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.

### Course: Calculus, all content (2017 edition)>Unit 2

Lesson 19: 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:
$\frac{d}{dx}\left[f\left(x\right)\cdot g\left(x\right)\right]=\frac{d}{dx}\left[f\left(x\right)\right]\cdot g\left(x\right)+f\left(x\right)\cdot \frac{d}{dx}\left[g\left(x\right)\right]$
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 $h\left(x\right)=\mathrm{ln}\left(x\right)\mathrm{cos}\left(x\right)$:

Problem 1
$f\left(x\right)={x}^{2}{e}^{x}$
${f}^{\prime }\left(x\right)=$

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

### Example 2

Suppose we are given this table of values:
$x$$f\left(x\right)$$g\left(x\right)$${f}^{\prime }\left(x\right)$${g}^{\prime }\left(x\right)$
$4$$-4$$13$$8$
$H\left(x\right)$ is defined as $f\left(x\right)\cdot g\left(x\right)$, and we are asked to find ${H}^{\prime }\left(4\right)$.
The Product rule tells us that ${H}^{\prime }\left(x\right)$ is ${f}^{\prime }\left(x\right)g\left(x\right)+f\left(x\right){g}^{\prime }\left(x\right)$. This means ${H}^{\prime }\left(4\right)$ is ${f}^{\prime }\left(4\right)g\left(4\right)+f\left(4\right){g}^{\prime }\left(4\right)$. Now let's plug the values from the table in the expression:
$\begin{array}{rl}{H}^{\prime }\left(4\right)& ={f}^{\prime }\left(4\right)g\left(4\right)+f\left(4\right){g}^{\prime }\left(4\right)\\ \\ & =\left(0\right)\left(13\right)+\left(-4\right)\left(8\right)\\ \\ & =-32\end{array}$

Problem 1
$x$$g\left(x\right)$$h\left(x\right)$${g}^{\prime }\left(x\right)$${h}^{\prime }\left(x\right)$
$-2$$2$$-1$$3$$4$
$F\left(x\right)=g\left(x\right)\cdot h\left(x\right)$
${F}^{\prime }\left(-2\right)=$

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

## Want to join the conversation?

• I'm having trouble with problem 1. The explanation says that the derivative of e^x is e^x, but wouldn't it be x*e^(x - 1) because of the power rule? Is it a special property of e? Could it be that the exponent is a variable? What am I not understanding?
• The Power Rule only works for powers of a variable. That is xⁿ, where n is a constant.
It does not work for for exponential functions ie n^x. In other words the exponent is a variable.

It is not a special property of e. It is - as you say - that "the exponent is a variable."
• This Product Rule Review page, located in the Derivative Rules unit, has examples and exercises that assume knowledge of how to find derivatives of exponential and logarithmic functions. However, those derivatives are not covered until the next unit, Advanced Derivatives.
• Is derivative of cos x sin x or -sin x?
• The derivative of cos(x) is -sin(x)dx but usually we find the derivative with respect to x meaning it is -sin(x) * dx/dx or just -sin(x).
• i am having trouble with a textbook question that reads (8x^2-3x)^3 which states i have to use the triple product rule to solve the problem. The answer is apparently 3x^2(8x-3)^2(16x-3). i am at a loss. please, what is going on here?
(1 vote)
• It's not as complicated as it looks at a glance! The trick is to use the chain rule. You have a composite function. Let's call the two parts of the function f(x) and g(x). Let f(x) = x^3 and g(x) = 8x^2-3x. Then f(g(x)) = f(8x^2-3x) = (8x^2-3x)^3. That's the function you have to differentiate.

To differentiate a composite function, you use the chain rule, which says that the derivative of f(g(x)) = f'(g(x))g'(x). In plain (well, plainer) English, the derivative of a composite function is the derivative of the outside function (here that's f(x)) evaluated at the inside function (which is (g(x)) times the derivative of the inside function.

We can apply the chain rule to your problem. The first step is to take the derivative of the outside function evaluated at the inside function. The derivative of f(x) is 3x^2, which we know because of the power rule. If we evaluate f'(x) at g(x), we get f'(g(x)) = 3
(g(x))^2. Expanding g(x), we get that f'(g(x)) = 3*(8x^2-3x)^2.

The next step is to find g'(x), the derivative of g. Since g(x) = 8x^2-3x, we know by the power rule that g'(x) = 16x-3.

According to the chain rule, as we saw above, the derivative of f(g(x)) = f'(g(x))g'(x). We have already found f'(g(x)) and g'(x) separately; now we just have to multiply them to find the derivative of the composite function. Multiplying our answers, we get 3(8x^2-3x)^2*(16x-3). This expands to 3072x^5 - 2880x^4 + 864x^3 - 81x^2.

I hope that helps!
• does 'differentiate' means 'get the derivative'?
• it means finding the gradient of the tangent to a point
• I'm a little bit study hard to find the derivative of y = tan(x)cos(2x), some of math sites called the answer is y' = 4cos^2(x) - sec^2(x) - 2. Anyone help me? Bunch of thanks for that!
(1 vote)
• d/dx (tan(x)*cos(2x) = tan(x) * d/dx cos(2x ) + cos(2x) * d/dx tan(x). I will leave it to you as this is precalculus now. Just substitute the quantities.
Hint: d/dx cos(2x) = -2sin(2x)