About this chapter
Pure 3 differentiation brings the chain, product and quotient rules, and with them the ability to differentiate almost any function on the course. The rules themselves are well known; marks go in the algebra afterwards, in missing statements, and in questions where x is given as a function of y.
This chapter covers the standard derivatives of e, ln and the trig functions, the three rules, finding dy/dx from dx/dy, tangents and normals with exact values, stationary points, increasing and decreasing functions, and rearranging f′(x) = 0 into a given form. One question ends by showing that π to the power e is less than e to the power π.
The ten questions
- Q1 (5 marks): the product rule on e to the power 3x times ln 2x, and the quotient rule on (2x + 1)/cos x.
- Q2 (5 marks): show a quotient-rule derivative has the form a/(x+2)², then a tangent.
- Q3 (5 marks): x = 2 tan 3y: find dy/dx in terms of y, then show it equals 2/(3(x² + 4)).
- Q4 (5 marks): the exact stationary point of x times e to the power −2x, and its maximum value.
- Q5 (5 marks): where ln(2x + 1) + 4/(2x + 1) is increasing.
- Q6 (10 marks): a product with a squared bracket and an exponential: turning points, a transformation, and when f(x) = k has three solutions.
- Q7 (10 marks): f(x) = 2x cos(x/2): rearrange f′(x) = 0, then a normal and the area of a triangle.
- Q8 (10 marks): show a quotient-rule derivative simplifies with double-angle identities, a tangent, and why there are no stationary points.
- Q9 (10 marks): the maximum of (ln x)/x, used to prove that π to the power e is less than e to the power π.
- Q10 (11 marks): x = y times e to the power 2y: dy/dx in terms of y, a tangent, and where the tangent is vertical.
Key skills tested
Standard derivatives. e to the power kx differentiates to k times itself, ln x to 1/x, sin kx to k cos kx, cos kx to −k sin kx, and tan kx to k sec²kx.
The chain rule. dy/dx = dy/du × du/dx. For example, ln(2x + 1) differentiates to 2/(2x + 1).
The product rule. For y = uv, dy/dx = u dv/dx + v du/dx. Factorise the answer, for example taking out the exponential.
The quotient rule. For y = u/v, dy/dx = (v du/dx − u dv/dx)/v². The order in the numerator matters.
x as a function of y. dy/dx = 1 ÷ dx/dy. Use an identity to write the answer in terms of x.
Tangents and normals. The normal's gradient is −1/m. Keep values exact, including powers of e.
Stationary points. Solve dy/dx = 0. An exponential factor is never zero, so only the other factors give solutions.
Increasing and decreasing. Find the sign of dy/dx and finish with a statement, such as "so g is increasing".
Using the derivative. Rearrange f′(x) = 0 into a given form, or use a maximum to compare two values.
Worked example
Question 4 from this chapter. The curve C has equation y = x times e to the power −2x. (a) Use calculus to find the exact coordinates of the stationary point. (b) Hence write down the maximum value of x times e to the power −2x.
(a) By the product rule, dy/dx = e to the power −2x plus x × (−2 e to the power −2x), which factorises to (1 − 2x) times e to the power −2x.
The exponential is never zero, so dy/dx = 0 only when 1 − 2x = 0, giving x = ½. Then y = ½ times e to the power −1, which is 1/(2e). The stationary point is (½, 1/(2e)).
(b) The maximum value is 1/(2e).
Seen on real papers
Every question in our booklets is original. These are the recent papers where each question type has appeared.
- June 2024, Q5(b): the quotient rule used to prove a function is increasing, as in Q5.
- June 2024, Q6(a): a normal to a curve found with the chain rule, as in Q7.
- June 2024, Q8(b): the product rule on a model, with the derivative set to zero and rearranged to a given form, as in Q7(b).
- June 2024, Q9(b) and (c): dy/dx found from dx/dy and written in terms of x, then a normal and a triangle's area, as in Q3 and Q10.
Where marks are lost
- No conclusion. In June 2024 many correct derivatives lost the final mark because no comment linked the sign of the derivative to the function being increasing.
- A sign error in the numerator. Multiplying out the second bracket of the quotient rule wrongly was a frequent slip in June 2024.
- A product rule with one term. A minority wrote only one of the two products, and could make no further progress.
- Answers left in y. When x is a function of y, the final derivative usually has to be written in terms of x using an identity.
Common questions
Product rule or quotient rule?
Either works for a quotient, since u/v is u times v to the power −1. The quotient rule is usually quicker to simplify.
How do I differentiate when x is given in terms of y?
Find dx/dy, then take its reciprocal to get dy/dx.
What does "use calculus" require?
Differentiate, set the derivative equal to zero, and show the working. Reading a turning point from a graph does not count.
Where this chapter leads
- Pure 3 Chapter 8: Numerical Methods: stationary points found by iteration when f′(x) = 0 cannot be solved exactly.
- Pure 4 Chapter 5: Differentiation: implicit and parametric differentiation, and rates of change.