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Differentiating x^y + y^x + (lnx)^x etc

  1. Apr 26, 2004 #1

    pig

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    How can this be done?

    I don't even know how I would begin.. How would you differentiate stuff like x^(y^(x^y))? Where y is a function of x, not a constant of course..
     
  2. jcsd
  3. Apr 26, 2004 #2

    arildno

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    The example:
    x^(y^(x^y))=x^(z(x))

    d/dx x^(z(x))=z(x)x^(z(x)-1)+x^(z(x))*ln(x)*z'(x)..
     
  4. Apr 26, 2004 #3

    pig

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    I don't understand this :(

    d/dx x^(z(x))=z(x)x^(z(x)-1)+...

    Where does the + come from? :confused:

    I know I should use the chain rule somehow but I can't seem to figure out how.. I'm having problems with differentiating both x^(f(x)) and f(x)^x..
     
  5. Apr 26, 2004 #4

    pig

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    Hmm if I do this:

    f(x) = x^y
    ln f(x) = ln x^y
    ln f(x) = ylnx
    (ln f(x))' = (ylnx)'
    f'(x)/f(x) = y'lnx + y/x
    f'(x) = f(x)*(y'lnx + y/x)

    (x^y)' = x^y*(y'lnx + y/x)

    Is this right? :confused:
     
  6. Apr 26, 2004 #5

    pig

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    (x^y)' = x^y*(y'lnx + y/x)

    (x^y)' = yx^(y-1)+x^y*lnx*y'

    I think I understand what you wrote after all.. Thanks arildno :)
     
    Last edited: Apr 26, 2004
  7. Apr 26, 2004 #6

    NateTG

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    Use the chain rule:

    [tex]\frac{d}{dx} f(g(x))=g'(x)f'(g(x))[/tex]

    So
    [tex]\frac{d}{dx} x^{(y^{(x^y)})}= \frac{d}{dx} e^{(\ln x \times y^{(x^y)})} = \frac{d}{dx} (\ln x \times y^{(x^y)}) \times x^{(y^{(x^y)})}[/tex]
    [tex]=(\frac{y^{(x^y)}}{x} + \ln x \times \frac{d}{dx} y^{(x^y)}) \times x^{(y^{(x^y)})}[/tex]
    [tex]=(\frac{y^{(x^y)}}{x} + \ln x \times \frac{d}{dx} e^{(y \times x^y)}) \times x^{(y^{(x^y)})}[/tex]
    [tex]=(\frac{y^{(x^y)}}{x} + \ln x \times \frac{d}{dx} {(y \times x^y)} \times y^{(x^y)}) \times x^{(y^{(x^y)})}[/tex]
    [tex]=(\frac{y^{(x^y)}}{x} + \ln x \times (y' x^y + y \frac{d}{dx} (x^y)) \times y^{(x^y)}) \times x^{(y^{(x^y)})}[/tex]
    [tex]=(\frac{y^{(x^y)}}{x} + \ln x \times (y' x^y + y \frac{d}{dx} (y \ln x ) \times x^y) \times y^{(x^y)}) \times x^{(y^{(x^y)})}[/tex]

    [tex]=(\frac{y^{(x^y)}}{x} + \ln x \times (y' x^y + y (\frac{y}{x} + xy' ) \times x^y) \times y^{(x^y)}) \times x^{(y^{(x^y)})}[/tex]
    Obviously, some regrouping is necessary. I somehow doubt that that's the same as what arnildo had.
     
  8. Apr 27, 2004 #7

    arildno

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    Well, I didn't bother to compute z'(x).
    Here's the simplest way to compute x^(z(x)):

    F(x,z)=x^(z), H(x)=F(x,z(x)).

    dH/dx=d^(p)F/dx+d^(p)F/dz*z'(x), where d^(p)/dx is the partial derivative with respect to x.
     
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