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Math Help - first and second order partial derivatives

  1. #1
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    first and second order partial derivatives

    Hi! I知 novice in the field and surely my question would be very simple for you. I知 now reading a textbook of Mathematics for biologists and try to understand the solved problems in the text. I have trouble with this one:
    Find all partial derivatives up to second order of the function: z=(x+y)/(x-y)

    The book says:
    first and second order partial derivatives-derivatives.jpg
    I am confused with ∂/∂y(∂z/∂x) and ∂/∂y(∂z/∂y). Trying to solve the problem, I get: first and second order partial derivatives-my_results.jpg
    I know that for the most functions of two variables ∂/∂y(∂z/∂x) = ∂/∂x(∂z/∂y) and probably the textbook isn稚 wrong, so I would be grateful if you point where I am wrong.

    Thank you very much in advance!
    Last edited by andrey; December 30th 2011 at 04:59 AM.
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  2. #2
    MHF Contributor alexmahone's Avatar
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    Re: first and second order partial derivatives

    Quote Originally Posted by andrey View Post
    Hi! I知 novice in the field and surely my question would be very simple for you. I知 now reading a textbook of Mathematics for biologists and try to understand the solved problems in the text. I have trouble with this one:
    Find all partial derivatives up to second order of the function: z=(x+y)/(x-y)

    The book says:
    Click image for larger version. 

Name:	derivatives.JPG 
Views:	10 
Size:	35.1 KB 
ID:	23164
    I am confused with ∂/∂y(∂z/∂x) and ∂/∂y(∂z/∂y). Trying to solve the problem, I get: Click image for larger version. 

Name:	my_results.JPG 
Views:	12 
Size:	21.0 KB 
ID:	23163
    I know that for the most functions of two variables ∂/∂y(∂z/∂x) = ∂/∂x(∂z/∂y) and probably the textbook isn稚 wrong, so I would be grateful if you point where I am wrong.

    Thank you very much in advance!
    \frac{\partial}{\partial{y}}(x-y)^2=-2(x-y)
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  3. #3
    Newbie MarceloFantini's Avatar
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    Re: first and second order partial derivatives

    I believe the error is here:

    \frac{\partial}{\partial y}((x-y)^2) = -2(x-y)

    And not 2(x-y) as you used. Try working those steps.
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  4. #4
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    Re: first and second order partial derivatives

    Hello, andrey!

    \text{Find all partial derivatives up to second order of the function: }\:z \:=\: \frac{x+y}{x-y}

    \frac{\partial z}{\partial x} \;=\;\frac{(x-y)(1) - (x+y)(1)}{(x-y)^2} \;=\;\frac{-2y}{(x-y)^2} \;=\;-2y(x-y)^{-2}

    \frac{\partial z}{\partial y} \;=\;\frac{(x-y)(1) - (x+y)(\text{-}1)}{(x-y)^2} \;=\;\frac{2x}{(x-y)^2} \;=\;2x(x-y)^{-2}


    \frac{\partial^2z}{\partial x^2} \;=\;-2y(\text{-}2)(x-y)^{-3}(1) \;=\;\frac{4y}{(x-y)^3}

    \frac{\partial^2z}{\partial y^2} \;=\;2x(\text{-}2)(x-y)^{-3}(\text{-}1) \;=\;\frac{4x}{(x-y)^3}


    \frac{\partial^2z}{\partial y\partial x} \;=\;\frac{(x-y)^2(\text{-}2) - (\text{-}2y)(2)(x-y)(\text{-}1)}{(x-y)^4}

    . . . . . =\;\frac{-2(x-y)\big[(x-y)+2y\big]}{(x-y)^4} \;=\; \frac{-2(x+y)}{(x-y)^3}

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  5. #5
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    Re: first and second order partial derivatives

    Thank you so much for your replies!
    The clarification that ∂/∂y((x-y)^2)=-2(x-y) was really important in my case.

    Soroban, thank you very much for the detailed solution!
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