Main content

## AP®︎/College Calculus AB

### Course: AP®︎/College Calculus AB > Unit 3

Lesson 7: Selecting procedures for calculating derivatives: multiple rules- Differentiating using multiple rules: strategy
- Differentiating using multiple rules: strategy
- Applying the chain rule and product rule
- Applying the chain rule twice
- Derivative of eᶜᵒˢˣ⋅cos(eˣ)
- Derivative of sin(ln(x²))
- Differentiating using multiple rules

© 2023 Khan AcademyTerms of usePrivacy PolicyCookie Notice

# Derivative of eᶜᵒˢˣ⋅cos(eˣ)

Sal differentiates eᶜᵒˢˣ⋅cos(eˣ) by applying both the product rule and the chain rule. Created by Sal Khan.

## Want to join the conversation?

- The derivative of e^x is equal to e^x. Then, why is the derivative of e^(cos(x)) equal to -sin(x)*(e^(cos(x)) and not to e^(cos(x)) ?(31 votes)
- That's because of the chain rule. In simple terms, when deriving e^A, you will get A'e^A, A' being the derivative of A. Since in the case of e^x, the derivative of x is 1, you simply get e^x. If it was e^2x however, then you would get 2e^2x, due to the derivative of 2x being 2.(84 votes)

- My book is full of functions like √(3x+1)(x-1)^2. My first approach was to use the chain rule on both then apply the product rule. I thought to do that because you clearly have compositions.

However you simply use the product rule to find the solution.

So my question is how do I recognize when to use EITHER the product rule OR the chain rule in function such as this?(13 votes)- The chain rule deals with a function of a function: d/dx [f (g(x))].

The product rule deals with two separate functions multiplied together. d/dx [f(x) * g(x)].

So you have to see whether your expression is a function of a function, like cos(e^x), or two functions multiplied together, like e^cosx * cos (e^x).(23 votes)

- can we factor out e^cosx from the solution?

so would e^cosx((-cos(e^x)*sinx)-e^x*sin(e^x)) be correct?(11 votes)- not sure why no one else has answered this, but yes that is correct(1 vote)

- Why we declare it as "e to the something with respect to something" at2:00. We normally say "x to the squared with respect to x" not "x to the squared with respect to 2". Shouldn't he have said that "e to the something with respect to e"?(4 votes)
- No. "With respect to" identifies the variable of differentiation, not where or how it might appear in the function. To find out, you look at the function definition. Is it f(x) or f(t). At this level of study, you're only dealing with one potential variable of differentiation, but later on you'll deal with more. For example:

f(x,y,z) = 3x²+cos(y) - e^z

When you differentiate, which variable is the variable of differentiation? What are you differentiating with respect to? x, y, or z? With functions like these, you get a different answer depending on what your variable of differentiation is. This should not be confused with differentiation in which one of the variables is defined as a function of the other.

But you don't need to worry about this just yet. For now, you just need to understand that "with respect to" references the variable in the f(of whatever) definition.(11 votes)

- for f(x)=xlnx-nx, x>0, find x in terms of n for f '(x)=0

it says to use the product rule to obtain f ' (x) = x * (1/x) + 1 * lnx-n

I cannot see how to obtain this result with the product rule, and I don't understand how the n fits into the process. Any and all help appreciated!(4 votes)- Let's start with the original equation, and I'll work through the math step by step.

f(x) = x * ln(x) - nx

We want to take the derivative of this thing with respect to x (aka f ' (x) ), so let's do that:

f ' (x) = d/dx (x * ln(x) - nx)

To solve this derivative, we can first separate out the two terms. Recall that the derivative of one thing minus another thing is just the same as the derivative of the first thing minus the derivative of the second thing. So we can re-write it like this:

f ' (x) = d/dx (x * ln(x)) - d/dx (nx)

Now, "n" is a constant. Remember that the derivative of any constant times x is just that constant. For instance, the derivative of "3x" is just 3, the derivative of "12x" is just 12, and so the derivative of "nx" is just n. So let's just replace "d/dx (nx)" with just "n":

f ' (x) = d/dx (x * ln(x)) - n

Now we just need to solve for the first part, "d/dx (x * ln(x))". We can do this with the product rule:

d/dx (x * ln(x)) = d/dx (x) * ln(x) + x * d/dx (ln(x))

The derivative of x with respect to x is just 1, and the derivative of ln(x) with respect to x is just 1/x. So we can re-write that again as:

1 * ln(x) + x * 1/x

And now, let's not forget to add back that "-n" from before:

f ' (x) = 1 * ln(x) + x * 1/x - n

And that is exactly the same thing that you wrote above (but in a slightly different order). I hope this helps.(8 votes)

- in the "Combining the product and chain rules" exercise i've encountered multiple problems where factors seem to arbitrarily cancel out. For example, the derivative of (x+1)/((e^x)+1) using the quotient rule is (1*((e^x)+1)-(x+1)(e^x))/((e^x)+1)^2. Somehow this simplifies to (1-(xe^x))/((e^x)+1)^2. What I don't understand is how ((e^x)+1)(1) =1 without canceling one of the squares in the denominator.(3 votes)
`d/dx((x + 1)/(e^x + 1))`

(d/dx(x + 1)•(e^x + 1) - (x + 1)•d/dx(e^x + 1))/(e^x + 1)^2

(d/dx(x + 1)•(e^x + 1) - (x + 1)•(d/dx(e^x) + d/dx(1)))/(e^x + 1)^2

(d/dx(x + 1)•(e^x + 1) - (x + 1)•(d/dx(e^x) + 0))/(e^x + 1)^2

(d/dx(x + 1)•(e^x + 1) - (x + 1)•(d/dx(e^x)))/(e^x + 1)^2

(d/dx(x + 1)•(e^x + 1) - (x + 1)•(e^x•d/dx(x)))/(e^x + 1)^2

(d/dx(x + 1)•(e^x + 1) - (x + 1)•(e^x•dx/dx))/(e^x + 1)^2

(d/dx(x + 1)•(e^x + 1) - (x + 1)•e^x)/(e^x + 1)^2

((d/dx(x) + d/dx(1))•(e^x + 1) - (x + 1)•e^x)/(e^x + 1)^2

((d/dx(x) + 0)•(e^x + 1) - (x + 1)•e^x)/(e^x + 1)^2

((d/dx(x))•(e^x + 1) - (x + 1)•e^x)/(e^x + 1)^2

((dx/dx)•(e^x + 1) - (x + 1)•e^x)/(e^x + 1)^2

((e^x + 1) - (x + 1)•e^x)/(e^x + 1)^2

((e^x + 1) - (e^x•x + e^x))/(e^x + 1)^2

(e^x + 1 - e^x•x - e^x)/(e^x + 1)^2

(1 - e^x•x)/(e^x + 1)^2(6 votes)

- Does the final result e^x and e^(cos(x)) could be combined together? so it would be e^(cos(2x)) ?(2 votes)
- No. e^(x) * e^(cos x) = e^{ x + cos x](3 votes)

- At5:35, doesn't (e^x)(e^cos(x)) simplify to e^(cos(x) + x) ?(1 vote)
- I didn't watch the video, but yes, you can write (e^x)(e^cos(x)) as e^(x + cos(x)).(2 votes)

- Would it be possible to simplify the find expression to:

d/dx = -e^(cos x) [(sin x)(cos e^x) + e^x(sin e^x)](1 vote)- you never put something EQUAL to d/dx. i believe you want to find the derivative of -e^(cos x) [(sin x)(cos e^x) + e^x(sin e^x)] . Isn't it? Yeah its possible. i'll do it for you.

(sin(X))^2cos(e^X)e^cos(X)+sin(X)sin(e^X)e^cos(X)+X-cos(X)cos(e^X)e^cos(X)+e^X sin(e^X)+e^2Xcos(e^X)(2 votes)

- How do you determine whether or not to use the chain rule with the product rule? Do you always use them together in problems such as

y=5x^2 (3x^2 - 1)?(1 vote)

## Video transcript

Let's now use what we know about the chain
rule and the product rule to take the derivative of
an even weirder expression. So, we're gonna take the derivative, we're
gonna take the derivative of either the cosine of x times the cosine of
e to the x. So, let's take the derivative of this. So, we can view this as the product of two
functions. So, the product rule tells us that this is
going to be the derivative with respect to x of e to
the cosine of x, e to the cosine of x times cosine times
cosine of e to the x plus, plus the first function, just e to
the cosine of x. E to the cosine of x, times the derivative
of the second function. Times the derivative with respect to x of
cosine of e to the x. Cosine of e to the x. And so, we just need to figure out what
these two derivatives are. And so you can imagine the chain rule
might be applicable here. So, let me make it clear. This, we got from the product rule. Product, product rule. But then, to evaluate each of these derivatives, we need to use the chain
rule. So, let's think about this a little bit. So, the derivative, let me copy and paste
this so I don't have to rewrite it. So, copy and paste. So, let's think about what the derivative
of e to the cosine of x is, e to the cosine of x. So, we can view our outer function as e to
the something ,as e to the something and the derivative
of e to the something with respect to something is
just going to be e to that something so it's going to be e to
the cosine of x. So, let me do that in that same blue
color. So, it's going to be e, actually I'm gonna
do it in that, actually I'm gonna do it in a new
color. Let me do it in magenta. So, the derivative of e to the something
with respect to something is just e to the
something. It's just e to the cosine of x, and we
have to multiply that times the derivative of the
something with respect to x. So, what's the derivative of cosine of x
with respect to x? Well, that's just negative sine of x. So, it's times negative sine of x. And so, we figured out this first
derivative. Let me make it clear. This right over here is the derivative of
e to the cosine of x. Derivative of e to the cosine of x, with
respect to, with respect to cosine of x. And this right over here, this right over
here is the derivative of cosine of x with respect,
with respect to x. And we just took the product of the two. That's what the chain rule tells us. Fair enough. Now, let's figure out this derivative out
here. So, we want to find the derivative with
respect to x of cosine e to the x. So, once again, let me copy and paste it. So, we need to figure out this thing right
over here. So first, just like we did, we're just
going to apply the chain rule again. We need to figure out the derivative of
cosine of something, in this case, e to the x, with respect to
that something. So, this is going to be equal to
derivative of cosine of something with respect to that something is equal to
the negative sine of that something. Negative sine of e to x, of e to the x. Once again, we can view this as the
derivative of cosine of e to the x, with respect to, with
respect to e to the x. And then we multiply that times the derivative of the something with respect
to x. So, let me do this, and this, and I'm
running out of colors. Let me do this in the screen color. So, times the derivative of e to the x
with respect to x is just e to the x. So that right over there is the derivative
of e to the x with respect, with respect to
x. And so, we're essentially done, we just
have to substitute what we found using the chain rule back into
our original expression. The derivative of this business up here is
going to be equal to, let me just copy and paste everything just to
make everything nice, nice and clean. So, copy and paste. So, that is going to be equal to, is going
to be equal to this times cosine e to the x, so this is
going to be, we'll see, we could put the e to the x out front, we
could put the negative out front, so we could write it
as negative e to the cosine x. e to the Cosine x, times Sine of x, times
Sine of x, times Cosine of e to the x, times Cosine
of e to the x. So, that's this first term here. Plus, plus e of the cosine x times all of
this stuff. And so, let's see, we can put the negative
out front again. So let's put that negative out front. So we have a negative, negative of e to
the cosine of x times e to the x. I can write it this way. e to the x times e to the cosine of x, and
you could simplify that, or combine it, since you're multiplying two things
with the same base, but I'll just leave it like this,
like this. e to the x, times e to the cosine of x,
times the sine. We already have the negative, so then we
have sine of e to the x. Sine of e to the x, right over here. Times sin, sin, of e to the x, we had
negative sin e to the x times e to the x, negative sin of
e to the x times e to the x, and then that
multiplied by e to the cosine of x so we have the exact same
thing right over here. And we're done.