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Math Help - implicit differentiation f(x,y,z)

  1. #1
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    implicit differentiation f(x,y,z)

    Let y be a function of x satisfying F(x,y, x+y)=0 where F(x,y,z) is a given function. find a formula for \frac{dy}{dx}

    F_{x} = (1, 0, 1), F_{y} = (0,1,1)
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  2. #2
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    Re: implicit differentiation f(x,y,z)

    I know \frac{dy}{dx} = \frac{-f_{x}}{{f_y}}

    but how does that change with 3 variables
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    Re: implicit differentiation f(x,y,z)

     \frac{\partial f}{\partial x} + \frac{\partial f}{\partial y}\frac{dy}{dx}+\frac{\partial f}{\partial x}  + \frac{\partial f}{\partial y}\frac{dy}{dx}

    \frac{-\frac{\partial f}{\partial x}(1+1)}{\frac{\partial f}{\partial y}(1+1)} = \frac{dy}{dx}


    or

    \frac{-\frac{\partial f}{\partial x}\frac{\partial f}{\partial z}}{\frac{\partial f}{\partial y}} = \frac{dy}{dx}


    ??
    Last edited by Jonroberts74; July 6th 2014 at 02:05 AM.
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  4. #4
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    Re: implicit differentiation f(x,y,z)

    If x, y, and z were independent variables, then z would be treated like a constant while calculating dy/dx.

    But if F(z, y, z)= F(x, y, x+ y)= 0, then z= x+y, and, differentiating with respect to x, \frac{\partial F}{\partial x}+ \frac{\partial F}{\partial y}\frac{dy}{dx}+ \frac{\partial F}{\partial z}\left(\frac{dy}{dx}+ 1\right)= 0.
    Last edited by HallsofIvy; July 6th 2014 at 06:49 AM.
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