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Math Help - Extremely hard calculus question.

  1. #1
    Member integral's Avatar
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    Extremely hard calculus question.

    Extremely hard for me at least!
    This problem is just something I came up with. for practicing properties, and how to use partial.

    I just want to know if I did this right.. Sorry, but no one I know can do this stuff, not even the cal teacher at my higschool .

    consider: t=x^{x^y}
    Find \frac{\partial t_x}{\partial x} and
    Then, say \frac{\partial t_x}{\partial x}=z_x

    Find \nabla z_x


    I found (hopefully correctly) that z_x=x^{x^y}\left [ yx^{y-1} \right ] ln(x)+x^{y-1}


    Then
    \nabla z=\frac{\partial z_x}{\partial x}\mathbf{i}+\frac{\partial z_y}{\partial y}\mathbf{j}}
    z=\left [x^{x^y} y\right]\left [x^{y-1}  ln(x)\right ]+x^{y-1}

    \frac{\partial z}{\partial x}=\frac{\partial}{\partial x}\left [x^{x^y} y\right]\left [x^{y-1}  ln(x)\right ]+\left [x^{x^y} y\right]\frac{\partial}{\partial x}\left [x^{y-1}  ln(x)\right ]+(y-1)x^{y-2}

    \frac{\partial z}{\partial x}=\left [ {yz_x} \right ]\left [ x^{y-1}ln(x) \right ]+ \left [ \left [ (y-1)x^{y-2}ln(x) \right ]+x^{y-2} \right ]yx^{x^y}+(y-1)x^{y-2}
    =\left [y \left [x^{x^y}\left [ yx^{y-1} \right ] ln(x)+x^{y-1} \right] \right ]\left [ x^{y-1}ln(x) \right ] + \left [ \left [ (y-1)x^{y-2}ln(x) \right ]+x^{y-2} \right ] yx^{x^y}+(y-1)x^{y-2}

    \frac{\partial z}{\partial x}\mathbf{i}=\left [\left [\left [x^{x^y}\left [ yx^{y-1} \right ] ln(x^y)+yx^{y-1}  \right] \right ]\left [ x^{y-1}ln(x) \right ] + \left [ \left [  (y-1)x^{y-2}ln(x) \right ]+x^{y-2} \right ] yx^{x^y}+(y-1)x^{y-2}\right ] \mathbf{i}
    Is that right?

    I can't figure out the partial with respect to y though.. Can someone help?
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  2. #2
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    h=x^{x^y}

    ln(h)=x^yln(x)

    \frac{1}{h}h'=(x^yln(x))ln(x)

    h'=x^{x^y}x^yln^2(x)

    h'=x^{x^y+y}ln^2(x)

    Regards.
    Last edited by p0oint; June 23rd 2010 at 12:49 PM.
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  3. #3
    A Plied Mathematician
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    I think p0oint might be a little confused. t is a function of x and y. It is not true that y=x^(x^y).

    I agree with integral's z_{x} computation. As for \frac{\partial x}{\partial x}, I get

    x^{-2 + x^y + y}\,\left( -1 + x^y + 2\,y + y\,\ln (x)\,\left( -1 + y + {x}^y\,\left( 2 + y\,\ln (x) \right)  \right)  \right). You can check to see if it matches up with your answer.

    Can you start on the y derivative?
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  4. #4
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    Sorry, it was notation problem.

    Now I changed it to h(x) which is the partial derivative of t with respect to y.

    Regards.
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