Pearson Edexcel International Advanced Level Mathematics: Product, quotient and chain rules (YMA01) – Revision Notes

These notes are aligned to the Pearson Edexcel International Advanced Subsidiary/Advanced Level in Mathematics, Further Mathematics and Pure Mathematics specification (Issue 3, April 2019). They cover Unit P3: Pure Mathematics 3, section 4 (Differentiation, 4.1 to 4.4). The specification states that P3 is compulsory for IAL Mathematics and Pure Mathematics and has no IAS weighting.

For full explanations and worked examples, read the study guide. When you are ready to practise, use the practice questions. The Edexcel A Level Mathematics hub and the printable checklist show the rest of the course, and the free 10-minute diagnostics show which topics need work first. Related earlier work is in differentiating powers of x.

4.1 Standard derivatives (radians)

f(x)	f′(x)
e^kx	k e^kx
ln kx	1/x
sin kx	k cos kx
cos kx	−k sin kx
tan kx	k sec² kx
sec x	sec x tan x
cosec x	−cosec x cot x
cot x	−cosec² x
a^x	a^x ln a

Differentiate sums and differences term by term. Constant multipliers stay.

Reminder. f(x) = 5e^(2x) − 3 sin 4x + 2 ln 3x gives f′(x) = 10e^(2x) − 12 cos 4x + 2/x.

4.2 The three rules

Rule	Use when	Formula
Product	two functions multiplied	y = uv, dy/dx = u′v + uv′
Quotient	one function divided by another	y = u/v, dy/dx = (v u′ − u v′) / v²
Chain	function of a function	dy/dx = dy/du × du/dx

Method in steps

1. Decide the outermost operation: sum, product, quotient or composition.
2. Name u and v (or the inner function) and write their derivatives separately.
3. Substitute into the rule, using brackets.
4. Factorise common factors, especially e^kx, which is never zero.
5. Check that every chain-rule factor has been included.

Quick chain rule patterns

y	dy/dx
(f(x))ⁿ	n (f(x))ⁿ⁻¹ f′(x)
e^f(x)	f′(x) e^f(x)
ln f(x)	f′(x) / f(x)
sin f(x)	f′(x) cos f(x)
cos f(x)	−f′(x) sin f(x)
tan f(x)	f′(x) sec² f(x)

Reminders.

y = x² cos 3x:  dy/dx = 2x cos 3x − 3x² sin 3x
y = (2x + 1)/e^x:  dy/dx = [2e^x − (2x + 1)e^x] / e^(2x) = (1 − 2x)/e^x
y = ln(cos x):  dy/dx = −sin x / cos x = −tan x
y = sec 3x:  dy/dx = 3 sec 3x tan 3x
y = tan² 2x:  dy/dx = 2 tan 2x × 2 sec² 2x = 4 tan 2x sec² 2x

4.3 Inverse relationship

dy/dx = 1 / (dx/dy)

1. Differentiate x with respect to y.
2. Take the reciprocal.
3. If asked for dy/dx in terms of x, rewrite the y-terms using x. A common route is 1 + tan² θ = sec² θ.

Reminder. x = y³ + 2y: dx/dy = 3y² + 2, so dy/dx = 1/(3y² + 2).

Reminder. x = tan 2y: dy/dx = 1/(2 sec² 2y) = 1/(2(1 + x²)).

4.4 Exponential models and a^x

• N = A e^(kt): A is the initial value (t = 0). k > 0 is growth and k < 0 is decay.
• dN/dt = kN: the rate is proportional to N.
• Half-life or doubling time: solve N = ½A or N = 2A, giving t = ln 2 / |k|.
• d/dx(a^x) = a^x ln a. For a^(kx), the chain rule gives k a^(kx) ln a.
• Large t: an exponential growth model grows without limit, and a decay model tends to zero. A model of the form N = P − Q e^(−kt) tends to the limit P.
• Evaluate a model by substituting a time and comparing with data. Refine it by changing the form so that it behaves sensibly for large t.

Reminder. For N = 300e^(−0.05t): initial value 300, rate at t = 0 is −15, half-life = ln 2 / 0.05 = 13.9 (3 s.f.).

Rebuilding results you forget

• tan x = sin x / cos x. The quotient rule gives (cos² x + sin² x) / cos² x = sec² x.
• sec x = (cos x)⁻¹. The chain rule gives sin x / cos² x = sec x tan x.
• cosec x = (sin x)⁻¹. The chain rule gives −cos x / sin² x = −cosec x cot x.
• cot x = cos x / sin x. The quotient rule gives −1 / sin² x = −cosec² x.
• ln x: put y = ln x, so x = e^y. Then dx/dy = e^y = x and dy/dx = 1/x.
• a^x = e^(x ln a), so the chain rule gives a^x ln a.

Worked reminder (quotient, with simplification). Differentiate y = sin x / (1 + cos x).

u = sin x, u′ = cos x;  v = 1 + cos x, v′ = −sin x
dy/dx = [cos x (1 + cos x) + sin x sin x] / (1 + cos x)²
      = [cos x + cos² x + sin² x] / (1 + cos x)²
      = (cos x + 1) / (1 + cos x)²  =  1 / (1 + cos x)

Show the use of sin² x + cos² x = 1 explicitly, so that a reader can follow the simplification.

Tangents, normals and stationary points

1. Differentiate to get dy/dx.
2. Substitute the x-value to get the gradient m. Substitute again into y to get the point.
3. Tangent: y − y₁ = m(x − x₁). Normal: gradient −1/m.
4. For stationary points, solve dy/dx = 0 after factorising.

Worked reminder. The curve y = x² ln x meets the line x = e at (e, e²).

dy/dx = 2x ln x + x  (product rule)
At x = e: gradient = 2e + e = 3e
Tangent: y − e² = 3e(x − e), so y = 3e x − 2e²
Normal:  y − e² = −(1/(3e))(x − e)

Worked reminder. Stationary points of y = x² e^(−2x):

dy/dx = 2x e^(−2x) − 2x² e^(−2x) = 2x(1 − x) e^(−2x)
dy/dx = 0 at x = 0 and x = 1, giving (0, 0) and (1, e^(−2))

Worked reminder (a model). A town’s population is P = 5000e^(0.04t), with t in years.

Rate at t = 10: dP/dt = 0.04 × 5000e^(0.4) = 200e^(0.4) = 298 people per year (3 s.f.)
Doubling time: e^(0.04t) = 2, so t = ln 2 / 0.04 = 17.3 years (3 s.f.)

After 10 years the model gives P = 7459, so about 7460 people. A comment on validity: unlimited growth is unrealistic over a long period, so a refined model might add an upper limit.

Checking your answer

• Test a derivative at an easy value. For y = x³ sin 2x, the derivative x²(3 sin 2x + 2x cos 2x) should give 0 at x = 0, which you can confirm by substituting.
• Check signs. A decay model must have a negative rate of change.
• Check the structure. Every term in a chain rule answer should carry one factor for each layer of the function.
• In a model question, finish with a sentence that answers the context, with units, rather than a bare number.

Must-know distinctions

• ln kx differentiates to 1/x, but e^kx differentiates to k e^kx.
• sin kx and tan kx give positive multipliers. cos kx gives a negative one.
• d/dx(sin² x) = 2 sin x cos x uses the chain rule. The expression sin x² means sin(x²), so its derivative is 2x cos x².
• Product rule is for multiplying. Chain rule is for composing. Some questions need both.
• dy/dx and dx/dy are reciprocals, but only at the same point.

Quick self-test

1. Differentiate e^(5x).
2. Differentiate ln 7x.
3. Differentiate sin(x/2).
4. Differentiate x e^(2x).
5. Differentiate (sin x)/x.
6. Differentiate cos(x²).
7. Differentiate tan² x.
8. Given x = y³ + 2y, find dy/dx in terms of y.
9. Differentiate 2^x.
10. N = 300e^(−0.05t). Find the initial value and the half-life to 3 s.f.

Answers

1. 5e^(5x)
2. 1/x
3. ½ cos(x/2)
4. (2x + 1)e^(2x)
5. (x cos x − sin x) / x²
6. −2x sin(x²)
7. 2 tan x sec² x
8. 1/(3y² + 2)
9. 2^x ln 2
10. 300; half-life = ln 2 / 0.05 = 13.9

Where marks are usually lost

• Using the wrong sign for the derivative of cos kx, or of cot x and cosec x.
• Forgetting to multiply by the derivative of the bracket, or of the angle, in a composite function.
• Mixing up the order of terms in the quotient rule numerator.
• Writing 1/(kx) or k/x for the derivative of ln kx.
• Differentiating a^x as x a^(x−1).
• Giving dx/dy when dy/dx is asked for, or not simplifying when the answer is wanted in terms of x.
• Using degrees for a trigonometric derivative or a gradient calculation.
• Giving a rate of change without its sign or units.
• Not commenting on a model’s behaviour for large t when asked to evaluate it.

Official syllabus

Pearson Edexcel International Advanced Subsidiary/Advanced Level in Mathematics, Further Mathematics and Pure Mathematics, Specification Issue 3, April 2019, published by Pearson Education Limited. Unit P3 section 4, topics 4.1 to 4.4.