Differentiate products while retaining both changing factors
Use two product-rule terms, combine them with the chain rule and check a time model's signed rate.
Section 2.4 · Differentiation
Exam scope and assessment notes
Pure Mathematics 2 (Paper 2, AS only) and Pure Mathematics 3 (Paper 3, A Level) · Mathematics 9709, 2026–2027. Paper 1 knowledge is assumed. The AS route using Papers 1 + 2 cannot be carried forward to A Level. Shared sections 2.4 and 3.4 cover exponential, natural logarithmic and trigonometric derivatives, composites, products and quotients, and first derivatives of parametric and implicit curves, including tangents and normals. Paper 3 additionally requires the derivative of inverse tangent; use its separate lesson for that method. Derivatives of inverse sine and inverse cosine are not required in section 3.4. Trigonometric differentiation uses radians. Keep original domains and check zero derivative denominators separately. Written practice is self-checked.
Before you begin
- Differentiate powers, natural functions and radian trigonometric functions.
- Apply the chain rule and keep positive logarithm arguments.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
Which expression is the derivative of u(x)v(x)?
What you will learn
- Apply (uv)′ = u′v + uv′ to polynomial, logarithmic, exponential and trigonometric products.
- Combine product and chain rules and interpret an evaluated rate on the original domain.
A product's derivative has two contributions
If both u(x) and v(x) change, the product rule is d(uv)/dx = u′v + uv′. In each term one factor is differentiated while the other remains unchanged. It is not u′v′. For y = x² ln x on x > 0, the two terms are (2x)ln x and x²(1/x), giving y′ = 2x ln x + x. The original logarithm still excludes zero and negative x even if part of the simplified derivative is defined there.
A product may contain a composite. For y = x e^(1 − x²), use u = x and v = e^(1 − x²). Then u′ = 1 and v′ = −2x e^(1 − x²). The product rule gives e^(1 − x²) − 2x²e^(1 − x²) = e^(1 − x²)(1 − 2x²), valid for every real x. A shared exponential factor makes the derivative's sign easier to inspect, but it does not replace either product contribution.
Check a result before using it as a gradient or rate
For y = (x² + 1)sin x with radian input, the derivative is 2x sin x + (x² + 1)cos x. At x = 0 the point is (0, 0) and the gradient is 1, so the tangent is y = x and the normal is y = −x. At a supplied point, evaluate the original product for the height and the derivative for the gradient; these are separate calculations.
A supplied normalised model M(t) = (t + 1)e^(−t) for t ≥ 0 hours has rate M′(t) = e^(−t) − (t + 1)e^(−t) = −t e^(−t) per hour. Its initial rate is zero, while for t > 0 it decreases because the exponential stays positive. At t = 1 the rate is −e^(−1) per hour. The initial time is a boundary of the model domain, so do not describe it as an interior stationary point merely because its rate is zero.
Pause and explain
For x² ln x, what is the second product-rule term x²(ln x)′?
Worked example
Differentiate y = x e^(1 − x²) and factor the result.
Show the worked solution
- Choose u = x and v = e^(1 − x²); then u′ = 1.
- Use the chain rule for v′ = −2x e^(1 − x²), and form u′v + uv′.
- The result e^(1 − x²) − 2x²e^(1 − x²) factors as e^(1 − x²)(1 − 2x²).
Answer y′ = e^(1 − x²)(1 − 2x²), for every real x.
Try a topic test
Choose Easy, Normal or Difficult. Save your answers and open worked review after finishing.
Test this topic →From guided practice to a new situation
Write your answer and reasoning first. Use a hint only if you are stuck. The model solution is for self-checking; your written work is not automatically marked.
Differentiate y = x² ln x and state its original domain.
Give me a hint
Treat the power and the logarithm as two changing factors.
Compare my reasoning
- Require x > 0 because the original logarithm is ln x.
- The product rule gives y′ = (2x)ln x + x²(1/x).
- Simplify to y′ = 2x ln x + x, retaining x > 0.
y′ = 2x ln x + x, for x > 0.
Look for these in your work
- Include both product-rule contributions.
- Keep the original log domain after simplifying the derivative.
Differentiate y = (x² + 1)sin x in radians, then find the tangent and normal at x = 0.
Give me a hint
Use both product contributions and evaluate the original function and its derivative separately.
Compare my reasoning
- The derivative is 2x sin x + (x² + 1)cos x.
- At x = 0 the original point is (0, 0) and the derivative gives gradient 1.
- The tangent through the origin is y = x, and its perpendicular normal is y = −x.
y′ = 2x sin x + (x² + 1)cos x; tangent y = x; normal y = −x at (0, 0).
Look for these in your work
- Use both factor derivatives with correct radian rules.
- Build both lines from the same curve point and the appropriate gradients.
A supplied normalised model is M(t) = (t + 1)e^(−t) for t ≥ 0 hours. Find its rate at t = 1 and determine whether it increases or decreases for t > 0.
Give me a hint
Use the product rule before simplifying the common exponential factor.
Compare my reasoning
- Differentiate to get M′(t) = e^(−t) − (t + 1)e^(−t) = −t e^(−t).
- At t = 1 the rate is −e^(−1) per hour.
- For t > 0 the factor t is positive and the exponential is positive, so M′ is negative and the model decreases.
M′(t) = −t e^(−t); rate −e^(−1) per hour at t = 1; decreasing for t > 0.
Look for these in your work
- Combine a product rule with the negative exponential inner rate.
- Use the derivative's sign on the supplied time domain.
Common mistakes
- The product rule uses two added terms, not the product of the two derivatives.
- A composite factor also needs its chain derivative.
- A zero rate at a model's boundary is not automatically an interior stationary point.
What can you explain now?
Differentiate y = x cos x for radian x.
Compare with the explanation
y′ = cos x − x sin x.
The first contribution differentiates x; the second differentiates cosine and introduces its negative sign.
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Make it stick
Explain the two contributions to a product derivative. Apply the rule to a log product and a composite exponential product, checking original domains and rate signs.
If you needed the solution, close it and solve the problem again from a blank page. A correct answer is stronger when you can explain the reason for each step.