The Product Rule

The derivative of a sum is the sum of the derivatives (topic 2.6), so it is tempting to guess that the derivative of a product is the product of the derivatives. Test the guess on $x\cdot x = x^2$: each factor has derivative 1, so the guess says the derivative is $1\cdot 1 = 1$, but the derivative of $x^2$ is $2x$. The correct rule has two terms, $f'g + fg'$, and the reason is visible in the area of a rectangle.

Why a product's derivative has two terms

Picture $f(x)\,g(x)$ as the area of a rectangle with width $f(x)$ and height $g(x)$. Now let $x$ increase by a small amount $\Delta x$. The width grows by $\Delta f$ and the height by $\Delta g$, and the rectangle gains three new pieces.

The change in area is

Divide by $\Delta x$ to get an average rate of change:

As $\Delta x \to 0$, the first two fractions become $f'$ and $g'$. In the last term, $\frac{\Delta g}{\Delta x}$ approaches $g'$, but it is multiplied by $\Delta f$, which shrinks to 0. The corner is small in both directions, so it disappears in the limit, and what remains is $g f' + f g'$. Each term is one factor's change while the other factor is held at its current value.

The same argument works with the limit definition. Adding and subtracting $f(x+h)\,g(x)$ in the numerator splits it into one piece where only $g$ changes and one where only $f$ changes:

As $h \to 0$ the fractions approach $g'(x)$ and $f'(x)$. The factor $f(x+h)$ approaches $f(x)$ because a function that is differentiable at $x$ is continuous there (topic 2.4). That continuity is the formal reason the corner term vanishes.

Because the two terms are added, their order doesn't matter, and you can write $fg' + f'g$ just as well. The rule agrees with what you already know. With a constant $k$ as one factor, it gives $\frac{d}{dx}\big[k\,f(x)\big] = 0\cdot f(x) + k\,f'(x) = k\,f'(x)$, the constant multiple rule. With $x\cdot x$ it gives $1\cdot x + x\cdot 1 = 2x$.

Using the rule

The product rule is needed whenever two factors both depend on $x$ and you can't simply multiply them out: $x^3\ln x$, $e^x\sin x$, $x^2\cos x$. Label the factors, find each derivative, then assemble. For $y = x^3\ln x$ with $x > 0$, the factors are $x^3$ (derivative $3x^2$) and $\ln x$ (derivative $\frac{1}{x}$), so

Factoring at the end isn't required for a derivative, but it is what you want when the next step is solving $\frac{dy}{dx} = 0$. Here, since $x^2 > 0$, the derivative is 0 exactly when $\ln x = -\frac{1}{3}$, that is, at $x = e^{-1/3}$.

A product of two polynomials can be expanded instead, and both routes give the same derivative. If you do use the product rule on polynomials, simplify carefully; expanding first is often less work.

Products of three factors

For $f\,g\,k$, treat $f\,g$ as one factor and apply the rule twice:

The pattern: one term for each factor, in which that factor alone is differentiated. For $y = x\,e^x\sin x$,

Products from tables and graphs

When $f$ and $g$ are given only by values, the product rule needs four numbers, all at the same input: $f(a)$, $f'(a)$, $g(a)$ and $g'(a)$. If $f(2) = 3$, $f'(2) = -1$, $g(2) = 4$ and $g'(2) = 5$, then the derivative of $f(x)\,g(x)$ at $x = 2$ is $(-1)(4) + (3)(5) = 11$. Notice that the function values matter as much as the derivatives: each derivative is weighted by the other factor's current value.

Worked examples

Where students lose points

On the exam

Product-rule questions on the exam can combine the rule with values from a table or graph, or with a follow-up such as a tangent line or a decision about whether a function is increasing. In a free-response answer, writing the general derivative (for example, $h'(x) = f'(x)g(x) + f(x)g'(x)$) before substituting shows the method. An unsimplified numeric answer such as $(-1)(-3) + (2)(4)$ is acceptable on no-calculator parts.