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How do you derive this x^(x/8)=4x

  1. Nov 13, 2005 #1
    How do you derivate this:

    Can anyone help?
  2. jcsd
  3. Nov 13, 2005 #2
    I don't think you can differentiate it, but I might be wrong
  4. Nov 13, 2005 #3
    Functions have derivatives. Equations do not.
  5. Nov 13, 2005 #4
    y = x^(x/8),
    ln y = x/8 ln x

    (1/y)(dy/dx) = (x/8)(1/x) + (1/8)ln x
    (dy/dx) = y(1/8 + (1/8)ln x)
    dy/dx = (1/8 + (1/8)ln x)x^(x/8)

    Im not sure if this is allowed. Certainly the power rule isn't allowed (eg. to say d/dx x^x = x*x^(x-1)) because the power is a function of x. But Im not sure if you can differentiate implicitly like this either...
  6. Nov 13, 2005 #5
    I think you misread the original post. But for the problem you gave, it looks good to me.
  7. Nov 13, 2005 #6


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    Theres only one variable in your equation. What do you want to differentiate and with respect to what?
  8. Nov 13, 2005 #7
    A derivative is a rate of change. A relative "growth" to something. You cannot possibly differentiate this, as there is only one variable, if you wish to see it that way. You COULD differentiate it with respect to itself, but i think you will find that that will not take long. I think there is an error in the problem.
  9. Nov 14, 2005 #8
    x^(x/8)=4x means as far as I can tell a y=X^(x/8)-4x so he's probably looknig for a min or a max of it? It's poorly expressed to say the least.
  10. Nov 14, 2005 #9


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    But there is no y variable. You could imply it to mean y=4x-x^(x/8) as well
  11. Nov 14, 2005 #10


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    I certainly wouln't assume that. I would assume the original post was asking how to differentiate both sides of the equation.
    MichaelW24 already did the hard part: if you let y= x^(x/8) then
    y'= (1/8 + (1/8)ln x)x^(x/8).
    That's the left side of the equation. Differentiating both sides of the equation x^(x/8)=4x gives simply (1/8 + (1/8)ln x)x^(x/8)= 4.
    IF "=" was a typo for "-" then the derivative of x^(x/8)- 4x is
    (1/8 + (1/8)ln x)x^(x/8)- 4.
  12. Nov 14, 2005 #11


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    By the way, if you have to differentiate
    [tex]y= f(x)^{g(x)}[/tex]
    There are two mistakes you might make:
    1. Treat the "g(x)" as if it were a constant and use the power law
    [tex]y'= (g(x))f(x)^{g(x)-1}f'(x)[/tex]
    2. Treat the "f(x)" as if it were a constant and use the exponential law
    [tex]y'= ln(f(x))f(x)^{g(x)}g'(x)[/tex]

    Of course, the correct answer is the sum of those:
    [tex]y'= (g(x))f(x)^{g(x)-1}f'(x)+ ln(f(x))f(x)^{g(x)}g'(x)[/tex]!
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