The Product Rule for Derivatives Explained (With Examples)

When you first learned limits, you discovered a beautiful, highly intuitive rule: the limit of a product is just the product of the limits. It is natural to assume derivatives work the exact same way. If you need to find the derivative of two functions multiplied together, shouldn’t you just multiply their individual derivatives?

No. Absolutely not.

Multiplying derivatives together is the single most common trap introductory calculus students fall into. Unlike limits, derivatives do not play perfectly nicely with multiplication. Instead, you must use a specific formula designed to handle intertwined functions: The Product Rule.

Here is exactly how the Product Rule works, a simple mnemonic to remember it, and several step-by-step examples showing how to use it safely.

What is the Product Rule?

The Product Rule dictates how to find the derivative of two distinct functions that are multiplied together.

Mathematically, if you have a function $h(x) = f(x)g(x)$ , the derivative is:
$h'(x) = f(x)g'(x) + g(x)f'(x)$
In plain English: The first times the derivative of the second, plus the second times the derivative of the first.

Many students memorize this by calling the first function “hi” and the second function “ho” (as in high and low). The mnemonic becomes: hi d-ho plus ho d-hi. Whichever way you remember it, the core concept is the same—each function takes turns being left completely alone while the other one gets differentiated.

5 Examples of the Product Rule in Action

Let’s look at how this rule works across polynomials, trigonometry, and exponentials.

Example 1: Multiplying Polynomials

Let’s find the derivative of two binomials multiplied together.

$h(x) = (x^2)(3x - 2)$
Let $f(x) = x^2$ and $g(x) = 3x - 2$ .
Their individual derivatives are $f'(x) = 2x$ and $g'(x) = 3$ .

Now, apply the Product Rule formula (first times derivative of second, plus second times derivative of first):
$h'(x) = (x^2)(3) + (3x - 2)(2x)$
Distribute and simplify the algebra:

$h'(x) = 3x^2 + 6x^2 - 4x$ $h'(x) = 9x^2 - 4x$
(Note: You could have expanded the original equation to $3x^3 - 2x^2$ and used the Power Rule to get $9x^2 - 4x$ , which proves that the Product Rule works perfectly!)

Example 2: Mixing Algebra and Trigonometry

The Product Rule becomes absolutely mandatory when you multiply functions from different mathematical families, because you cannot combine them algebraically.

$h(x) = x^2 \sin(x)$
Identify your two pieces:
First function: $x^2$
Second function: $\sin(x)$

Apply the rule:

$h'(x) = [x^2] \cdot \frac{d}{dx}[\sin(x)] + [\sin(x)] \cdot \frac{d}{dx}[x^2]$
Take the derivatives:

$h'(x) = x^2 \cos(x) + \sin(x) \cdot (2x)$
Clean up the formatting to get your final answer:

$h'(x) = x^2 \cos(x) + 2x \sin(x)$

Example 3: Working with Exponentials

The natural exponential function, $e^x$ , is famous because its derivative is simply itself. Let’s see how that behaves in the Product Rule.

$h(x) = 5x^3 e^x$
Your two functions are $5x^3$ and $e^x$ . Apply the pattern:

$h'(x) = (5x^3) \cdot \frac{d}{dx}[e^x] + (e^x) \cdot \frac{d}{dx}[5x^3]$
Evaluate the derivatives:

$h'(x) = 5x^3 e^x + e^x(15x^2)$
You can leave it like this, or factor out the common terms to make it cleaner:

$h'(x) = 5x^2 e^x (x + 3)$

Example 4: Multiplying Two Abstract Curves

Sometimes, neither function can be simplified with the Power Rule. What happens if you multiply an exponential by a trigonometric function?

$h(x) = e^x \cos(x)$
Set up the Product Rule:

$h'(x) = (e^x) \cdot \frac{d}{dx}[\cos(x)] + (\cos(x)) \cdot \frac{d}{dx}[e^x]$
Take the individual derivatives (remembering that the derivative of cosine is negative sine):

$h'(x) = e^x (-\sin(x)) + \cos(x)(e^x)$
Factor out the $e^x$ to clean up the final expression:

$h'(x) = e^x(\cos(x) - \sin(x))$

Example 5: Finding a Specific Value with Unknown Functions

Calculus exams love to test your knowledge of the rule itself by hiding the equations from you and just giving you data points.

Suppose $h(x) = f(x)g(x)$ .
You are given the following values at $x = 3$ :
$f(3) = 4$
$f'(3) = -1$
$g(3) = 2$
$g'(3) = 5$

Find $h'(3)$ .

Instead of panicking about missing functions, just write out the Product Rule formula for $x = 3$ :

$h'(3) = f(3)g'(3) + g(3)f'(3)$
Now, simply plug in the numbers provided:

$h'(3) = (4)(5) + (2)(-1)$ $h'(3) = 20 - 2 = 18$

While it requires a bit more writing than the simple Sum Rule, the Product Rule is an incredibly elegant solution to a complex problem. Without it, finding the slope of interacting systems—like a sound wave’s frequency being dampened by exponential decay—would be nearly impossible. By taking turns and evaluating one piece at a time, the Product Rule ensures that you never lose track of how multiple changing functions influence one another.

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The Product Rule for Derivatives Explained (With Examples)

What is the Product Rule?