# Derivative of an expoential within an exponential

#### sprint

[SOLVED] derivative of an expoential within an exponential

help. i need help on finding the derivative of an exponential within an exponential

1. The problem statement, all variables and given/known data

d/dx of e^(e^4x)

2. Relevant equations

d/dx of e^(e^4x)

3. The attempt at a solution

d/dx of e^(e^4x)

i don't know how to attempt this cause the function im interested is in the power of e

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#### hotcommodity

We know that the derivative of an exponential is simply the exponential times the derivative of power term, right?

$$\frac{d}{du} e^u = e^uu'$$

You're going to have to apply the chain rule.

Does that help?

#### sprint

We know that the derivative of an exponential is simply the exponential times the derivative of power term, right?

$$\frac{d}{du} e^u = e^uu'$$

You're going to have to apply the chain rule.

Does that help?
i think we are not looking at the same problem. my problem is an exponential within an exponential

d/dx e^(e^4x)

#### hotcommodity

It's not the exact equation that you need to apply, but that's where you start. If I let $$u = e^{4x}$$, then I must first find the derivative of $$e^u$$.

I know $$\frac{d}{du} e^u = e^uu'$$.

So you know what "u" is, now you must find the derivative of "u" and plug it into the above equation.

#### sprint

oooooh.

i see. i guess just use substitution letting u = e^(4x)

by any chance, is the correct answer 4e^(4x) times e^(4x) = 4e^(8x)

?

#### hotcommodity

Not quite, you want to think of "u" as separate from everything else and only plug it in at the end. If $$u = e^{4x}$$, what's the derivative, u' ?

#### sprint

thats exactly what i did. i made "u" separate.

and the derivative of u is 4e^(4x), to answer your question.

#### hotcommodity

Right, so you'd have $$4e^{e^{4x}}e^{4x}$$. I didn't see you type in the $$e^{e^{4x}}$$ part.

#### sprint

wow. totally confusing.

but i think the answer is what i stated earlier...

4e^(4x) times e^(4x)

i get this by sticking to the basics and using the chain rule and substitution

Last edited:

#### hotcommodity

I'm not trying to confuse you, haha, but I'd hope that you walk away from this understanding what's going on. Let's say I wanted to find the derivative of $$e^{e^x}$$. I'd use the chain rule once again letting $$u = e^x$$. So I'd have:

$$\frac{d}{du} e^u = e^uu'$$

Find u':

$$u' = e^x * 1$$

And plug u and u' into the first equation to get the derivative:

$$e^{e^x}*e^x*1$$

Right?

#### sprint

check it. i was wrong. the answer should be

4e^(4x) times e^(e^(4x))

which is

u' times e^u

via the chainrule

#### sprint

I'm not trying to confuse you, haha, but I'd hope that you walk away from this understanding what's going on. Let's say I wanted to find the derivative of $$e^{e^x}$$. I'd use the chain rule once again letting $$u = e^x$$. So I'd have:

$$\frac{d}{du} e^u = e^uu'$$

Find u':

$$u' = e^x * 1$$

And plug u and u' into the first equation to get the derivative:

$$e^{e^x}*e^x*1$$

Right?
i follow you. you are right.

#### sprint

and how do you type in the equations like that? it looks clean...

#### hotcommodity

Haha, you just use the "tex" tags. You can click on the equation itself to see how you would type it in. I think there's a tutorial somewhere here on the site.

ok. thanks. lol.

#### hotcommodity

You're quite welcome :)

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