both functions of x:

dy/dx = [v (du/dx) − u (dv/dx)] / v²

Worked Example 7

Differentiate y = x²/(x + 1).

u = x², v = x + 1
du/dx = 2x, dv/dx = 1

dy/dx = [(x + 1)(2x) − x²(1)]/(x + 1)² = (2x² + 2x − x²)/(x + 1)²

= (x² + 2x)/(x + 1)²

Worked Example 8

Differentiate y = sin x/cos x (= tan x) using the quotient rule.

u = sin x, v = cos x

dy/dx = [cos x · cos x − sin x · (−sin x)]/cos²x = (cos²x + sin²x)/cos²x = 1/cos²x

= sec²x

This confirms that the derivative of tan x is sec²x.

Implicit Differentiation

When y is defined implicitly (not explicitly as y = f(x)), differentiate each term with respect to x, treating y as a function of x.

Key rule: d/dx [f(y)] = f’(y) × dy/dx

Worked Example 9

Find dy/dx for x² + y² = 25.

Differentiate: 2x + 2y(dy/dx) = 0

2y(dy/dx) = −2x

dy/dx = −x/y

Worked Example 10

Find dy/dx for x³ + 3xy + y³ = 5.

Differentiate: 3x² + 3y + 3x(dy/dx) + 3y²(dy/dx) = 0

3x(dy/dx) + 3y²(dy/dx) = −3x² − 3y

dy/dx(3x + 3y²) = −3(x² + y)

dy/dx = −(x² + y)/(x + y²)

Worked Example 11

Find the gradient of the curve x² + xy − y² = 7 at the point (3, 1).

Differentiate: 2x + y + x(dy/dx) − 2y(dy/dx) = 0

dy/dx(x − 2y) = −2x − y

dy/dx = (−2x − y)/(x − 2y) = (−6 − 1)/(3 − 2)

= −7

Connected Rates of Change

Use the chain rule to connect rates:

dy/dt = (dy/dx) × (dx/dt)

Worked Example 12

A sphere has radius r that increases at 0.5 cm/s. Find the rate of change of volume when r = 4 cm.

V = (4/3)πr³ → dV/dr = 4πr²

dV/dt = dV/dr × dr/dt = 4π(16) × 0.5

= 32π ≈ 100.5 cm³/s

Exam Tips

Chain rule: always identify the “inner” and “outer” functions.
Product rule: remember “first times derivative of second plus second times derivative of first”.
Quotient rule: “bottom times derivative of top minus top times derivative of bottom, all over bottom squared”.
For implicit differentiation, every time you differentiate a y-term, multiply by dy/dx.
Connected rates of change questions often involve geometry — know your volume and area formulae.

A-Level Deep Dive: Differentiation Techniques

Spec mapping

AQA 7357 specification, Paper 2 — Pure Mathematics, Section G (Differentiation), Year 2 sub-strands: chain rule for composite functions, product and quotient rules, implicit differentiation, parametric differentiation, and connected rates of change. The Year 1 power rule $\frac{d}{dx}x^n = nx^{n-1}$ remains the engine — every Year 2 technique reduces to applying it inside a wrapper. The AQA formula booklet lists the chain, product and quotient rules explicitly, but candidates are still expected to recall and apply them without prompting. Differentiation techniques surface throughout the spec: section H (integration, as the inverse process), section J (numerical methods, where Newton–Raphson uses $f'(x)$ ), section L (vectors, via parametric curves) and across Paper 3 statistics and mechanics whenever a rate of change is modelled.

Worked example with full mark scheme

Question (8 marks): Let $f(x) = x^2 \ln(2x + 1)$ for $x > -\tfrac{1}{2}$ .

(a) Find $f'(x)$ , simplifying your answer. (4)

(b) Show that $f$ has a stationary point at $x = 0$ , and find the $x$ -coordinate of any other stationary point in the interval $x > 0$ , giving your answer to 3 significant figures. (4)

Solution with mark scheme:

(a) Step 1 — identify the structure. $f(x) = u \cdot v$ where $u = x^2$ and $v = \ln(2x + 1)$ . Use the product rule $f' = u'v + uv'$ .

Step 2 — differentiate each factor. $u' = 2x$ . For $v = \ln(2x + 1)$ , apply the chain rule with inner function $g(x) = 2x + 1$ : $v' = \dfrac{g'(x)}{g(x)} = \dfrac{2}{2x + 1}$ .

M1 — correct identification of product structure and attempt at product rule.

M1 — correct chain-rule derivative of $\ln(2x + 1)$ , namely $\dfrac{2}{2x + 1}$ .

Step 3 — assemble.

$f'(x) = 2x \ln(2x + 1) + x^2 \cdot \dfrac{2}{2x + 1} = 2x \ln(2x + 1) + \dfrac{2x^2}{2x + 1}$

A1 — both terms correct.

Step 4 — simplify by factoring $2x$ .

$f'(x) = 2x\left[\ln(2x + 1) + \dfrac{x}{2x + 1}\right]$

A1 — clean factored form. The factorisation matters for part (b).

(b) Step 1 — set $f'(x) = 0$ . From the factored form:

$2x\left[\ln(2x + 1) + \dfrac{x}{2x + 1}\right] = 0$

Either $2x = 0 \Rightarrow x = 0$ , or the bracket equals zero.

M1 — equating to zero and identifying $x = 0$ as a root.

Step 2 — verify $x = 0$ lies in the domain. The domain requires $2x + 1 > 0$ , i.e. $x > -\tfrac{1}{2}$ . Since $0 > -\tfrac{1}{2}$ , $x = 0$ is admissible. At $x = 0$ : $\ln(1) + 0 = 0$ , confirming the bracket also vanishes (so $x = 0$ is a double root of $f'$ ).

A1 — $x = 0$ confirmed as a stationary point with reasoning.

Step 3 — search for further roots in $x > 0$ . Define $h(x) = \ln(2x + 1) + \dfrac{x}{2x + 1}$ . For $x > 0$ , both $\ln(2x + 1) > 0$ and $\dfrac{x}{2x + 1} > 0$ , so $h(x) > 0$ for all $x > 0$ . Hence the bracket has no zero in $x > 0$ , and the only stationary point with $x \geq 0$ is $x = 0$ .

M1 — sign analysis of the bracket on $x > 0$ .

A1 — conclusion: no further stationary point exists in $x > 0$ (the question's "if any" is honoured).

Total: 8 marks (M3 A4 plus 1 communication).

Specimen question modelled on the AQA 7357 Paper 2 format

Question (6 marks): A curve is defined parametrically by $x = t^2 + 1$ , $y = t^3 - 3t$ for $t \in \mathbb{R}$ .

(a) Find $\dfrac{dy}{dx}$ in terms of $t$ . (2)

(b) Find the coordinates of the points on the curve where the tangent is parallel to the $x$ -axis. (4)

Mark scheme decomposition by AO:

(a)

M1 (AO1.1a) — using $\dfrac{dy}{dx} = \dfrac{dy/dt}{dx/dt}$ with $\dfrac{dx}{dt} = 2t$ and $\dfrac{dy}{dt} = 3t^2 - 3$ .
A1 (AO1.1b) — $\dfrac{dy}{dx} = \dfrac{3t^2 - 3}{2t} = \dfrac{3(t^2 - 1)}{2t}$ .

(b)

M1 (AO1.1b) — setting numerator to zero: $3(t^2 - 1) = 0$ , giving $t = \pm 1$ .
A1 (AO1.1b) — both values found, with the implicit check $t \neq 0$ (else $dy/dx$ undefined).
M1 (AO2.1) — substituting back into $x$ and $y$ : at $t = 1$ , $(x, y) = (2, -2)$ ; at $t = -1$ , $(x, y) = (2, 2)$ .
A1 (AO2.5) — both coordinate pairs stated cleanly.

Total: 6 marks split AO1 = 4, AO2 = 2. Parametric questions reward candidates who keep $t$ as the working variable until the very last step — premature elimination of $t$ to produce a Cartesian equation usually multiplies the algebra unnecessarily.

Synoptic links

Connects to:

Year 1 basic differentiation (Section G): every Year 2 technique reduces to the power rule applied inside a wrapper. The chain rule $\frac{d}{dx}f(g(x)) = f'(g(x)) \cdot g'(x)$ is just the power rule "with bookkeeping". Confidence with $\frac{d}{dx}x^n$ for fractional and negative $n$ is a prerequisite — without it, differentiating $\sqrt{2x + 1}$ stalls.
Section H — Integration as inverse: every chain-rule pattern has a reversal. $\frac{d}{dx}\ln(2x + 1) = \dfrac{2}{2x + 1}$ tells you immediately that $\int \dfrac{1}{2x + 1}\,dx = \tfrac{1}{2}\ln|2x + 1| + C$ . Recognition of "the chain rule run backwards" is the core skill of integration by inspection and substitution.
Modelling rates of change (Section O of applied papers): connected rates use the chain rule $\dfrac{dV}{dt} = \dfrac{dV}{dr} \cdot \dfrac{dr}{dt}$ . A spherical balloon inflating at a known volume rate yields a radius rate via $V = \tfrac{4}{3}\pi r^3 \Rightarrow \dfrac{dV}{dr} = 4\pi r^2$ . Every "rates" problem in mechanics, physics or biology obeys the same chain.
Tangents and normals: gradient at a point requires $f'(x_0)$ . For implicit curves like $x^2 + xy + y^2 = 7$ , implicit differentiation gives $\dfrac{dy}{dx}$ as an expression in both $x$ and $y$ , and the tangent line at $(x_0, y_0)$ uses that mixed gradient directly.
Second derivative and stationary-point classification: once $f'(x) = 0$ is solved, classifying the stationary point uses $f''(x)$ . $f''(x_0) > 0$ gives a local minimum; $f''(x_0) < 0$ gives a local maximum; $f''(x_0) = 0$ requires a sign-change test on $f'$ either side. Differentiating a product or quotient twice multiplies the bookkeeping — clean factoring of $f'$ pays off here.

Mark-scheme literacy

Differentiation-technique questions on 7357 split AO marks fairly evenly between AO1 and AO2:

AO	Typical share	Earned by
AO1 (knowledge / procedure)	55–65%	Stating and applying the chain, product, quotient rules; differentiating standard forms ( $e^{kx}$ , $\ln(kx)$ , $\sin(kx)$ ) correctly
AO2 (reasoning / interpretation)	25–35%	Choosing between product and quotient rules sensibly; factoring $f'(x)$ to expose stationary points; justifying sign of $f'$ on an interval
AO3 (problem-solving)	5–15%	Modelling-led questions: connected rates, optimisation in real contexts, parametric trajectories

Examiner-rewarded phrasing: "by the product rule with $u = \ldots$ and $v = \ldots$ "; "by the chain rule with inner function $g(x) = \ldots$ "; "differentiating implicitly with respect to $x$ , treating $y$ as a function of $x$ "; "since $\dfrac{dx}{dt} \neq 0$ at this value of $t$ ". Phrases that lose marks: jumping straight to $f'(x) = \ldots$ without stating which rule is used (loses the "method seen" mark when a slip later in the working makes the rule unrecoverable); writing $\dfrac{d}{dx}\sin(2x) = \cos(2x)$ (missing chain factor of 2); confusing the product rule sign $u'v + uv'$ with the quotient rule sign $\dfrac{u'v - uv'}{v^2}$ .

A specific AQA pattern to watch: implicit-differentiation questions often ask for $\dfrac{dy}{dx}$ at a specific point, not as a general expression. Plugging in numbers immediately (rather than carrying symbolic $y$ through) shortens the algebra and reduces slip risk. Read the question to spot when a numerical evaluation is requested.

Grade-band model answers

3-mark question

Question: Differentiate $y = \sin(3x^2 + 1)$ with respect to $x$ .

Grade C response (~140 words):

Differentiation Techniques

Differentiation Techniques

The Chain Rule

Worked Example 1

Worked Example 2

Worked Example 3

Worked Example 4

The Product Rule

Worked Example 5

Worked Example 6

The Quotient Rule

Worked Example 7

Worked Example 8

Implicit Differentiation

Worked Example 9

Worked Example 10

Worked Example 11

Connected Rates of Change

Worked Example 12

Exam Tips

A-Level Deep Dive: Differentiation Techniques

Spec mapping

Worked example with full mark scheme

Specimen question modelled on the AQA 7357 Paper 2 format

Synoptic links

Mark-scheme literacy

Grade-band model answers

3-mark question

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