Product Rule for Derivatives

The product rule is the differentiation rule you use whenever a function is written as the product of two separate functions of \(x\). If you have something like \(f(x) = x^2 \sin(x)\) or \(f(x) = (3x+1)(x^2-4)\), you cannot differentiate each factor on its own and simply multiply the results together. The product rule tells you exactly how to combine the two pieces correctly.

If \(f(x) = u(x) \cdot v(x)\), where \(u\) and \(v\) are both differentiable functions of \(x\), then:

\( f'(x) = u'(x)v(x) + u(x)v'(x) \)

In words: take the derivative of the first function, multiply it by the second function unchanged, then add the first function unchanged multiplied by the derivative of the second function. Some textbooks write this using Leibniz notation as \( \frac{d}{dx}\big[u v\big] = \frac{du}{dx}v + u\frac{dv}{dx} \), which says exactly the same thing.

A very common mistake is guessing that \((uv)' = u'v'\). A quick check shows why that fails. Let \(u(x) = x\) and \(v(x) = x\), so \(f(x) = x \cdot x = x^2\). We know \(f'(x) = 2x\) directly from the power rule. But \(u'(x)v'(x) = 1 \cdot 1 = 1\), which is not equal to \(2x\) at all. The product rule's extra structure, adding two terms instead of multiplying one, is what makes it match the true derivative.

Differentiate \(f(x) = x^2 \sin(x)\).

Let \(u(x) = x^2\) and \(v(x) = \sin(x)\). Then \(u'(x) = 2x\) and, using the standard result for derivative trigonometry, \(v'(x) = \cos(x)\).

Applying the formula:

\( f'(x) = 2x \sin(x) + x^2 \cos(x) \)

That's the full answer, left as a sum of two terms rather than simplified further, since there is no common factor that reduces the expression cleanly here.

Differentiate \(g(x) = (3x^2 - 1)e^x\).

Let \(u(x) = 3x^2 - 1\), so \(u'(x) = 6x\). Let \(v(x) = e^x\); the exponential function is one of the rare functions that is its own derivative, as covered in the lesson on the inverse of e^x, so \(v'(x) = e^x\).

Applying the product rule:

\( g'(x) = 6x \cdot e^x + (3x^2-1)e^x \)

Since \(e^x\) appears in both terms, you can factor it out for a tidier final form:

\( g'(x) = e^x(6x + 3x^2 - 1) \)

If a function is a product of three factors, \(f(x) = u(x)v(x)w(x)\), the rule extends by differentiating one factor at a time while holding the others fixed:

\( f'(x) = u'vw + uv'w + uvw' \)

Each term differentiates exactly one factor and leaves the remaining two untouched, then all three terms are added together.

Students often confuse the product rule with the derivative of 5x style composite-function rule. Use the product rule when two separate functions are multiplied together, such as \(x^2 \sin(x)\). Use the chain rule instead when one function is nested inside another, such as \(\sin(x^2)\), where \(x^2\) sits inside the sine function rather than being multiplied by it. Some problems, such as differentiating \(x^2 \sin(3x)\), actually require both rules together: the product rule to split the two factors, and the chain rule to differentiate \(\sin(3x)\) itself.

Watch out for these frequent errors when applying the product rule:

• Multiplying \(u'\) and \(v'\) directly instead of adding the two cross terms.
• Forgetting to differentiate one of the factors and simply copying it down unchanged in both terms.
• Applying the product rule to a sum instead of a product, for example treating \(x^2 + \sin(x)\) as if it needed the rule, when ordinary term-by-term differentiation is all that's required there.
• Not simplifying or factoring the final answer when a common factor, like \(e^x\) in Worked Example 2, would make it cleaner.

The product rule builds directly on earlier ideas about what a derivative represents, such as those developed in the lesson on the slope and equation of tangent line. Once you're comfortable applying it, it becomes a building block for more advanced techniques, including finding second and third derivatives of products in higher order derivatives.

Related lessons

• Calculus Help