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Math Help - partial derivative

  1. #1
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    partial derivative

    I encountered the following implication in a text book:

    <br />
g(x,y) = - \frac{\partial}{\partial y} f(x,y)\ \Longrightarrow\ <br />
\underbrace{\frac{\partial}{\partial y} \frac{1}{2} f^2(x,y)}_{(a)} =<br />
g(x,y) \int_x^y g(x,u) du<br />

    which I'm strugling to understand...

    Being (a) = f(x,y), it can be derived integrating the first equation:

    <br />
\int_x^y g(x,u) du = \int_x^y -\frac{\partial}{\partial u} f(x,u) = -f(x,u) \vert_x^y = -f(x,y)<br />

    That's because f(x,x) := 0 by definition.

    Therefore the initial implication does not make sense to me.

    Does it sense to some you guys?
    If yes, would you pls help me to understand/derive it?

    Thanks!
    Last edited by paolopiace; May 7th 2008 at 02:40 PM. Reason: adding "in a text book"
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  2. #2
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    Hello,

    Quote Originally Posted by paolopiace View Post
    I encountered the following implication in a text book:

    <br />
g(x,y) = - \frac{\partial}{\partial y} f(x,y)\ \Longrightarrow\ <br />
\underbrace{\frac{\partial}{\partial y} \frac{1}{2} f^2(x,y)}_{(a)} =<br />
g(x,y) \int_x^y g(x,u) du<br />

    which I'm strugling to understand...

    Being (a) = f(x,y), it can be derived integrating the first equation:

    <br />
\int_x^y g(x,u) du = \int_x^y -\frac{\partial}{\partial u} f(x,u) = -f(x,u) \vert_x^y = -f(x,y)<br />

    That's because f(x,x) := 0 by definition.

    Therefore the initial implication does not make sense to me.

    Does it sense to some you guys?
    If yes, would you pls help me to understand/derive it?

    Thanks!
    Well, \frac{\partial}{\partial y} \frac{1}{2} f^2(x,y){\color{magenta}=}\frac 12 \cdot \ \left(2 \cdot {\color{red}\frac{\partial}{\partial y}} {\color{blue}f(x,y)}\right) \cdot f(x,y) by using the chain rule for differentiation.

    {\color{magenta}=}{\color{red}-g(x,y)} \cdot {\color{blue}f(x,y)}

    You have proved that \int_x^y g(x,u) du=-f(x,y) --> {\color{blue}f(x,y)}=-\int_x^y g(x,u) du

    {\color{magenta}=}-g(x,y) \cdot (-\int_x^y g(x,u) du)=g(x,y) \cdot \int_x^y g(x,u) du

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