If you're seeing this message, it means we're having trouble loading external resources on our website.

If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked.

## AP®︎/College Calculus AB

### Course: AP®︎/College Calculus AB>Unit 2

Lesson 9: The product rule

# Product rule review

Review your knowledge of the Product rule for derivatives, and use it to solve problems.

## What is the Product rule?

The Product rule tells us how to differentiate expressions that are the product of two other, more basic, expressions:
start fraction, d, divided by, d, x, end fraction, open bracket, f, left parenthesis, x, right parenthesis, dot, g, left parenthesis, x, right parenthesis, close bracket, equals, start fraction, d, divided by, d, x, end fraction, open bracket, f, left parenthesis, x, right parenthesis, close bracket, dot, g, left parenthesis, x, right parenthesis, plus, f, left parenthesis, x, right parenthesis, dot, start fraction, d, divided by, d, x, end fraction, open bracket, g, left parenthesis, x, right parenthesis, close bracket
Basically, you take the derivative of f multiplied by g, and add f multiplied by the derivative of g.

## What problems can I solve with the Product rule?

### Example 1

Consider the following differentiation of h, left parenthesis, x, right parenthesis, equals, natural log, left parenthesis, x, right parenthesis, cosine, left parenthesis, x, right parenthesis:
\begin{aligned} &\phantom{=}h'(x) \\\\ &=\dfrac{d}{dx}(\ln(x)\cos(x)) \\\\ &=\dfrac{d}{dx}(\ln(x))\cos(x)+\ln(x)\dfrac{d}{dx}(\cos(x))&&\gray{\text{Product rule}} \\\\ &=\dfrac{1}{x}\cdot \cos(x)+\ln(x)\cdot (-\sin(x))&&\gray{\text{Differentiate }\ln(x)\text{ and }\cos(x)} \\\\ &=\dfrac{\cos(x)}{x}-\ln(x)\sin(x)&&\gray{\text{Simplify}} \end{aligned}

Problem 1
• Current
f, left parenthesis, x, right parenthesis, equals, x, squared, e, start superscript, x, end superscript
f, prime, left parenthesis, x, right parenthesis, equals

Want to try more problems like this? Check out this exercise.

### Example 2

Suppose we are given this table of values:
xf, left parenthesis, x, right parenthesisg, left parenthesis, x, right parenthesisf, prime, left parenthesis, x, right parenthesisg, prime, left parenthesis, x, right parenthesis
4minus, 413space, space, space, 08
H, left parenthesis, x, right parenthesis is defined as f, left parenthesis, x, right parenthesis, dot, g, left parenthesis, x, right parenthesis, and we are asked to find H, prime, left parenthesis, 4, right parenthesis.
The Product rule tells us that H, prime, left parenthesis, x, right parenthesis is f, prime, left parenthesis, x, right parenthesis, g, left parenthesis, x, right parenthesis, plus, f, left parenthesis, x, right parenthesis, g, prime, left parenthesis, x, right parenthesis. This means H, prime, left parenthesis, 4, right parenthesis is f, prime, left parenthesis, 4, right parenthesis, g, left parenthesis, 4, right parenthesis, plus, f, left parenthesis, 4, right parenthesis, g, prime, left parenthesis, 4, right parenthesis. Now let's plug the values from the table in the expression:
\begin{aligned} H'(4)&=f'(4)g(4)+f(4)g'(4) \\\\ &=(0)(13)+(-4)(8) \\\\ &=-32 \end{aligned}