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Second Derivative of e^t + t^e ?

  1. Nov 17, 2011 #1
    1. The problem statement, all variables and given/known data

    Find the second derivative of e^t + t^e :


    2. Relevant equations

    (e^u) = u' * (e^u)


    3. The attempt at a solution

    e^t + t^e

    1* (e^t) + e*[t^(e-1)]

    ^^^first derivative

    (e^t) + e*e[t^{(e-1)-1}]

    ^^^ 2nd derivative

    Answer?: (e^t) + e^2[t^(e-2)]
     
  2. jcsd
  3. Nov 17, 2011 #2
    you're quite close. after taking the first derivative, the exponent of t changes, you have not accounted for that.
     
  4. Nov 17, 2011 #3
    I'm not sure I follow. Doesn't the exponent of t become (e-2) ?
     
  5. Nov 17, 2011 #4
    d/dt (t^e) = e*t^(e-1)

    d/dt (e*t^(e-1)) = ??

    EDIT/HINT: Your exponent is correct, it is the constant in front of the exponent that isn't. It's quite a common error, though you probably won't be making it in the future.
     
    Last edited: Nov 17, 2011
  6. Nov 17, 2011 #5
    scratch that... it didn't even look almost right
     
    Last edited: Nov 17, 2011
  7. Nov 17, 2011 #6
    [tex]t^e[/tex]
    [tex]t^u = u' f'(u)[/tex]

    What is u'?
     
  8. Nov 17, 2011 #7
    u' = e

    ?

    :/
     
  9. Nov 17, 2011 #8
    Wrong, derivative of e^x is e^x; don't confuse that with "e"

    hint: e itself is equal to what?
     
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