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## AP®︎ Calculus AB (2017 edition)

### Course: AP®︎ Calculus AB (2017 edition)>Unit 3

Lesson 6: Product rule

# Product rule review

AP.CALC:
FUN‑3 (EU)
,
FUN‑3.B (LO)
,
FUN‑3.B.1 (EK)
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}

Problem 1
• Current
xg, left parenthesis, x, right parenthesish, left parenthesis, x, right parenthesisg, prime, left parenthesis, x, right parenthesish, prime, left parenthesis, x, right parenthesis
minus, 22minus, 134
F, left parenthesis, x, right parenthesis, equals, g, left parenthesis, x, right parenthesis, dot, h, left parenthesis, x, right parenthesis
F, prime, left parenthesis, minus, 2, right parenthesis, equals

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).
• does 'differentiate' means 'get the derivative'?
• it means finding the gradient of the tangent to a point
• what is a product?
(1 vote)
• A product is what you get when you multiply two things. For example, 4 is the product of 2•2.
• 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)
• 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!
• How would I be able to do this with p=xcos(x)? I'm having trouble understanding this.
(1 vote)
• p'= x* d(cosx)/dx + cosx* d(x)/dx
= -xsin(x)+cos(x) [since d(x)/dx = 1]
(1 vote)
• What special properties of e^x make d/dx(e^x)=e^x? Is it the definition of e (e=(1+1/x)^x as x approaches infinity)?
(1 vote)
• Would it be possible to make this and the Quotient rule available during quizzes and tests? I find I've had to search it up during a quiz/test multiple times per test.
(1 vote)
• The end result is that you'd have these rules in your brain so that you prepare yourself for more serious quizzes and tests, so the Khan Academy system doesn't usually like giving you a formula. The product rule is more straightforward to memorize, but for the quotient rule, it's commonly taught with the sentence "Low de High minus High de Low, over Low Low". "Low" is the function that is being divided by the "High".
Additionally, just take some time to play with the formulas and see if you can understand what they're doing. Try looking up a proof (Khan Academy has one!) and just working with the derivation rules in practice. With enough time, they'll be second-nature.
(1 vote)