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Differentiate Exponential Functions

  1. Sep 2, 2007 #1
    this is an even problem of my homework so there is no answer

    i think i did it right, i would just like a thumbs up/down ... thanks

    [tex]y=2^{3^{x^{2}}}}[/tex]

    [tex]\ln{y}=\ln2^{3^{x^{2}}}}[/tex]

    [tex]\ln{y}=3^{x^{2}}}\times\ln2[/tex]

    [tex]\frac{y'}{2^{3^{x^{2}}}}}=\ln2\times3^{x^{2}}}\times\ln3\times2x[/tex]

    [tex]y'=2x\times2^{3^{x^{2}}}}\times3^{x^{2}}}\ln2\times\ln3[/tex]
     
    Last edited: Sep 2, 2007
  2. jcsd
  3. Sep 2, 2007 #2

    rock.freak667

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    thumbs up..it is correct
     
  4. Sep 2, 2007 #3

    olgranpappy

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    ...there is no answer? That's deep, man.
     
  5. Sep 2, 2007 #4
    thumbs up

    but a faster method is to do the chain rule and remember that:

    [tex]
    y=a^x[/tex]
    [tex]y'=a^xlna[/tex]
     
  6. Sep 2, 2007 #5

    olgranpappy

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    hmm... that's not actually too useful here, is it?

    Perhaps more useful is something along the lines of:

    If
    [tex]
    y(x)=a^{f(x)}
    [/tex]

    then
    [tex]
    y'(x)=a^{f(x)}\ln(a)f'(x)
    [/tex]
     
  7. Sep 2, 2007 #6
    i like that
     
  8. Sep 2, 2007 #7

    olgranpappy

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  9. Sep 3, 2007 #8

    Hurkyl

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    Of course, you are doing exactly what bob said to do.
     
  10. Sep 3, 2007 #9
    so i was looking over my hw and found a few mistakes on some of them (even problems)

    i'm suppose to simplify this problem.

    1st one

    [tex]e^{x+\ln{x}}[/tex]

    [tex]e^{x}\times e^{\ln{x}}[/tex]

    [tex]xe^{x}[/tex] should this be my final answer?

    [tex]e^{x^{2}}[/tex] can i do this step or can i not assume it is was a power of?
     
    Last edited: Sep 3, 2007
  11. Sep 3, 2007 #10

    olgranpappy

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    No. His suggestion was a special case of mine having f(x)=x.

    ...
    ......
     
  12. Sep 3, 2007 #11

    olgranpappy

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    Stop there! The next line does not follow

    Nooooo. that is not a property of the exponential. You are confusing the properties of exponentials and logs...
     
  13. Sep 3, 2007 #12

    Hurkyl

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    He said to use the chain rule. What you wrote is what you get when you combine the chain rule with (a^x)' = a^x ln a
     
  14. Sep 3, 2007 #13

    olgranpappy

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    So what? You said "exactly", but it was not "exactly what bob said to do".

    I'm just pointing that out. Don't you agree?
     
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