The Product Rule: When Two Functions Multiply

Here's a question that seems like it should have an obvious answer: if you have two functions multiplied together, shouldn't the derivative just be the product of their derivatives?

No.

And that's not a technicality. It's genuinely wrong. If f(x) = x² and g(x) = x³, then the derivative of their product is not f'(x)·g'(x) = (2x)(3x²) = 6x³.

The actual derivative is 5x⁴.

Why? Because when two things are changing at the same time, you have to account for how each change affects the other. The product rule captures that interaction.

d/dx [f·g] = f'·g + f·g'

First function times the derivative of the second, plus the derivative of the first times the second function. Both terms matter. You can't skip either one.

This rule shows up everywhere: economics (revenue = price × quantity, both changing), physics (momentum = mass × velocity, both variable), probability (joint distributions), and anywhere you're multiplying functions that depend on the same variable.

Let's prove why it works, see it in action, and understand why the naive approach (just multiply the derivatives) fails so spectacularly.

The Intuition: Area of a Changing Rectangle

Imagine a rectangle with width f(x) and height g(x). The area is A = f(x)·g(x).

Now nudge x by a tiny amount h. The width changes to f(x+h) and the height changes to g(x+h). The new area is A' = f(x+h)·g(x+h).

How much did the area change? You have:

• The original rectangle: area = f(x)·g(x)
• A thin vertical strip added on the right: area ≈ g(x)·[f(x+h) - f(x)] = g(x)·Δf
• A thin horizontal strip added on top: area ≈ f(x)·[g(x+h) - g(x)] = f(x)·Δg
• A tiny corner square where both strips overlap: area ≈ Δf·Δg

The total change in area is: ΔA = g·Δf + f·Δg + Δf·Δg

Divide by h: ΔA/h = g·(Δf/h) + f·(Δg/h) + (Δf/h)·Δg

As h → 0:

• Δf/h → f'(x)
• Δg/h → g'(x)
• Δg → 0

The last term (Δf/h)·Δg vanishes because Δg goes to zero.

What remains: dA/dx = f'·g + f·g'

That's the product rule. The area grows in two ways: the width changes while the height stays fixed (f'·g), and the height changes while the width stays fixed (f·g'). You add both contributions.

The corner square (Δf·Δg) is second-order small—it involves the product of two tiny quantities—so it disappears in the limit.

The Formal Proof

Start with the limit definition of the derivative:

d/dx [f(x)·g(x)] = lim (h→0) [f(x+h)·g(x+h) - f(x)·g(x)] / h

The trick is to add and subtract f(x+h)·g(x) in the numerator:

= lim (h→0) [f(x+h)·g(x+h) - f(x+h)·g(x) + f(x+h)·g(x) - f(x)·g(x)] / h

Factor the first two terms: = lim (h→0) [f(x+h)·(g(x+h) - g(x)) + g(x)·(f(x+h) - f(x))] / h

Split into two limits: = lim (h→0) [f(x+h)·(g(x+h) - g(x))/h] + lim (h→0) [g(x)·(f(x+h) - f(x))/h]

Recognize the derivatives: = f(x)·g'(x) + g(x)·f'(x)

(We use the fact that f is continuous, so f(x+h) → f(x) as h → 0.)

Reorder: = f'(x)·g(x) + f(x)·g'(x)

That's the product rule.

Why You Can't Just Multiply Derivatives

Let's see what happens if you ignore the product rule and just multiply derivatives.

Example: f(x) = x², g(x) = x³, so h(x) = f·g = x⁵

The correct derivative: h'(x) = d/dx [x⁵] = 5x⁴

Using the product rule: h'(x) = f'·g + f·g' = (2x)·(x³) + (x²)·(3x²) = 2x⁴ + 3x⁴ = 5x⁴ ✓

What if you just multiply the derivatives? f'·g' = (2x)·(3x²) = 6x³ ✗

Wrong. Off by a full power of x.

Why does this fail? Because when you multiply two changing quantities, the change in the product is not the product of the changes. The changes interact. You get cross-terms.

Algebraically: (f + Δf)(g + Δg) = fg + f·Δg + g·Δf + Δf·Δg

The middle terms (f·Δg and g·Δf) are the ones the product rule captures. If you just take Δf·Δg, you're ignoring the dominant contributions.

The Product Rule in Action

Example 1: h(x) = (x² + 3)(x³ - 2x)

Let f(x) = x² + 3 and g(x) = x³ - 2x.

Derivatives:

• f'(x) = 2x
• g'(x) = 3x² - 2

Product rule: h'(x) = f'·g + f·g' = (2x)·(x³ - 2x) + (x² + 3)·(3x² - 2) = 2x⁴ - 4x² + 3x⁴ - 2x² + 9x² - 6 = 5x⁴ + 3x² - 6

You could also expand h(x) first and then differentiate: h(x) = (x² + 3)(x³ - 2x) = x⁵ - 2x³ + 3x³ - 6x = x⁵ + x³ - 6x

Differentiate: h'(x) = 5x⁴ + 3x² - 6 ✓

Same answer. The product rule saves you from expanding.

Example 2: h(x) = (sin x)(x²)

Let f(x) = sin x, g(x) = x².

Derivatives:

• f'(x) = cos x
• g'(x) = 2x

Product rule: h'(x) = (cos x)(x²) + (sin x)(2x) = x² cos x + 2x sin x

You can't simplify further. That's the answer.

Example 3: h(x) = (eˣ)(x³)

Let f(x) = eˣ, g(x) = x³.

Derivatives:

• f'(x) = eˣ
• g'(x) = 3x²

Product rule: h'(x) = (eˣ)(x³) + (eˣ)(3x²) = eˣ(x³ + 3x²)

Factor out eˣ: h'(x) = eˣ(x³ + 3x²)

Example 4: h(x) = (x + 1)(x + 2)(x + 3)

This is a product of three functions. Use the product rule twice.

First, treat (x+1)(x+2) as one function: Let u(x) = (x+1)(x+2) and v(x) = x+3.

Find u'(x) using the product rule: u'(x) = (1)(x+2) + (x+1)(1) = x+2 + x+1 = 2x+3

Now apply the product rule to h = u·v: h'(x) = u'·v + u·v' = (2x+3)(x+3) + (x+1)(x+2)(1) = (2x+3)(x+3) + (x+1)(x+2)

Expand: = 2x² + 6x + 3x + 9 + x² + 3x + 2 = 3x² + 12x + 11

Or expand h(x) first: h(x) = (x+1)(x+2)(x+3) = (x² + 3x + 2)(x+3) = x³ + 3x² + 3x² + 9x + 2x + 6 = x³ + 6x² + 11x + 6

Differentiate: h'(x) = 3x² + 12x + 11 ✓

The Product Rule for Three or More Functions

For three functions: d/dx [f·g·h] = f'·g·h + f·g'·h + f·g·h'

Each function gets differentiated once while the others stay fixed, and you sum all the terms.

For n functions: d/dx [f₁·f₂·...·fₙ] = Σ [f₁·...·fᵢ₋₁·fᵢ'·fᵢ₊₁·...·fₙ]

This is called the generalized product rule or Leibniz rule.

Example: h(x) = x·sin x·eˣ

Apply the three-function product rule: h'(x) = (1)·sin x·eˣ + x·(cos x)·eˣ + x·sin x·eˣ = eˣ sin x + x eˣ cos x + x eˣ sin x = eˣ (sin x + x cos x + x sin x) = eˣ [(1+x) sin x + x cos x]

Common Mistakes

Mistake 1: Multiplying the derivatives.

Wrong: d/dx [(x²)(x³)] = (2x)(3x²) = 6x³ Right: d/dx [(x²)(x³)] = (2x)(x³) + (x²)(3x²) = 2x⁴ + 3x⁴ = 5x⁴

Mistake 2: Forgetting one of the terms.

Wrong: d/dx [(x²)(sin x)] = (2x)(sin x) Right: d/dx [(x²)(sin x)] = (2x)(sin x) + (x²)(cos x)

Mistake 3: Confusing the product rule with the chain rule.

The product rule is for f(x)·g(x) (two functions multiplied). The chain rule is for f(g(x)) (one function inside another).

Wrong: d/dx [sin(x²)] = cos x · 2x (that's mixing rules) Right: d/dx [sin(x²)] = cos(x²) · 2x (chain rule)

Mistake 4: Using the product rule when you don't need it.

If one factor is a constant, you don't need the product rule—just use the constant multiple rule.

d/dx [5·x³] = 5·3x² = 15x² (no product rule needed)

The product rule would give: (0)·x³ + 5·3x² = 15x², which is correct but overcomplicated.

The Product Rule and Economics: Marginal Revenue

In economics, revenue R is price p times quantity q: R = p·q

If both price and quantity depend on some variable (like advertising spend, or time), then the derivative dR/dx requires the product rule:

dR/dx = (dp/dx)·q + p·(dq/dx)

The first term is "how does revenue change when price changes, holding quantity constant?" The second term is "how does revenue change when quantity changes, holding price constant?"

You need both. If you raise the price (dp/dx > 0), you might sell fewer units (dq/dx < 0). The net effect on revenue depends on which term dominates.

Example: Suppose p(x) = 100 - x and q(x) = x (where x is units sold).

Revenue: R(x) = (100 - x)·x = 100x - x²

Marginal revenue (derivative): dR/dx = 100 - 2x

Using the product rule: dR/dx = (-1)·x + (100-x)·1 = -x + 100 - x = 100 - 2x ✓

Marginal revenue is zero when 100 - 2x = 0, so x = 50. That's the revenue-maximizing quantity.

The Product Rule and Physics: Angular Momentum

In physics, angular momentum L is the product of moment of inertia I and angular velocity ω: L = I·ω

If both I and ω are changing (say, a spinning ice skater pulls in their arms, reducing I and increasing ω), then:

dL/dt = (dI/dt)·ω + I·(dω/dt)

The first term: change in inertia times current angular velocity. The second term: current inertia times angular acceleration.

For a system with no external torque, L is conserved, so dL/dt = 0. This gives:

(dI/dt)·ω + I·(dω/dt) = 0

So: I·(dω/dt) = -(dI/dt)·ω

If the skater reduces I (pulls in arms), dI/dt < 0, which means dω/dt > 0—the angular velocity increases.

The product rule makes this relationship explicit.

The Quotient Rule: The Product Rule in Disguise

The quotient rule for d/dx [f/g] can be derived from the product rule.

Rewrite f/g as f·(1/g), then differentiate:

d/dx [f·g⁻¹] = f'·g⁻¹ + f·d/dx [g⁻¹]

Use the chain rule on g⁻¹: d/dx [g⁻¹] = -g⁻²·g' = -g'/g²

So: d/dx [f/g] = f'/g - f·g'/g² = (f'g - fg')/g²

That's the quotient rule. It's the product rule plus the chain rule.

The Product Rule and Logarithmic Differentiation

When you have a complicated product, like:

h(x) = (x²+1)(x³-2)(x⁴+5)

You could use the product rule multiple times. Or you could use logarithmic differentiation:

Take the natural log of both sides: ln h = ln(x²+1) + ln(x³-2) + ln(x⁴+5)

Differentiate both sides: h'/h = [2x/(x²+1)] + [3x²/(x³-2)] + [4x³/(x⁴+5)]

Multiply by h: h' = h·{[2x/(x²+1)] + [3x²/(x³-2)] + [4x³/(x⁴+5)]}

This avoids expanding the product. The product rule is still there, hidden in the log derivative, but the algebra is cleaner.

The Product Rule in Reverse: Integration by Parts

The reverse of the product rule is integration by parts:

∫ f·g' dx = f·g - ∫ f'·g dx

This comes from rearranging the product rule: d/dx [f·g] = f'·g + f·g'

Integrate both sides: f·g = ∫ f'·g dx + ∫ f·g' dx

Solve for ∫ f·g' dx: ∫ f·g' dx = f·g - ∫ f'·g dx

This is one of the most important integration techniques. The product rule in reverse.

Why the Product Rule Matters

Because multiplication is everywhere. Any time two variable quantities are combined multiplicatively, you need the product rule to differentiate.

• Area: A = length × width
• Volume: V = area × height
• Work: W = force × distance
• Power: P = voltage × current
• Kinetic energy: KE = (1/2)·mass × velocity²
• Probability: P(A and B) = P(A) × P(B) (if independent)

When both factors are functions of the same variable, the product rule is unavoidable.

The Big Sentence

When two things change together, you can't ignore how each change affects the other—the product rule makes that interaction explicit.

Further Reading

• Stewart, James. Calculus: Early Transcendentals (8th ed.). Cengage Learning, 2015.
  • Chapter 3 covers the product rule with extensive examples and applications.
• Spivak, Michael. Calculus (4th ed.). Publish or Perish, 2008.
  • Rigorous proof of the product rule using the limit definition, with geometric interpretation.
• Strang, Gilbert. Calculus (3rd ed.). Wellesley-Cambridge Press, 2017.
  • Clear presentation of the product rule and its connection to integration by parts.
• Thomas, George B., et al. Thomas' Calculus (14th ed.). Pearson, 2017.
  • Applications of the product rule in physics, economics, and engineering.
• Apostol, Tom M. Calculus, Volume 1 (2nd ed.). Wiley, 1991.
  • Proof-oriented approach with treatment of the generalized product rule (Leibniz rule).

Part 5 of the Calculus Derivatives series.

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