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## AP®︎/College Calculus AB

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

Lesson 9: The product rule# Differentiating products

We explore the product rule by finding the derivative of eˣcos(x). We identify eˣ as the first function and cos(x) as the second, then apply the product rule to calculate the derivative, simplifying our result for a clearer understanding.

## Want to join the conversation?

- Could someone help explain to me (or point me towards a video) why d/dx e^x = e^x? I haven't heard of this before :) Thanks!(9 votes)
- what if you have the product of 3 functions? like sinx*cosx*lnx.. etc(6 votes)
- Just apply the product rule twice:

(𝑓 ∙ 𝑔 ∙ ℎ)' = 𝑓 ' ∙ (𝑔 ∙ ℎ) + 𝑓 ∙ (𝑔 ∙ ℎ)' =

= 𝑓 ' ∙ 𝑔 ∙ ℎ + 𝑓 ∙ (𝑔' ∙ ℎ + 𝑔 ∙ ℎ') =

= 𝑓 ' ∙ 𝑔 ∙ ℎ + 𝑓 ∙ 𝑔' ∙ ℎ + 𝑓 ∙ 𝑔 ∙ ℎ'(17 votes)

- How do we know that v(x) = Cos x is equal to v'(x) = -Sin x? As in, is there a rule of which that tells the derivative of Cos, Sin or Tan?(2 votes)
- The results are given in this video:

https://www.khanacademy.org/math/calculus-home/taking-derivatives-calc/ex-and-lnx-derivatives-calc/v/derivatives-of-sin-x-cos-x-tan-x-e-x-and-ln-x

Proofs are available in other videos - I'll leave you to practice your search skills to find those.(6 votes)

- Can someone show me how to differentiate

y= 3x^2(4x+1)^3

Thanks(2 votes)- You have to use product rule, because it is the product of two functions. You can check out the product rule at https://www.khanacademy.org/math/ap-calculus-ab/ab-differentiation-1-new/modal/v/applying-the-product-rule-for-derivatives

Simply stated, the product rule states that h'(x)=f'(x)g(x)+f(x)g'(x) where h(x) = [f(x)g(x)]

using f(x)=3x^2 and g(x)=(4x+1)^3, you have to find the derivatives of each of these functions

f'(x) can be found easily using power rule

g'(x) needs chain use, since you have a function inside another function. you can check out the chain rule at https://www.khanacademy.org/math/ap-calculus-ab/ab-differentiation-2-new/modal/v/chain-rule-introduction

the chain rule states that H'(x) = F'(G(x))G'(x) where H(x)=F(G(x))

in this example, F(x) = x^3 and G(x)=4x+1

so you will need to use a combination of chain rule and power rule to find g'(x), and then plug it in to the product rule.(4 votes)

- Why is
**d/dx**of**u(x)v(x**) equal to**u'(x)v(x)+u(x)v'(x**) and not**u'(x)v'(x**)?(2 votes)- The reasons are explained here - the process is called the Product Rule:

https://www.khanacademy.org/math/differential-calculus/product-quotient-chain-rules-dc#product-rule-dc(4 votes)

- When using the product rule, do both functions need to be functions of x (when I'm taking the derivative with respect to x)?(2 votes)
- Wow - the short answer to your question is no, the two functions do not need to both be functions of x. But if they are not, things get a bit more complicated so Khan does not address that until a bit further in the playlist. That way you can be firm in your grasp of the basics first. :D(2 votes)

- So....product rule=power rule?

I'm totally confused. whether which method i use i should get the same answer right? but using the power rule you get e^x*-Sin(x), using product rule you get thing at4:12

confused because now i don't know which method is right to use now(2 votes)- Using the power rule gets you nothing, because the power rule only applies to polynomials. The power rule says that d/dx(xⁿ)=n·xⁿ⁻¹.

There is no rule that gives e^x·(-sin(x)) as the derivative of e^x·cos(x).(2 votes)

- Why is the 1st derivative of ln(x) = 1/x?(2 votes)
- Let's say that y=ln(x). Instead of taking the derivative of both sides here, we can re-write and differentiate implicitly. y=ln(x)>>>>x=e^y.

Hence, x=e^y

d/dx[x]=d/dx[e^y]

1=e^y*dy/dx (Chain rule)

dy/dx=1/e^y

dy/dx=1/x (Because x=e^y)

I hope this helped.(1 vote)

- how does the derivative of tan^2x be shown in a graph(1 vote)
- Doi you mean something like this?

http://www.wolframalpha.com/input/?i=graph+of+derivative+of+tan%5E2(x)

Wolfram alpha is a good site for generating quick graphs like that.(3 votes)

- I'm again confused -- again. I thought d/dx (e^ x) was [x e^x-1] -- NO?(1 vote)
- no, there is a different property for e^x, you do not use the power rule in this case.

d/dx[e^x] = e^x

This might seem a little weird but this comes out of the definition of "e" and this is what makes it so special.(3 votes)

## Video transcript

- [Voiceover] So let's see if we can find the derivative with respect to x, with either x times the cosine of x. And like always, pause this video and give it a go on your own before we work through it. So when you look at this you might say, "well, I know how to find "the derivative with e to the x," that's infact just e to the x. And let me write this down. We know a few things. We know the derivative
with respect to x of e to the x. E to the x is e to the x. We know how to find the
derivative cosine of x. The derivative with
respect to x of cosine of x is equal to negative sine of x. But, how do we find the
derivative of their product? Well as you can imagine,
this might involve the product rule. And let me just write down the product rule generally first. So if we take the
derivative with respect to x of the first expression in terms of x, so this is, we could
call this u of x times another expression that involves x. So u times v of x. This is going to be equal to, and I'm color-coding it so we can really keep track of things. This is going to be
equal to the derivative of the first expression. So I could write that as u prime of x times just the second expression
not the derivative of it, just the second expression. So times v of x and then we have plus the first expression, not its derivative, just the first expression. U of x times the derivative
of the second expression. Times the derivative of
the second expression. So the way you remember it is, you have these two things here, you're going to end up
with two different terms. In each of them, you're going to take the derivative of one of
them, but not the other one, and then the other one
you'll take the derivative of the other one, but not the first one. So, the derivative of u
times v is u prime times v, plus u times v prime. When you just look at it like that, it seems a little bit abstract and that might even be a little bit confusing, but that's why we have
a tangible example here and I color-coded intentionally. We could say that u of x
is equal to e to the x. And v of x is equal to the cosine of x. So v of x is equal to cosine of x. And if u of x is equal to e to the x, we know that the derivative of that with respect to x is still e to the x. That's one of the most
magical things in mathematics. One of the things that makes e so special. So u prime of x is still
equal to e to the x. And v prime of x, we know as negative sine of x. Negative sine of x, and so, what's this going to be equal to? This is going to be equal to the derivative of the first expression. So, the derivative of e
to the x which is just, e to the x, times the second expression, not taking it's derivative,
so times cosine of x. Plus the first expression, not taking its derivative, so e to the x, times the derivative of
the second expression. So, times the derivative of cosine of x which is negative sine. Negative sine of x. And it might be a little bit confusing, because e to the x is its own derivative. This right over here, you can view this as this was the derivative as e to the x which happens to be e to the x. That's what's exciting
about that expression, or that function. And then this is just e
to the x without taking it's derivative - they are
of course, the same thing. But anyway, now we can just simplify it. This is going to be equal to... We could write either as e
to the x times cosine of x, times cosine of x minus e to the x. E to the x times sine of x. Times sine of x. Or, if you want, you could factor out an e to the x. This is the same thing as e to the x times cosine
of x minus sine of x. Cosine of x minus sine of x. So hopefully this makes the product rule a little bit more tangible. And once you have this in your tool belt, there's a whole broader class of functions and expressions that we
can start to differentiate.