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Derivative of an exponential function

  1. Jan 24, 2009 #1
    1. The problem statement, all variables and given/known data
    What is the derivative of ex3? also what is the derivative of (ln1/x)2


    2. Relevant equations



    3. The attempt at a solution
    is it 3x2e3x2?
    2(ln1/x)(x)?
     
    Last edited: Jan 24, 2009
  2. jcsd
  3. Jan 24, 2009 #2

    jgens

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    Gold Member

    Re: Derivative

    Your solution for d(e^(x^3))/dx is incorrect. Remember d(e^u)/dx = u'e^u where u is a function of x. For the other derivative, let u = ln(1/x) and then apply the chain rule.
     
  4. Jan 24, 2009 #3

    Mark44

    Staff: Mentor

    Re: Derivative

    No. No.
    Both problems require the chain rule.
    For the second problem, d/dx((ln(1/x))^2) = 2* ln(1/x) * d/dx(1/x). You made a mistake in your derivative of 1/x.
     
    Last edited: Jan 24, 2009
  5. Jan 24, 2009 #4
    Re: Derivative

    so the second is 2(ln1/x)(-x^-2)?
    and the first one is x^3e^x^3?
     
  6. Jan 24, 2009 #5

    Mark44

    Staff: Mentor

    Re: Derivative

    Right for the second one, but I would write it as 2 ln(1/x) (-1/x^2) or even better as
    [tex]2 ln(\frac{1}{x}) \frac{-1}{x^2}[/tex]
    Not right for the first. As jgens said, d/dx(e^u) = du/dx * e^u (where u = x^3).
     
  7. Jan 24, 2009 #6

    jgens

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    Re: Derivative

    Your first derivative is still incorrect. d(e^u)/dx = u'e^u, not ue^u.
    Your second derivative is also incorrect. Mark gave you an incorrect expression to differentiate. You need to find 2*ln(x)*d(ln(1/x))/dx which also requires the chain rule.
     
  8. Jan 24, 2009 #7

    jgens

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    Re: Derivative

    Don't forget d(ln(1/x))/dx, which certainly is not -1/x^2!
     
  9. Jan 24, 2009 #8
    Re: Derivative

    so its 3x2ex3?
     
  10. Jan 24, 2009 #9

    jgens

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    Re: Derivative

    Correct.
     
  11. Jan 24, 2009 #10
    Re: Derivative

    d(ln1/x) is x^2?
     
  12. Jan 24, 2009 #11

    jgens

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    Re: Derivative

    No, let u = 1/x then you need to find d(lnu)/dx which will be (1/u)(du/dx).
     
  13. Jan 24, 2009 #12
    Re: Derivative

    Might be easier if you just look at it as:

    [tex]\frac{d}{dx}\;(-ln(x))[/tex]

    Then it's just straight forwardly obvious.
     
  14. Jan 24, 2009 #13

    jgens

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    Re: Derivative

    That certainly should make it simpler for the original poster!
     
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