Educerie · IB Diploma · Mathematics: analysis and approaches
Topic 5 Calculus · 5.6 Standard derivatives, and the chain, product and quotient rules
What you must be able to do
| You must be able to | Level | What it looks like in the exam |
|---|---|---|
| Differentiate xⁿ for rational n, and sin x, cos x, eˣ, ln x | SL, HL | "Find f′(x)" for f(x) = 4√x − 3 sin x + 2eˣ (2 to 3 marks) |
| Differentiate sums and multiples of these | SL, HL | Term by term, as in 5.3 |
| Use the chain rule for composite functions | SL, HL | "Differentiate sin(3x − 1)"; "Find the derivative of ex² + 2" (2 to 3 marks) |
| Use the product rule | SL, HL | "Show that the derivative of x³e2x is x²e2x(3 + 2x)" (3 marks) |
| Use the quotient rule | SL, HL | "Show that f′(x) = (1 − ln x) ÷ x²" (3 marks) |
| Choose the right rule, and combine rules | SL, HL | Product with a chain inside; quotient with a chain inside |
| Use the derivative afterwards: exact gradients, tangents, stationary points | SL, HL | Paper 1 follow-on parts (2 to 4 marks) |
Before you start
You need 5.3 (the power rule, term by term) and 5.4 (tangents), and composite functions from 2.5, because the chain rule is about a function of a function. You need the exact values of sin and cos from 3.5, and the laws of logarithms from 1.7, which often simplify a function before you differentiate it. The formula booklet gives all five standard derivatives and all three rules, so the marks are for choosing and applying them, not for recalling them. Your calculator must be in radians for any calculus with trigonometric functions.
1The idea in one paragraph
Five functions have derivatives you should know by sight: xⁿ gives nxⁿ⁻¹ for any rational n, sin x gives cos x, cos x gives −sin x, eˣ gives itself, and ln x gives 1/x. Everything else in the course is built from these by four operations. Sums you already handle term by term. For the other three there is one rule each. A function of a function, like sin(4x + 1), needs the chain rule: differentiate the outside, keep the inside, multiply by the derivative of the inside. A product of two functions, like x²eˣ, needs the product rule. A quotient, like x ÷ (x² + 1), needs the quotient rule. The skill is recognising which operation is on the outside, and then being patient with the algebra.
2The standard derivatives
d/dx xⁿ = nxⁿ⁻¹ (n ∈ ℚ) · d/dx sin x = cos x · d/dx cos x = −sin x · d/dx eˣ = eˣ · d/dx ln x = 1/x. (Formula booklet; x in radians.)
Rational powers. The power rule from 5.3 works unchanged for fractional powers, so roots are just powers in disguise. Rewrite first.
sin and cos. Figure 1 shows why the derivative of sin x is cos x: where sin x is steepest going up (x = 0), its gradient is 1, the value of cos 0; at the top (x = π/2) the gradient is 0 = cos(π/2); going down through π the gradient is −1 = cos π. The gradient of cos x, similarly, is −sin x: the minus sign is there because cos x starts by falling. These results rely on x being in radians; in degrees they would carry an extra factor of π/180, which is one reason calculus is always done in radians.
eˣ. The exponential function is the one function whose gradient at every point equals its height (Figure 2). That property is what makes e the natural base: it is how e is defined.
ln x. The natural logarithm, defined for x > 0, has gradient 1/x (Figure 3): 1 at x = 1, 1/2 at x = 2, and so on, always positive but shrinking, so the curve keeps rising more and more slowly.
Sums and multiples work term by term, exactly as before:
A logarithm law can save work. ln(5x) = ln 5 + ln x, and ln 5 is a constant, so d/dx ln(5x) = 1/x. Likewise ln(x³) = 3 ln x, whose derivative is 3/x.
3The chain rule
A composite function is a function of a function (2.5): in y = sin(4x + 1), first x is turned into u = 4x + 1 (the inner function), then u is turned into y = sin u (the outer function). Figure 4 draws the two stages.
A small change in x is multiplied by 4 on its way into u, because du/dx = 4. That change in u is then multiplied by cos u on its way into y, because dy/du = cos u. Rates through a chain multiply:
Chain rule: if y = g(u) and u = f(x), then dy/dx = dy/du × du/dx. (Formula booklet.) In words: differentiate the outside, leave the inside alone, then multiply by the derivative of the inside.
Once you are used to it, skip the u and say it aloud: "derivative of sin is cos, of the same thing, times the derivative of the inside".
Three patterns come up so often that they are worth knowing as shortcuts, each a special case of the chain rule:
| Function | Derivative | Example |
|---|---|---|
| f(ax + b), a linear inside | a f′(ax + b) | d/dx cos(3x) = −3 sin(3x) |
| eg(x) | g′(x) eg(x) | d/dx e−0.5x = −0.5e−0.5x |
| ln g(x) | g′(x) ÷ g(x) | d/dx ln(x² + 1) = 2x ÷ (x² + 1) |
Notice the difference between sin(x²), where the inside is x² and the answer is 2x cos(x²), and sin²x, which is (sin x)², where the outside is the square and the answer is 2 sin x cos x. Read the notation carefully before choosing inner and outer.
4The product rule
When a function is a product of two functions of x, such as x²eˣ, you cannot differentiate each factor and multiply: d/dx(x²eˣ) is not 2x × eˣ. The correct rule has two terms.
Product rule: if y = uv, then dy/dx = u dv/dx + v du/dx. (Formula booklet.) In words: first times derivative of second, plus second times derivative of first.
Figure 5 shows why there are two terms. Think of uv as the area of a rectangle with sides u and v. If x changes a little, u grows by Δu and v by Δv, and the area grows by two strips, u × Δv and v × Δu, plus a tiny corner Δu × Δv. Divide by Δx and let it shrink: the corner vanishes in the limit and the two strips give the two terms.
Set out the four pieces before combining them. It is slower by ten seconds and much safer.
More examples:
A product with a chain inside. For y = x(2x + 1)⁴, v = (2x + 1)⁴ needs the chain rule for dv/dx.
Factorising at the end is not decoration. A "show that" question will print the factorised form, and a stationary-point question needs it: (2x + 1)³(10x + 1) = 0 is easy to solve, while the unfactorised version is not.
5The quotient rule
For a quotient of two functions of x:
Quotient rule: if y = u/v, then dy/dx = (v du/dx − u dv/dx) ÷ v². (Formula booklet.) The order on top matters: bottom times derivative of top, minus top times derivative of bottom, all over bottom squared.
Because of the minus sign, swapping the two terms on top changes the sign of the whole answer. The product rule is a sum, so its order does not matter; the quotient rule is not.
tan x. tan x is not on the standard list at standard level, but the quotient rule gives it: tan x = sin x ÷ cos x, so
Sometimes you do not need it. If the denominator is a single power of x, split the fraction instead, as in 5.3: (x² + 3x − 1) ÷ x = x + 3 − x⁻¹ is quicker than any quotient rule. And a constant on top is a chain-rule job: 5 ÷ (2x + 1)³ = 5(2x + 1)⁻³, with derivative −30(2x + 1)⁻⁴. Both methods give the same answer; the quicker one leaves less room for error.
6Choosing the rule
Figure 6 is the decision. Ask what you would do last if you were evaluating the function at, say, x = 2.
- x²eˣ at x = 2: work out 4, work out e², then multiply. Last operation ×: product rule.
- ex² − 3x: work out x² − 3x = −2, then take e to that power. Last operation is applying a function to a result: chain rule.
- x ÷ (x² + 1): work out both, then divide. Quotient rule.
- A number times a function, such as 5 sin x, is not a product in this sense: the 5 is a constant, and the constant stays in front.
When rules nest, apply the outer rule first and bring in the inner rule when a piece needs it, as with x(2x + 1)⁴ above: product rule outside, chain rule inside dv/dx.
7Using the derivative: exact values, tangents, stationary points
Paper 1 rarely stops at differentiating. Three follow-ons are typical.
An exact gradient. Find the gradient of y = x² ln x at x = e.
The answer 3e is exact. Writing 8.15 on Paper 1 when the exact value is available can cost the accuracy mark.
A tangent. The tangent to y = e2x at x = 0: y = e⁰ = 1, dy/dx = 2e2x, which is 2 at x = 0. So y − 1 = 2(x − 0), that is y = 2x + 1.
Stationary points. For y = xe−x: dy/dx = x × (−e−x) + e−x × 1 = e−x(1 − x). Since e−x > 0 for every x, the derivative is zero only when 1 − x = 0, so x = 1, at the point (1, e⁻¹). Taking out the exponential factor and noting that it is never zero is a standard move worth an R mark.
Rates of change. A cup of tea cools so that its temperature after t minutes is T = 20 + 60e−0.05t °C. Then dT/dt = 60 × (−0.05)e−0.05t = −3e−0.05t. At t = 10, dT/dt = −3e−0.5 ≈ −1.82: after 10 minutes the tea is cooling at about 1.82 °C per minute.
On Paper 2 you can check any derivative numerically: compare your f′(a) with the GDC's numerical derivative at the same a, in radians.
8Where marks are lost
Differentiating a product factor by factor. d/dx(x²eˣ) is not 2xeˣ. Two functions of x multiplied need the product rule.
Forgetting the derivative of the inside. d/dx sin(4x + 1) is 4 cos(4x + 1), not cos(4x + 1). Every chain-rule answer ends with "× derivative of the inside".
Swapping the terms in the quotient rule. v du/dx − u dv/dx, bottom first. The other order gives the negative of the right answer.
Sign of the cosine derivative. d/dx cos x = −sin x. And d/dx(−cos x) = sin x.
Differentiating ln(5x) as 5/x or 1/(5x). By the chain rule it is 5 × 1/(5x) = 1/x.
Confusing sin(x²) with sin²x. The first has x² inside; the second is (sin x)². They have different derivatives.
Degrees on the calculator. Every calculus result for sin and cos assumes radians. A GDC in degrees gives wrong gradients silently.
Leaving an unfactorised answer on a "show that". If the target is x²e2x(3 + 2x), your line 3x²e2x + 2x³e2x must be followed by the factorisation, or the final mark is lost.
9Work it right
- Rewrite first: roots as fractional powers, 1/xⁿ as x⁻ⁿ, and use log laws where they simplify.
- Identify the last operation: sum, chain, product or quotient.
- For a product or quotient, write u, du/dx, v, dv/dx on separate lines before combining.
- For a chain, name the inside, differentiate the outside with the inside unchanged, and multiply by the inside's derivative.
- Substitute into the rule exactly as the formula booklet writes it.
- Simplify by taking out common factors, especially exponentials and brackets.
- On "show that", write every line until your expression matches the printed one.
- For exact values on Paper 1, keep e, π, ln and surds; use ln e = 1, e⁰ = 1 and exact trigonometric values.
10Try it
Marks in brackets. Q1 to Q4 are Paper 1 style, no calculator. Q5 is Paper 2 style, with a GDC.
Q1. Differentiate with respect to x:
(a) f(x) = 5eˣ − 4 ln x + 6√x 3 marks
(b) g(x) = sin(3x² − 1) 2 marks
Q2. Let y = x³e2x.
(a) Show that dy/dx = x²e2x(3 + 2x). 3 marks
(b) Hence find the values of x for which dy/dx = 0. 2 marks
Q3. Let f(x) = ln x ÷ x, for x > 0.
(a) Show that f′(x) = (1 − ln x) ÷ x². 3 marks
(b) Find the exact value of x for which f′(x) = 0. 2 marks
Q4. Find the equation of the tangent to the curve y = √(2x + 5) at the point where x = 2. Give your answer in the form ax + by + d = 0, where a, b, d ∈ ℤ. 5 marks
Q5. The temperature of a bowl of soup t minutes after it is served is T = 20 + 60e−0.05t degrees Celsius.
(a) Find an expression for dT/dt. 2 marks
(b) Find the value of dT/dt when t = 10, and interpret your answer. 2 marks
(c) Find the time at which the soup is cooling at a rate of 1 °C per minute. 2 marks
11In one breath
Know five derivatives on sight: xⁿ gives nxⁿ⁻¹ for any rational n (so rewrite roots as powers first), sin x gives cos x, cos x gives −sin x, eˣ gives eˣ and ln x gives 1/x, all in radians. Sums and multiples go term by term. For a function of a function, use the chain rule, dy/dx = dy/du × du/dx: differentiate the outside, keep the inside, multiply by the inside's derivative, so sin(4x + 1) gives 4 cos(4x + 1) and eg(x) gives g′(x)eg(x). For a product use u dv/dx + v du/dx, and for a quotient (v du/dx − u dv/dx) ÷ v², bottom first because of the minus sign. Choose the rule by the last operation you would do to evaluate the function, write u, v and their derivatives on separate lines, then factorise the result, because the next part will want exact gradients, tangents or the stationary points where your factorised derivative is zero.
Answers
Q1. (a) f′(x) = 5eˣ − 4/x + 3x−1/2 (or 3/√x). A1 for each term.
(b) g′(x) = 6x cos(3x² − 1). M1 for cos(3x² − 1) multiplied by a derivative of the inside, A1 for 6x cos(3x² − 1).
Q2. (a) u = x³, du/dx = 3x²; v = e2x, dv/dx = 2e2x. dy/dx = x³ × 2e2x + e2x × 3x² = 2x³e2x + 3x²e2x = x²e2x(3 + 2x). A1 for dv/dx = 2e2x (chain rule), M1 for the product rule with their derivatives, A1 for a correct unfactorised expression leading to the given answer. The factorisation must be seen, since the answer is given.
(b) e2x > 0 for all x, so x²(3 + 2x) = 0, giving x = 0 or x = −3/2. A1 for each value. Stating e2x ≠ 0 is expected.
Q3. (a) u = ln x, du/dx = 1/x; v = x, dv/dx = 1. f′(x) = (x × (1/x) − ln x × 1) ÷ x² = (1 − ln x) ÷ x². A1 for du/dx = 1/x, M1 for the quotient rule in the correct order, A1 for simplifying x × (1/x) = 1 to reach the given answer.
(b) 1 − ln x = 0, so ln x = 1 and x = e. M1 for setting the numerator to 0, A1 for x = e. A decimal 2.72 is A0 when the exact value is asked for.
Q4. At x = 2, y = √9 = 3. dy/dx = (1/2)(2x + 5)−1/2 × 2 = 1 ÷ √(2x + 5), so the gradient is 1/3. Tangent: y − 3 = (1/3)(x − 2), so 3y − 9 = x − 2, and x − 3y + 7 = 0. A1 for y = 3, M1 for the chain rule, A1 for the gradient 1/3, M1 for the point-gradient form with their values, A1 for the integer form (any non-zero multiple).
Q5. (a) dT/dt = −3e−0.05t. M1 for the chain rule on the exponential, A1 for −3e−0.05t.
(b) dT/dt = −3e−0.5 = −1.82 (3 s.f.). Ten minutes after it is served, the soup's temperature is decreasing at 1.82 °C per minute. A1 for −1.82, A1 for the interpretation with "decreasing", units and t = 10.
(c) 3e−0.05t = 1, so e−0.05t = 1/3 and t = 20 ln 3 = 22.0 minutes (3 s.f.). M1 for setting dT/dt = −1 (or its size equal to 1), A1 for 22.0. Setting dT/dt = +1 has no solution and scores M0.
Educerie · written from the published IB Diploma Programme Mathematics: analysis and approaches guide, first assessment 2021, section 5.6 Standard derivatives, and the chain, product and quotient rules. Original text, examples and questions. Diagrams drawn by Educerie. Last reviewed 25 September 2026.