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Using chain rule to differentiate x^x

  1. Oct 7, 2009 #1
    1. The problem statement, all variables and given/known data

    Use the chain rule to find (d/dx)(xx) by using the function f(y,z)=yz.

    2. Relevant equations

    Chain rule: [tex]\frac{dz}{dt}[/tex] = [tex]\frac{\partial z}{\partial x}[/tex] [tex]\frac{dx}{dt}[/tex] + [tex]\frac{\partial z}{\partial y}[/tex] [tex]\frac{dy}{dt}[/tex]

    3. The attempt at a solution

    I honestly have no clue on how to use chain rule in this problem. I have spent about an hour trying to set up a chain rule equation and I have given up hope on finding one.

    The partial derivatives of f(y,z) are:
    [tex]\frac{\partial x}{\partial y}[/tex] = zyz-1
    [tex]\frac{\partial x}{\partial z}[/tex] = yzln(y)

    I have a strong feeling that I need to set up parametric equations somewhere to cancel out and get dy/dx, but I can't seem to find any possible way.

    I know how to find the derivative using one-variable calculus, and it comes to out be [dy/dz = xx(1+ln(x))]. Any hints on how to setup a chain rule equation would be greatly appreciated, thank you.
    Last edited: Oct 7, 2009
  2. jcsd
  3. Oct 7, 2009 #2
    Let y(t) = t and let z(t) = t. What you're looking for is then df/dt, which the fits the chain rule you posted nicely.
  4. Oct 8, 2009 #3
    Alright so taking it a step further:

    [tex]\frac{df}{dt} = \frac{\partial f}{\partial y} \frac{dy}{dt} + \frac{\partial f}{\partial z} \frac{dz}{dt}[/tex]



    [tex]\frac{dy}{dt} = 1[/tex]

    [tex]\frac{dz}{dt} = 1[/tex]

    [tex]\frac{df}{dt} = \frac{\partial f}{\partial y}[/tex] + [tex]\frac{\partial f}{\partial z}[/tex]

    [tex]\frac{df}{dt} = zy^z-1 + y^z ln(y)[/tex]

    [tex]\frac{df}{dt} = \frac{zy^z}{y} + y^z ln(y)[/tex]

    [tex]\frac{df}{dt} = y^z (\frac{z}{y} + ln(y))[/tex]



    [tex]\frac{dy}{dx} = x^x (\frac{x}{x} + ln(x))[/tex]

    [tex]\frac{dy}{dx} = x^x (1+ln(x))[/tex]

    Thank you very much for your help.
  5. Oct 8, 2009 #4
    The answer to your problem: (d/dx)(x^x) = y(x^x)....z is to be cancelled out along the operations..try differentiating both sides of the equations.

    First let y=x^x and
    f(y,z)=u=y^z--->from here solve for y in terms of z, you can do that by getting the ln of both sides.
    You will arrive at having: y=e^lnu.

    Hope this could be of any help.
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