# Product Rule for Derivatives: Formula, Examples and Proof

> The product rule: the derivative of f·g is f'g + fg'. The formula, worked examples, the three-factor version, a short proof and common mistakes.

Source: https://practicecalculus.com/rules/product-rule
Updated: 2026-11-03

$$ \frac{d}{dx}\,[f(x)\,g(x)] = f'(x)\,g(x) + f(x)\,g'(x) $$

## At a glance

- **Applies to:** A product of two differentiable functions, both depending on $x$
- **In words:** Derivative of the first times the second, plus the first times the derivative of the second
- **Three factors:** $(fgh)' = f'gh + fg'h + fgh'$
- **Not needed when:** One factor is a constant (use the constant multiple rule)
- **Integral version:** Integration by parts

## Key takeaways

- The derivative of $fg$ is $f'g + fg'$, not $f'g'$.
- Each term differentiates one factor and leaves the other alone.
- Skip the product rule when one factor is a constant, and consider expanding simple polynomial products.

The **product rule** tells you how to differentiate two functions multiplied together. The derivative of a product is **not** the product of the derivatives. Instead, you differentiate one factor at a time and add the results.

## The product rule formula

If $f$ and $g$ are differentiable, then

$$\frac{d}{dx}\,[f(x)\,g(x)] = f'(x)\,g(x) + f(x)\,g'(x)$$

A common way to say it: **derivative of the first times the second, plus the first times the derivative of the second.** In shorthand, $(fg)' = f'g + fg'$.

For three factors the pattern continues, with one differentiated factor per term:

$$(fgh)' = f'gh + fg'h + fgh'$$

## Worked examples

**Example 1: a power times a trig function.** Let $f = x^2$ and $g = \sin x$, so $f' = 2x$ and $g' = \cos x$:

$$\frac{d}{dx}\,x^2\sin x = 2x\sin x + x^2\cos x$$

**Example 2: a power times an exponential.** With $f = x$ and $g = e^x$:

$$\frac{d}{dx}\,x e^x = 1 \cdot e^x + x\,e^x = (x + 1)\,e^x$$

**Example 3: a logarithm.** With $f = x$ and $g = \ln x$:

$$\frac{d}{dx}\,x\ln x = 1 \cdot \ln x + x \cdot \frac{1}{x} = \ln x + 1$$

**Example 4: two polynomials, checked by expanding.** For $(3x + 1)(x^2 - 2)$:

$$3(x^2 - 2) + (3x + 1)(2x) = 3x^2 - 6 + 6x^2 + 2x = 9x^2 + 2x - 6$$

Expanding first gives $3x^3 + x^2 - 6x - 2$, whose derivative is also $9x^2 + 2x - 6$.

**Example 5: three factors.** For $x\,e^x\sin x$:

$$\frac{d}{dx}\,x e^x \sin x = e^x\sin x + x e^x\sin x + x e^x\cos x$$

## When to use the product rule

Use it when two (or more) factors **both depend on $x$**:

| Expression | Use the product rule? |
| --- | --- |
| $x^3\cos x$ | Yes |
| $e^x \ln x$ | Yes |
| $5\sin x$ | No, $5$ is a constant: [constant multiple rule](/rules/constant-multiple-rule) |
| $\sin(x^2)$ | No, that is a composition: [chain rule](/rules/chain-rule) |
| $\frac{x^2}{x + 1}$ | Use the [quotient rule](/rules/quotient-rule), or the product rule on $x^2(x+1)^{-1}$ |

Many derivatives need the product rule and the chain rule together. For $x^2 e^{3x}$, the product rule splits it up and the chain rule handles $e^{3x}$: the result is $2x e^{3x} + 3x^2 e^{3x}$.

## Why the product rule works

Start from the [limit definition of the derivative](/blog/limit-definition-of-the-derivative) and add and subtract $f(x+h)g(x)$ in the numerator:

$$\frac{f(x+h)g(x+h) - f(x)g(x)}{h} = f(x+h)\,\frac{g(x+h) - g(x)}{h} + g(x)\,\frac{f(x+h) - f(x)}{h}$$

As $h \to 0$, $f(x+h) \to f(x)$ because differentiable functions are continuous, and the two difference quotients become $g'(x)$ and $f'(x)$. That leaves $f(x)g'(x) + g(x)f'(x)$.

Picture a rectangle with sides $f$ and $g$. When both sides grow a little, the area gains a strip of size $f'g$ along one side and a strip of size $fg'$ along the other.

## Common mistakes

1. **Multiplying the derivatives.** $\frac{d}{dx}\,x^2\sin x \ne 2x\cos x$. You need two terms.
2. **Differentiating both factors in the same term.** Each term has exactly one primed factor.
3. **Using it on a constant multiple.** It works but wastes time. $\frac{d}{dx}\,4x^3$ is just $12x^2$.
4. **Forgetting the chain rule inside a factor.** In $x\sin(2x)$, the derivative of $\sin(2x)$ is $2\cos(2x)$, so the answer is $\sin(2x) + 2x\cos(2x)$.

Integrating the product rule gives [integration by parts](/rules/integration-by-parts-rule). For more examples and a side-by-side comparison with the quotient rule, read the full guide to [the product and quotient rules](/blog/product-and-quotient-rules).

## Frequently asked questions

### What is the product rule in calculus?

The product rule says $\frac{d}{dx}[f(x)g(x)] = f'(x)g(x) + f(x)g'(x)$. You differentiate one factor at a time and add the two results.

### Does order matter in the product rule?

No. Addition is commutative, so $f'g + fg'$ and $fg' + f'g$ are equal. Order only matters in the quotient rule, which has a subtraction.

### How do you use the product rule with three functions?

Differentiate one factor at a time and add: $(fgh)' = f'gh + fg'h + fgh'$. Each term has exactly one differentiated factor.

### When should I use the product rule instead of expanding?

Expand when the product is two short polynomials. Use the product rule when a factor is a trig, exponential or log function, or when expanding would be long.