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How to take this derivative?

  1. Jul 1, 2012 #1
    I promise guys, no homework here, just curiosity.

    I am trying to find dy/dx for the equation y = x^x - C, where C is any arbitrary constant. I've found two ways that SHOULD be ok to take this derivative, but they produce different answers, I was wondering which method is correct and which method is incorrect. Also, why? It seems to me that both of these methods should be ok. Anyways, here they are:

    Method 1:
    [itex]y = x^x - C[/itex]
    [itex]ln(y) = ln(x^x - C)[/itex]
    [itex]ln(y) = \frac{x ln(x)}{ln(C)}[/itex]
    Now take the derivative:
    [itex]\frac{\frac{dy}{dx}}{y}=\frac{1}{ln(C)}(x ln(x))'[/itex]
    Using the Product Rule, it can be seen that [itex](x ln(x))' = ln(x) + 1[/itex]. Therefore:
    [itex]\frac{dy}{dx}=\frac{y}{ln(C)}(ln(x) + 1)[/itex]
    [itex]\frac{dy}{dx}=\frac{x^x}{ln(C)}(ln(x) + 1)[/itex]

    Method 2:
    [itex]y = x^x - C[/itex]
    [itex]y' = (x^x)' - C'[/itex]
    [itex]y' = (x^x)'[/itex]
    [itex](x^x)'[/itex] can be evaluated using method 1 for the equation [itex]y = x^x[/itex]
    [itex]\frac{dy}{dx} = x^x(ln(x) + 1)[/itex]

    Method one seems a bit less hand wavy, so I'm more confident in it; however, the derivative shouldn't depend on C, so that makes me lean more toward Method 2.

    Anybody have any input they'd be willing to share?
     
  2. jcsd
  3. Jul 1, 2012 #2

    jedishrfu

    Staff: Mentor

    rewrite x as e^ln(x) to e^(xln(x)) and then use the rule d/dx (e^u) = e^u * u'
     
  4. Jul 1, 2012 #3

    Ibix

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    [itex]\ln(a-b)\neq\ln a/\ln b[/itex]. If I were you I'd just try to rewrite the [itex]x^x[/itex] as [itex]e^{f(x)}[/itex].
     
    Last edited: Jul 1, 2012
  5. Jul 1, 2012 #4
    Rewrite which x? Both? In the original or in the derivative?

    Are you saying to do this?:

    [itex]y = x^x - C[/itex]
    [itex]e^y = e^{x^x - C} = e^{x^x}/e^C[/itex]

    And then differentiate?

    Oh right!! I was getting the rule backwards it seems.

    So would method 2 be the one that is correct then?

    y = x^x can be differentiated pretty easily with implicit differentiation.
     
  6. Jul 1, 2012 #5

    eumyang

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    Homework Helper

    jedishrfu is saying that
    [itex]x = e^{\ln x}[/itex]
    so
    [itex]x^x = \left(e^{\ln x}\right)^x = e^{x\ln x}[/itex]

    So what you need to take the derivative of is this:
    [itex]y = e^{x\ln x} - C[/itex]
    and you won't need to worry about implicit differentiation.
     
  7. Jul 1, 2012 #6
    Oh I see, well let's try:

    [itex]y = e^{x\ln x} - C[/itex]
    [itex]\frac{dy}{dx} = (e^{x\ln x})'[/itex]
    [itex]\frac{dy}{dx} = e^{x\ln x}(ln(x) + 1)[/itex]
    [itex]\frac{dy}{dx} = x^x(ln(x) + 1)[/itex]

    Thanks for the help guys. I think the only problem I was having was mixing up my logarithm properties :)
     
  8. Jul 1, 2012 #7

    HallsofIvy

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    Staff Emeritus
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    The derivative of an added or subtracted constant is 0 so the derivative of
    [itex]y= x^x- C[/itex] is exactly the same as the derivative of [itex]y= x^x[/itex]. Now, y= x ln(x) so dy/dx= ln(x)- (x/x)= ln(x)- 1.
     
  9. Jul 1, 2012 #8

    haruspex

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    I think you meant: ln(y)= x ln(x) so (dy/dx)/y = ln(x) + (x/x)= ln(x)+ 1 (Note: +, not -).
    This leads to the same answer as in the OP by Method 2.
    Method 1 went wrong after the second line. ln(A-B) is not ln(A)/ln(B). Must be thinking of ln(A/B) = ln(A) - ln(B) (or, equivalently, exp(A-B) = exp(A)/exp(B)).
     
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