Use the chain rule for composite power functions
Identify the outer power and inner rule, multiply their derivatives and preserve the original real domain.
Section 1.7 · Differentiation
Pure Mathematics 1 (Paper 1) · AS & A Level Mathematics 9709, 2026–2027. Use an informal understanding of limits; formal general differentiation from first principles and points of inflexion are not required in section 1.7. Written practice is self-checked.
Before you begin
- Differentiate rational powers and elementary polynomials.
- Understand function composition and real-domain restrictions.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
In y = (3x + 2)⁴, what is the inner derivative?
What you will learn
- Differentiate composite functions whose outer function is a rational power.
- Use an inner derivative for a gradient or rate rather than treating a composite as a simple x power.
Differentiate the outside and multiply by the inside rate
If y = uⁿ and u = g(x), then dy/dx = n[g(x)]^(n − 1)g′(x), where the functions are differentiable. The inner derivative measures how fast its input changes with x. For y = (3x + 2)⁴, the outer derivative is 4u³ and the inner derivative is 3, so y′ = 12(3x + 2)³. Omitting 3 gives an incorrect gradient.
The inner rule need not be linear. For y = (x² + 1)³, the outer derivative is 3u² and g′(x) = 2x, giving y′ = 6x(x² + 1)². You can verify this by expanding the original polynomial and differentiating term by term. Do not differentiate each factor of the result as if a product were a sum; this lesson's method is the chain rule for the original composite.
A root or reciprocal adds a domain check
For y = √(x² + 4), rewrite the outer rule as u^(1/2). Since x² + 4 is always positive, y′ = (1/2)(x² + 4)^(−1/2)(2x) = x/√(x² + 4) for every real x. For y = √(5x + 1), the function is real for x ≥ −1/5 but its finite derivative 5/[2√(5x + 1)] requires x > −1/5.
For y = (4x + 3)^(−2), the function excludes x = −3/4. The derivative is −2(4x + 3)^(−3) × 4 = −8/(4x + 3)³ on that same domain. If the independent variable is time, the identical chain reasoning gives an instantaneous rate with the output units divided by seconds. State the variable explicitly instead of changing symbols halfway through a calculation.
Pause and explain
What is the derivative of (3x + 2)⁴?
Worked example
Differentiate y = (x² + 1)³ and check by expanding the original polynomial.
Show the worked solution
- Set u = x² + 1; the outer derivative is 3u² and the inner derivative is 2x.
- Multiply them to get y′ = 6x(x² + 1)².
- Expanding y gives x⁶ + 3x⁴ + 3x² + 1, whose derivative 6x⁵ + 12x³ + 6x equals the chain-rule result.
Answer y′ = 6x(x² + 1)² = 6x⁵ + 12x³ + 6x.
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 = (2x − 1)⁵.
Give me a hint
Keep the inner bracket intact while reducing the outer power.
Compare my reasoning
- Set u = 2x − 1, so dy/du = 5u⁴.
- The inner derivative is du/dx = 2.
- Multiply to obtain dy/dx = 10(2x − 1)⁴, valid for every real x.
y′ = 10(2x − 1)⁴.
Look for these in your work
- Identify the outer power and inner derivative separately.
- Multiply their rates without losing the inner bracket.
Differentiate y = 1/(4x + 3)² and state its excluded input.
Give me a hint
Rewrite the reciprocal as a negative power before using the chain rule.
Compare my reasoning
- Write y = (4x + 3)^(−2), with x ≠ −3/4.
- The outer derivative is −2(4x + 3)^(−3), and the inner derivative is 4.
- Thus y′ = −8/(4x + 3)³; the original excluded input remains excluded.
y′ = −8/(4x + 3)³, with x ≠ −3/4.
Look for these in your work
- Apply a negative outer exponent correctly.
- Retain the domain exclusion from the original reciprocal.
A supplied square-side model is L(t) = 2t² + 1 metres for t ≥ 0 seconds. Find the area rate at t = 2 seconds.
Give me a hint
The area is a composite A(t) = [L(t)]²; differentiate the side as well as the outer square.
Compare my reasoning
- A(t) = (2t² + 1)² square metres.
- The chain rule gives dA/dt = 2(2t² + 1)(4t) = 8t(2t² + 1).
- At t = 2, this is 8(2)(9) = 144 m²/s, not the side's rate 8 m/s.
dA/dt = 8t(2t² + 1); at t = 2, the area increases at 144 m²/s.
Look for these in your work
- Model area as the square of the supplied side function.
- Use both chain factors and distinguish area-rate units from side-rate units.
Common mistakes
- The inner derivative is a multiplier, not an optional extra.
- A root's real domain and finite-derivative domain may have different endpoints.
- Preserve the independent variable and rate units in a time model.
What can you explain now?
Differentiate y = √(5x + 1) and state where the finite derivative is valid.
Compare with the explanation
y′ = 5/[2√(5x + 1)], for x > −1/5.
The factor 5 is the inner derivative; at the root boundary the derivative denominator is zero.
After trying it yourself, choose your next review. This is your self-assessment.
Your review choice appears on Today. Sign in to sync it across devices.
Make it stick
Explain the chain rule as an outside rate multiplied by an inside rate, then apply it to a root, a reciprocal and a non-linear inner function.
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.