The chain rule differentiates a function of a function, such as (2x + 1)⁵ or √(x² + 1). Differentiate the outside, keeping the inside unchanged, then multiply by the derivative of the inside. In symbols, if y is a function of u and u is a function of x, then dy/dx = dy/du × du/dx.
The chain rule
- dy/dx = dy/du × du/dx
- y = (ax + b)ⁿ gives dy/dx = an(ax + b)ⁿ⁻¹
- d/dx f(g(x)) = f′(g(x)) × g′(x)
- dy/dx = 1 / (dx/dy)
Outside first, then multiply by the derivative of the inside.
Powers of a bracket
For a power of a linear bracket, bring the power down, reduce it by one, and multiply by the derivative of the inside. For example, (2x + 5)⁴ differentiates to 4(2x + 5)³ × 2 = 8(2x + 5)³. That final multiplication is the chain rule at work.
Expanding a bracket like (2x + 5)⁴ by hand would be slow and error-prone; the chain rule is the point.
The general chain rule
When the inside is not linear, set u equal to it, write y in terms of u, and use dy/dx = dy/du × du/dx. The idea is the same: dy/du is the outside, du/dx is the inside. With practice you can do this in one line without writing u.
Roots and reciprocals
Rewrite roots and reciprocals as powers first: √(g(x)) = (g(x))^(1/2) and 1/g(x) = (g(x))^(−1). Then the power rule handles the fractional or negative index, and you still multiply by g′(x). Tidy the answer back into a root or a fraction at the end if the question asks.
Using the chain rule for gradients and tangents
Once you can differentiate a composite, the earlier skills carry over unchanged. The gradient at x = a is f′(a), a tangent uses y − y₁ = m(x − x₁), and stationary points solve f′(x) = 0. Differentiate with the chain rule first; the rest is exactly as before.
dy/dx = 1/(dx/dy)
If x is given in terms of y, you do not need to rearrange. Differentiate x with respect to y to get dx/dy, then take the reciprocal. This result comes straight from the chain rule and is used again in implicit and parametric differentiation.
The chain rule with eˣ, ln and trig
The same idea applies to the standard functions. sin(g(x)) gives cos(g(x)) × g′(x), e^(g(x)) gives e^(g(x)) × g′(x), and ln(g(x)) gives g′(x)/g(x). Forgetting to multiply by g′(x) is the most common chain rule slip.
Worked examples
Example 1. Differentiate y = (2x + 5)⁴.
- Bring the power down and reduce it: 4(2x + 5)³.
- Multiply by the derivative of the inside, which is 2.
Answer: dy/dx = 8(2x + 5)³.
Marks: M1 for 4(2x + 5)³, A1 for 8(2x + 5)³.
Example 2. Differentiate y = (x² + 4)³.
- Let u = x² + 4, so y = u³.
- dy/du = 3u² = 3(x² + 4)² and du/dx = 2x.
- Multiply: 3(x² + 4)² × 2x.
Answer: dy/dx = 6x(x² + 4)².
Marks: M1 for 3(x² + 4)², M1 for × 2x, A1 for 6x(x² + 4)².
Example 3. Differentiate y = √(3x − 2).
- Rewrite: y = (3x − 2)^(1/2).
- dy/dx = ½(3x − 2)^(−1/2) × 3.
Answer: dy/dx = 3 / (2√(3x − 2)).
Marks: M1 for ½(3x − 2)^(−1/2), M1 for × 3, A1 for the simplified fraction.
Example 4. Find the equation of the tangent to y = (x² − 3)² at x = 2.
- f′(x) = 2(x² − 3) × 2x = 4x(x² − 3).
- f′(2) = 8 × 1 = 8, and f(2) = 1² = 1, so the point is (2, 1).
- Tangent: y − 1 = 8(x − 2).
Answer: y = 8x − 15.
Marks: M1 for the derivative, M1 for the point, A1 for the equation.
Example 5. Given x = y³ + 2y, find dy/dx at the point where y = 1.
- dx/dy = 3y² + 2.
- At y = 1, dx/dy = 5.
- dy/dx = 1/5.
Answer: dy/dx = 1/5.
Marks: M1 for dx/dy, A1 for 1/5.
Common mistakes
- Forgetting the derivative of the inside. (2x + 5)⁴ differentiates to 8(2x + 5)³, not 4(2x + 5)³.
- Changing the inside. Differentiate the outside with the inside left exactly as it is.
- Not rewriting roots as powers. Write √(3x − 2) as (3x − 2)^(1/2) before differentiating.
- Expanding unnecessarily. Use the chain rule rather than multiplying out a high power.
- Rearranging when x is given in terms of y. Differentiate x with respect to y and flip: dy/dx = 1/(dx/dy).
How the chain rule is examined
The chain rule appears throughout Pure 3 (WMA13) of Edexcel International A-Level, and in Year 2 of Edexcel A-Level (9MA0). Questions ask you to:
- differentiate a power of a bracket, a root or a reciprocal of a function
- find the gradient of a composite curve at a point, or a tangent or normal
- explain why a curve has no stationary points
- use dy/dx = 1/(dx/dy) when x is given in terms of y
The Pure 3 differentiation exam questions include chain rule questions with mark schemes, and the differentiation worksheet adds more. The A-Level differentiation revision engine gives unlimited practice.
The chain rule is used inside the product rule and quotient rule, and reversed in the reverse chain rule for integration.
Frequently asked questions
What is the chain rule?
A rule for differentiating a function of a function: dy/dx = dy/du × du/dx. Differentiate the outside, then multiply by the derivative of the inside.
How do you know when to use the chain rule?
When one function is inside another, such as a bracket raised to a power, a root of an expression, or sin, e or ln of something other than x.
Why is dy/dx = 1/(dx/dy)?
By the chain rule, dy/dx × dx/dy = dy/dy = 1, so the two derivatives are reciprocals, provided neither is zero.