Practice Maths

The Product Rule

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Key Terms

The product rule applies when a function is the product of two differentiable functions: y = u(x) × v(x).
It states: d/dx[u × v] = u′v + uv′ (differentiate the first and multiply by the second, plus the first multiplied by the derivative of the second).
You do not simply differentiate each factor independently: d/dx[uv] ≠ (du/dx)(dv/dx).
When expanding first is simpler (e.g., two linear factors), you may choose to expand and use the power rule instead — but the product rule always works.
📚 QCAA Formula
Product rule: d/dx[f(x)g(x)] = f′(x)g(x) + f(x)g′(x)
Equivalently, if y = uv: dy/dx = u′v + uv′
y = u, v dy/dx = u′v + uv′
x(x + 3)u = x, v = x + 31 · (x + 3) + x · 1 = 2x + 3
x²(2x − 1)u = x², v = 2x − 12x(2x − 1) + x²(2) = 4x² − 2x + 2x² = 6x² − 2x
x³√x = x7/2u = x³, v = x1/23x² · x1/2 + x³ · ½x−1/2 = (7/2)x5/2
y = u · v product → differentiate u u'v (first term) + differentiate v uv' (second term) dy/dx = u'v + uv'
Hot Tip A useful memory aid: “derivative of first times second, plus first times derivative of second.” Always simplify your result by collecting like terms after applying the rule.

Worked Example 1 — Polynomial × Polynomial

Question: Differentiate y = (x² + 1)(3x − 2).

Identify: u = x² + 1 ⇒ u′ = 2x. v = 3x − 2 ⇒ v′ = 3.

Apply: dy/dx = u′v + uv′ = 2x(3x − 2) + (x² + 1)(3)

= 6x² − 4x + 3x² + 3 = 9x² − 4x + 3

Verify by expanding first: (x² + 1)(3x − 2) = 3x³ − 2x² + 3x − 2. d/dx = 9x² − 4x + 3 ✓

Worked Example 2 — Polynomial × Power

Question: Differentiate y = (2x + 1)√x.

Rewrite: y = (2x + 1)x1/2. u = 2x + 1 ⇒ u′ = 2. v = x1/2 ⇒ v′ = ½x−1/2.

Apply: dy/dx = 2 × x1/2 + (2x + 1) × ½x−1/2

= 2x1/2 + (2x + 1)/(2x1/2) = [4x + 2x + 1]/(2√x) = (6x + 1)/(2√x)

Worked Example 3 — When to expand instead

Question: Find d/dx[(x + 2)(x − 3)].

Simpler method: Expand first: (x + 2)(x − 3) = x² − x − 6. d/dx = 2x − 1.

Product rule (verify): u = x + 2, u′ = 1; v = x − 3, v′ = 1. dy/dx = 1(x − 3) + (x + 2)(1) = x − 3 + x + 2 = 2x − 1 ✓

Both methods are correct. For two linear factors, expanding is generally quicker.

Why Does the Product Rule Exist?

When a function is expressed as the product of two separate functions, you cannot differentiate each factor independently. Consider y = x² × x³ = x5, which gives dy/dx = 5x4. But differentiating each factor and multiplying gives 2x × 3x² = 6x³ — which is wrong. The product rule corrects this.

The rule is derived from the definition of the derivative by considering [u(x+h)v(x+h) − u(x)v(x)]/h and adding and subtracting u(x+h)v(x). The result is the product rule: d/dx[uv] = u′v + uv′.

When to Use the Product Rule

Use the product rule whenever a function is written (or can be written) as the product of two functions that each depend on x. Common cases include:

  • Polynomial × polynomial: e.g., x³(2x + 1)
  • Polynomial × power (including roots): e.g., x²√x
  • Any product where expanding would be messy or impossible: e.g., x²(x + 1)5

If two simple linear or constant factors can be expanded easily, expanding first is equally valid and often quicker.

Step-by-Step Method

  1. Identify u and v — label the two factor functions clearly.
  2. Differentiate each: find u′ = du/dx and v′ = dv/dx.
  3. Apply the formula: dy/dx = u′v + uv′.
  4. Simplify by expanding and collecting like terms.
Worked Example 1 — Product of two polynomials

Differentiate y = x²(3x − 1).

Step 1: u = x², v = 3x − 1.

Step 2: u′ = 2x, v′ = 3.

Step 3: dy/dx = 2x(3x − 1) + x²(3) = 6x² − 2x + 3x².

Step 4: dy/dx = 9x² − 2x.

Worked Example 2 — Product with a square root

Differentiate y = x³√x.

Rewrite: y = x³ × x1/2. u = x³, v = x1/2.

Derivatives: u′ = 3x², v′ = ½x−1/2.

Product rule: dy/dx = 3x² × x1/2 + x³ × ½x−1/2 = 3x5/2 + ½x5/2 = (7/2)x5/2.

Verify: y = x7/2, dy/dx = (7/2)x5/2 ✓

Worked Example 3 — Product rule with a composite factor

Differentiate y = x²(x + 1)4.

u = x², u′ = 2x. v = (x + 1)4, v′ = 4(x + 1)³ (chain rule).

dy/dx = 2x(x + 1)4 + x² × 4(x + 1)³

= 2x(x + 1)4 + 4x²(x + 1)³

= 2x(x + 1)³[(x + 1) + 2x] = 2x(3x + 1)(x + 1)³

Combining Product Rule with Chain Rule

In more complex problems, one or both factors may themselves require the chain rule to differentiate. Always apply chain rule to individual factors first, then combine using the product rule formula.

💡 Key Reminder: The product rule generates two terms. Always check that you have collected all like terms at the end. Factoring the final answer (especially for tangent and turning point problems) makes further work much easier.

Summary

Product rule: if y = uv, then dy/dx = u′v + uv′. Steps: label u and v, differentiate each, apply formula, simplify. Use when expanding is impractical. May be combined with the chain rule for composite factors.

Mastery Practice

See Answers ➔
  1. Fluency

    Differentiate each function using the product rule.

    1. (a) y = x(x + 1)
    2. (b) y = x²(3x − 1)
    3. (c) y = (x + 2)(x² − x + 3)
    4. (d) y = x³√x
  2. Fluency

    Differentiate each function. Simplify your answer fully.

    1. (a) y = (2x + 3)(x − 4)
    2. (b) y = x(x² + 5)
    3. (c) y = (x + 1)(x + 2)(x + 3)   [Hint: multiply two brackets first, then apply the product rule to the result and the remaining bracket]
  3. Fluency

    Differentiate these functions involving surds. Leave answers with positive indices.

    1. (a) y = x²√x
    2. (b) y = (2x + 1)√x
    3. (c) y = x√(x + 1)   [set u = x, v = (x + 1)1/2]
  4. Understanding

    For each function, find f′(x) using the product rule, then evaluate at the given point.

    1. (a) f(x) = x(x + 4), at x = 2
    2. (b) f(x) = x²(x − 3), at x = 1
    3. (c) f(x) = (x + 1)(x² + 2), at x = 0
  5. Understanding

    Find the equation of the tangent to each curve at the given point.

    Method: Differentiate using the product rule, find the gradient at the given x-value, then use y − y1 = m(x − x1).
    1. (a) y = x(x + 3) at x = 1
    2. (b) y = x²(2x − 1) at x = 2
    3. (c) y = (x + 2)(x² + 1) at x = 0
  6. Understanding

    Show that the following derivative results are correct using the product rule, and verify by expanding first.

    1. (a) If y = (x + 1)(x + 2), show that dy/dx = 2x + 3.
    2. (b) If y = x²(x + 1), show that dy/dx = 3x² + 2x.
  7. Understanding

    Product rule combined with chain rule.

    Find dy/dx for each function.

    1. (a) y = x(2x + 1)³
    2. (b) y = x²(x² + 1)4
  8. Problem Solving

    Finding stationary points using the product rule.

    For the curve y = x²(x − 3):

    1. (a) Find dy/dx using the product rule.
    2. (b) Find all x-values where dy/dx = 0.
    3. (c) Classify each stationary point as a maximum or minimum, using a sign diagram.
    4. (d) Find the coordinates of each stationary point.
  9. Problem Solving

    Real-world application — product rule and rates.

    Challenge. The area of a rectangle at time t is A(t) = l(t) × w(t), where l(t) = 2t + 1 cm and w(t) = t² + 3 cm.
    1. (a) Write an expression for A(t).
    2. (b) Use the product rule to find dA/dt.
    3. (c) Find the rate of change of area when t = 2 seconds.
    4. (d) At what time t is the rate of change of area equal to 20 cm²/s?
  10. Problem Solving

    Product rule with fractional powers and combined simplification.

    For f(x) = (x² − 1)√(x + 2):

    1. (a) Identify u and v, and find u′ and v′.
    2. (b) Apply the product rule to find f′(x).
    3. (c) Simplify your answer by writing over a common denominator of √(x + 2).
    4. (d) Find the gradient of the curve at x = 2.