Product Rule: Formula, Proof and Step-by-Step Examples

The product rule states a formula. It gives the derivative of two multiplied, differentiable functions, u(x) and v(x). The formula is d/dx[u · v] = u’v + uv’. In words, it equals the derivative of the first function multiplied by the second. Then you add the first function multiplied by the derivative of the second.

You need the product rule when two variable expressions are multiplied. Also, you cannot simplify them into a single power term. Examples include x²sin x, xe^x, and x³ln x. This rule is a foundational differentiation rule. It works alongside the quotient rule and the chain rule. Also, it serves as the foundation for integration by parts.

The product rule formula and standard notation

You can write the product rule in different ways. Depending on the textbook or context, you will see either Lagrange notation (prime notation) or Leibniz notation (fractional notation).

• Lagrange notation: (uv)’ = u’v + uv’
• Leibniz notation: d/dx[u · v] = (du/dx)v + u(dv/dx)

Both formulations state the same principle. You differentiate each factor in turn. Also, you keep the other factor fixed. Then, you add the results. Because addition is commutative, u’v + uv’ is mathematically identical to uv’ + u’v. However, a consistent order helps you build reliable habits. This is useful when moving between the product rule and the quotient rule.

Why does multiplying individual derivatives fail?

You might guess that a product’s derivative is just the product of the derivatives. You would write (uv)’ = u’v’. However, test this idea on a simple function. This immediately shows why it fails.

Consider the function f(x) = x · x = x². The true derivative is 2x, according to the power rule. If you incorrectly multiply the individual derivatives, set u = x and v = x. This gives u’ = 1 and v’ = 1. As a result, u’v’ = 1 · 1 = 1, which is incorrect. However, applying the product rule gives (1)(x) + (x)(1) = 2x. This matches the true derivative.

This failure occurs because two quantities change simultaneously. The rate of change of their product depends on both individual values. It also depends on their respective rates of growth. Therefore, a correct differentiation rule must account for both simultaneous changes.

Geometric intuition: the expanding rectangle

Visualize why the product rule creates two distinct terms. Imagine u(x) and v(x) are a rectangle’s width and height. Its area is A = u · v.

When x increases by a small step Δx, the width expands by Δu. The height also expands by Δv.

Area = (u + Δu)(v + Δv) = uv + vΔu + uΔv + ΔuΔv

The total change in area, Δ(uv), consists of three separate regions:

• Therefore, a side strip has area vΔu. This is height times the change in width.
• Also, a top strip has area uΔv. This is width times the change in height.
• A small corner box with area ΔuΔv

To find the rate of change with respect to x, divide the entire expression by Δx:

Δ(uv)/Δx = v(Δu/Δx) + u(Δv/Δx) + (Δu/Δx)Δv

Taking the limit as Δx approaches zero causes Δv to vanish. The corner box shrinks to zero, leaving only the two main strips: v(du/dx) + u(dv/dx). This geometric breakdown illustrates why each factor takes turns holding steady while the other expands.

Proof of the product rule using limits

The formal proof uses the limit definition of the derivative. First, start with the difference quotient for the product u(x)v(x).

d/dx[u(x)v(x)] = lim_{h → 0} [u(x+h)v(x+h) – u(x)v(x)] / h

Separate the effects of changing u and changing v. You can add and subtract u(x+h)v(x) in the numerator. This equals zero. Therefore, it does not change the expression’s value:

lim_{h → 0} [u(x+h)v(x+h) – u(x+h)v(x) + u(x+h)v(x) – u(x)v(x)] / h

You can split this quotient into two separate fractions. Then factor out common terms.

lim_{h → 0} [ u(x+h) · (v(x+h) – v(x))/h + v(x) · (u(x+h) – u(x))/h ]

Evaluate the limits individually as h approaches zero:

• Because u(x) is differentiable, it is continuous, meaning lim_{h → 0} u(x+h) = u(x).
• lim_{h → 0} [v(x+h) – v(x)] / h = v'(x)
• lim_{h → 0} [u(x+h) – u(x)] / h = u'(x)

You can combine these limit results. As a result, you get the product rule formula: u(x)v'(x) + v(x)u'(x) = u’v + uv’.

How to apply the product rule in 4 steps

1. Identify the factors. First, split the expression into two clear functions. Set one as u and the other as v.
2. Differentiate separately. Find u’ and v’ independently. Also, use the chain rule if a factor contains an inner function.
3. Assemble the formula. Plug all four pieces into u’v + uv’.
4. Simplify the expression. You can factor out common terms. Also, combine like terms. This simplifies the final derivative.

Step-by-step worked examples

Example 1: x²sin x

First, set u = x² and v = sin x. You can differentiate these. This gives u’ = 2x and v’ = cos x. Now, assemble the pieces.

d/dx[x²sin x] = 2x sin x + x²cos x

Example 2: xe^x

Set u = x and v = e^x. Differentiating gives u’ = 1 and v’ = e^x:

d/dx[xe^x] = (1)e^x + x e^x = e^x(1 + x)

Factoring out e^x provides the standard factored form.

Example 3: x³ln x

Set u = x³ and v = ln x. Differentiating gives u’ = 3x² and v’ = 1/x (see derivative of ln x):

d/dx[x³ln x] = 3x²ln x + x³(1/x). So, this equals 3x²ln x + x². As a result, this is x²(3ln x + 1).

Example 4: e^xcos x

Set u = e^x and v = cos x. Differentiating yields u’ = e^x and v’ = -sin x:

d/dx[e^xcos x] = e^xcos x + e^x(-sin x) = e^x(cos x – sin x)

Example 5: (x² + 1)(3x – 2)

Set u = x² + 1 and v = 3x – 2. Then u’ = 2x and v’ = 3:

d/dx[(x² + 1)(3x – 2)] = 2x(3x – 2) + 3(x² + 1) = 6x² – 4x + 3x² + 3 = 9x² – 4x + 3

Alternatively, expanding first gives (3x³ – 2x² + 3x – 2), which yields 9x² – 4x + 3 via the power rule. Both methods yield identical results.

Example 6: x ln x – x

Apply the product rule to the first term and differentiate the second term directly:

d/dx[x ln x – x] = [(1)ln x + x(1/x)] – 1 = ln x + 1 – 1 = ln x

This confirms that x ln x – x is an antiderivative of ln x. So, it matches the result in integral of ln x.

Example 7: Finding critical points of x²e^x

First, find the first derivative using the product rule:

d/dx[x²e^x] = 2xe^x + x²e^x = xe^x(x + 2)

Set the derivative equal to zero. This finds critical points: xe^x(x + 2) = 0. Since e^x is never zero, the solutions are x = 0 and x = -2. You can evaluate the function at these points. This shows a local maximum of 4e^-2 ≈ 0.54 at x = -2. It also shows a local minimum of 0 at x = 0.

Example 8: x²sin(3x) (combining product rule and chain rule)

Set u = x² and v = sin(3x). Differentiating v requires the chain rule: v’ = 3cos(3x):

d/dx[x²sin(3x)] = 2x sin(3x) + x²[3cos(3x)] = 2x sin(3x) + 3x²cos(3x)

Real-world application: calculating marginal revenue

In economics, total revenue equals price times quantity sold: R = p · q. Because both price and market demand change over time, finding the rate of revenue growth requires the product rule.

Suppose a store sells an item for $20, and price is increasing by $1 per month. Monthly sales currently stand at 500 units but are dropping by 10 units per month. Is total revenue increasing or decreasing?

Assign rates of change with respect to time: p = 20, p’ = 1, q = 500, and q’ = -10. Apply the product rule:

R’ = p’q + pq’ = (1)(500) + (20)(-10) = 500 – 200 = 300

Revenue is increasing at $300 per month. The term p’q ($500) represents added revenue. This comes from higher prices on current sales. Conversely, pq’ (-$200) measures revenue lost from reduced volume. The product rule balances these two opposing effects.

Extending the product rule to three or more factors

When you multiply three functions (u, v, and w), differentiate one factor at a time. Leave the other two unchanged. Then add the products.

(uvw)’ = u’vw + uv’w + uvw’

For example, to differentiate f(x) = xe^xsin x:

f'(x) = (1)e^xsin x + x(e^x)sin x + xe^x(cos x) = e^x(sin x + x sin x + x cos x)

For n functions, this pattern expands to n terms, where each term contains the derivative of exactly one factor.

Finding second and higher-order derivatives

You can calculate the second derivative of a product (uv)”. First, differentiate u’v + uv’ a second time. Also, use the product rule on both terms.

(uv)” = (u”v + u’v’) + (u’v’ + uv”) = u”v + 2u’v’ + uv”

The coefficients (1, 2, 1) match the expansion of (a + b)². Higher-order derivatives follow the General Leibniz Rule, using binomial coefficients from Pascal’s triangle.

For f(x) = xe^x, set u = x and v = e^x. Since u” = 0, v’ = e^x, and v” = e^x:

f”(x) = (0)e^x + 2(1)e^x + x e^x = e^x(2 + x)

When to use the product rule versus other techniques

Common product rule mistakes and fixes

• Multiplying derivatives directly: Writing (uv)’ = u’v’ gives wrong results. Always use u’v + uv’.
• Dropping a term: Every product rule calculation must produce two terms before simplification.
• Forgetting the chain rule: Inside composite factors like sin(3x), remember to multiply by the inner derivative.
• Using the product rule on constant multiples: For expressions like 5x³, the constant multiple rule is faster. You do not need to treat 5 as a separate function.
• Leaving expressions unfactored: Leaving 2xe^x + x²e^x unsimplified makes solving f'(x) = 0 unnecessarily difficult. Always factor out common terms.

Where the product rule is used in calculus and physics

The product rule is essential across advanced mathematics and physics:

• Quotient Rule Proof: You can write u/v as u · v^-1. This allows you to derive the quotient rule directly. You use the product and chain rules to do this.
• Implicit Differentiation: Expressions containing product terms like xy yield y + x(dy/dx) when differentiated with respect to x.
• Integration by Parts: You can integrate both sides of the product rule. This yields the integration by parts formula. It is ∫ u dv = uv – ∫ v du.
• Physics Momentum: Newton’s second law applies to systems with changing mass and velocity. It uses the product rule. For example, dp/dt = d(mv)/dt = m(dv/dt) + v(dm/dt).

Try it yourself

The derivative calculator breaks down product rule problems step by step so you can check your work.

Practice problems

1. d/dx [x⁴e^x]
2. Next, d/dx [sin x cos x]
3. d/dx [√x ln x]
4. Also, d/dx [x²cos x]
5. d/dx [e^xln x]
6. Then, d/dx [x tan x]
7. Find the second derivative of x²e^x

Answers:

1. 4x³e^x + x⁴e^x = x³e^x(4 + x)

2. cos²x – sin²x = cos(2x)

3. (ln x)/(2√x) + 1/√x = (ln x + 2)/(2√x)

4. 2x cos x – x²sin x

5. e^xln x + e^x/x = e^x(ln x + 1/x)

6. tan x + x sec²x

7. e^x(x² + 4x + 2)

Frequently asked questions

Does order matter in the product rule?

No. Multiplication and addition are both commutative, so u’v + uv’ is identical to uv’ + vu’.

How do I know when to use the product rule?

Use the product rule whenever two expressions containing the variable x are multiplied together. If one factor is a constant number, use the constant multiple rule instead.

Product rule or chain rule?

Use the product rule when two separate functions are multiplied side by side. For example, use it with x sin x. However, use the chain rule when one function is nested inside another. An example is sin(x²).

Is there a product rule for three or more functions?

Yes. Differentiate one factor at a time while leaving the remaining factors unchanged, then sum all resulting terms.

What is the integration equivalent of the product rule?

The integration equivalent is integration by parts. You derive this directly by integrating both sides of the product rule formula.

Further reading

• Product and Quotient Rule (Paul’s Online Math Notes) — extra examples, including an applied rate problem.
• Differentiation Rules (OpenStax Calculus Volume 1, Section 3.3) — the textbook proof of the product rule with practice exercises.
• Product rule (Wikipedia) — alternative proofs, the general Leibniz rule for higher derivatives, and history.