OxfordAQA International AS and A-level Mathematics: Product, quotient and chain rules; e^x, ln x, trig (9660) – Revision Notes

These notes condense the first part of specification section P2.6 (Differentiation) of the OxfordAQA International AS and A-level Mathematics (9660) specification, Version 5.2. The specification states that this content is part of International A-level (A2) Unit P2 (Pure Maths) and is not examined at International AS. For full explanations and fully worked examples, read the study guide. When you are ready to test yourself, use the practice questions. Related pages are the P1 differentiation notes, the course hub and the printable checklist. To find your weak spots, try the free diagnostics.

Standard results (x in radians)

y	dy/dx
e^x	e^x
ln x	1/x
sin x	cos x
cos x	−sin x
tan x	sec²x = 1/cos²x
e^(kx)	k e^(kx)
sin(kx)	k cos(kx)
cos(kx)	−k sin(kx)
tan(kx)	k sec²(kx)
ln(ax + b)	a/(ax + b)

Sums, differences and constant multiples are differentiated term by term. The derivative of ln(kx) is 1/x, because ln(kx) = ln k + ln x.

The three rules

Rule	Statement	Use when
Chain	dy/dx = dy/du × du/dx	One function inside another
Product	If y = uv then dy/dx = u v′ + v u′	Two x-functions multiplied
Quotient	If y = u/v then dy/dx = (v u′ − u v′)/v²	Two x-functions divided
Inverse	dy/dx = 1/(dx/dy)	x given as a function of y

Method in steps: product or quotient rule

1. Name u and v.
2. Write down u′ and v′ (use the chain rule here if needed).
3. Substitute into the rule, with the quotient numerator in the order v u′ − u v′.
4. Take out common factors. Check whether the result simplifies.

Method in steps: chain rule

1. Identify the inner function u (the bracket, the power, the angle, the exponent).
2. Differentiate the outer function with respect to u.
3. Multiply the two derivatives together, using du/dx.
4. Write the result in terms of x only.

Method in steps: inverse rule

1. Differentiate x with respect to y.
2. Substitute the given y-value (or keep the expression).
3. Flip the fraction.

Small worked reminders

Product with a trig function: y = x² cos x.

dy/dx = x²(−sin x) + cos x (2x) = 2x cos x − x² sin x

Quotient: y = (5 − x)/(2x + 3).

dy/dx = (−1(2x + 3) − (5 − x)(2))/(2x + 3)² = −13/(2x + 3)²

Chain: y = e^(x²) gives dy/dx = 2x e^(x²). Also y = tan(4x − 1) gives dy/dx = 4 sec²(4x − 1).

Product with a linear exponent: y = x e^(4x).

dy/dx = x(4e^(4x)) + e^(4x)(1) = e^(4x)(4x + 1)

Inverse: x = y² + 7y at the point where y = 2 (so x = 18).

dx/dy = 2y + 7 = 11, so dy/dx = 1/11

Must-know distinctions

• d/dx of e^(3x) is 3e^(3x), while d/dx of e^(x³) is 3x² e^(x³). The inner function decides the factor.
• sin²x means (sin x)². Its derivative uses the chain rule: 2 sin x cos x.
• The derivative of a product is not the product of the derivatives.
• The derivative of a quotient is not the quotient of the derivatives.
• dy/dx = 1/(dx/dy) gives the reciprocal of the gradient value. It does not say dy/dx = −1/(dx/dy).
• ln(x²) and (ln x)² differ: d/dx ln(x²) = 2x/x² = 2/x, but d/dx (ln x)² = 2 ln x / x.
• If a question asks for a value, put x in radians for trigonometric functions.

Linear combinations and constants

Differentiate each term separately, and write each constant factor outside before you differentiate. Take y = 4e^(x/3) − 6 cos 2x + 9 ln x − tan 5x.

dy/dx = 4(1/3)e^(x/3) − 6(−2 sin 2x) + 9(1/x) − 5 sec²(5x)
      = (4/3)e^(x/3) + 12 sin 2x + 9/x − 5 sec²(5x)

Two signs interact here: the minus in front of the cosine term and the minus in the derivative of cos. They multiply to give a plus.

Which rule?

Function	Rule	Result
sin(x²)	Chain	2x cos(x²)
cos²x	Chain (power of a function)	−2 sin x cos x
√x e^(−x)	Product	e^(−x)(1 − 2x)/(2√x)
x²/(1 + e^x)	Quotient	(2x(1 + e^x) − x² e^x)/(1 + e^x)²
x ln(3x + 1)	Product, with chain inside	ln(3x + 1) + 3x/(3x + 1)

The last row is a typical combined question. The gradient at x = 1 is 3/4 + ln 4. Leave it exact unless the question asks for decimals.

How the results link together

You do not need to memorise every result separately.

• Writing tan x as a fraction, sin x over cos x, and using the quotient rule produces sec²x.
• If y = ln x then x = e^y. The inverse rule gives dy/dx = 1/e^y = 1/x.
• The derivative of e^(kx) comes from the chain rule with u = kx.
• The derivative of ln(ax + b) is the chain rule with u = ax + b and dy/du = 1/u.

If you forget a result in an exam, derive it from a rule you do remember and check it with a quick test value.

Checking your answer

1. Substitute a simple value such as x = 0 or x = 1 into your derivative. Does the sign of the gradient agree with what you expect from the original function?
2. Check the form. A product-rule answer has two terms, and a quotient-rule answer has a squared denominator.
3. Check that the power of e or the angle is unchanged in the answer, since only the factor outside changes.
4. If the question asks you to show a given result, work towards it and finish on the printed form.

Using the derivative

Two short patterns use these derivatives again and again.

Gradient equal to zero. Find the x-coordinate where y = e^(2x) − 8e^x has zero gradient.

dy/dx = 2e^(2x) − 8e^x = 2e^x(e^x − 4)
e^x is never zero, so e^x = 4 and x = ln 4
y = 16 − 32 = −16

Rate of change in context. The mass of a sample is M = 500e^(−0.02t) grams after t days.

dM/dt = 500(−0.02)e^(−0.02t) = −10e^(−0.02t)
At t = 30: dM/dt = −10e^(−0.6) = −5.49 g per day (3 s.f.)

The negative sign shows that the mass is decreasing. Include the unit in your answer.

Quick self-test

Differentiate each of these with respect to x, unless told otherwise.

1. y = 5e^(−x)
2. y = cos 5x
3. y = ln(4x − 1)
4. y = tan 3x
5. y = x² sin x
6. y = e^x/x
7. y = sin(e^x)
8. A curve has equation x = 3y² − 2y + 4. Find dy/dx where y = 1.
9. Find the gradient of y = (x + 1)/(x − 2) at x = 3.
10. y = ln(sin x)
11. Find the value of the gradient of y = e^(2x) ln x at x = 1.
12. y = x e^x − e^x. Simplify your answer.

Answers

1. −5e^(−x)
2. −5 sin 5x
3. 4/(4x − 1)
4. 3 sec²(3x)
5. 2x sin x + x² cos x
6. e^x(x − 1)/x²
7. e^x cos(e^x)
8. dx/dy = 6y − 2 = 4 at y = 1, so dy/dx = 1/4 (the point is (5, 1))
9. dy/dx = −3/(x − 2)², which is −3 at x = 3
10. cos x / sin x
11. 2e² ln 1 + e² × 1 = e²
12. x e^x

Where marks are usually lost

• Missing a minus sign on cos x, or on the second term of a quotient numerator.
• Omitting the factor from the inner function in e^(3x), sin(2x) or (4x − 1)⁵.
• Writing the product-rule answer as u′v′.
• Leaving the answer unsimplified when a factor such as e^(−2x) can be taken out, then failing to spot the zero.
• Not showing u, v, u′ and v′, so one slip hides the method.
• Substituting degrees instead of radians into a trigonometric derivative.
• Forgetting to flip dx/dy at the end of an inverse-rule question.
• Treating e^x − 4 = 0 as if e^x could be zero, or cancelling a common factor that may be zero, instead of factorising and reading off the solutions.
• Losing a bracket when a minus sign sits in front of a product, so that only the first term changes sign.
• Using the product rule when only a constant multiplies the function, which wastes time and invites slips.
• Giving a rounded decimal when the question implies an exact answer such as 3√3/4 or e².

Official syllabus

OxfordAQA International AS and A-level Mathematics (9660), Version 5.2, published by OxfordAQA: section 3.3, International A-level Unit P2 (Pure Maths), content P2.6 Differentiation (differentiation of e^x, ln x, sin x, cos x and tan x, and the product, quotient and chain rules, and dy/dx = 1/(dx/dy)).