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Homework Help: Derivative of an expoential within an exponential

  1. Nov 4, 2007 #1
    [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
  2. jcsd
  3. Nov 4, 2007 #2
    We know that the derivative of an exponential is simply the exponential times the derivative of power term, right?

    [tex] \frac{d}{du} e^u = e^uu' [/tex]

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

    Does that help?
  4. Nov 4, 2007 #3
    i think we are not looking at the same problem. my problem is an exponential within an exponential

    d/dx e^(e^4x)
  5. Nov 4, 2007 #4
    It's not the exact equation that you need to apply, but that's where you start. If I let [tex] u = e^{4x} [/tex], then I must first find the derivative of [tex] e^u [/tex].

    I know [tex] \frac{d}{du} e^u = e^uu' [/tex].

    So you know what "u" is, now you must find the derivative of "u" and plug it into the above equation.
  6. Nov 4, 2007 #5

    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)

  7. Nov 4, 2007 #6
    Not quite, you want to think of "u" as separate from everything else and only plug it in at the end. If [tex] u = e^{4x} [/tex], what's the derivative, u' ?
  8. Nov 4, 2007 #7
    thats exactly what i did. i made "u" separate.

    and the derivative of u is 4e^(4x), to answer your question.
  9. Nov 4, 2007 #8
    Right, so you'd have [tex] 4e^{e^{4x}}e^{4x}[/tex]. I didn't see you type in the [tex] e^{e^{4x}} [/tex] part.
  10. Nov 5, 2007 #9
    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: Nov 5, 2007
  11. Nov 5, 2007 #10
    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 [tex] e^{e^x}[/tex]. I'd use the chain rule once again letting [tex] u = e^x [/tex]. So I'd have:

    [tex] \frac{d}{du} e^u = e^uu' [/tex]

    Find u':

    [tex] u' = e^x * 1 [/tex]

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

    [tex] e^{e^x}*e^x*1 [/tex]

  12. Nov 5, 2007 #11
    check it. i was wrong. the answer should be

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

    which is

    u' times e^u

    via the chainrule
  13. Nov 5, 2007 #12
    i follow you. you are right.
  14. Nov 5, 2007 #13
    and how do you type in the equations like that? it looks clean...
  15. Nov 5, 2007 #14
    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.
  16. Nov 5, 2007 #15
    ok. thanks. lol.
  17. Nov 5, 2007 #16
    You're quite welcome :)
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