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Math Help - Partial Derivatives

  1. #1
    Bar0n janvdl's Avatar
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    Partial Derivatives

    Find \frac{ \partial z }{ \partial y } where z is defined implicitly as a function of x and y by:

    sin(xyz) = x + 2y + 3z

    Thanks in advance
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  2. #2
    Member pberardi's Avatar
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    \frac{\partial}{\partial y}\ [sin(xyz) = x+2y+3z]

    cos(xyz)xz \frac{\partial z}{\partial y}\ = 0+2+3\frac{\partial z}{\partial y}\

    \frac{\partial z}{\partial y}\ (xzcos(xyz)-3) = 2

    \frac{\partial z}{\partial y}\ = \frac{2}{xzcos(xyz)-3}


    Could someone please check my work?
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  3. #3
    Flow Master
    mr fantastic's Avatar
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    Quote Originally Posted by pberardi View Post
    \frac{\partial}{\partial y}\ [sin(xyz) = x+2y+3z]

    cos(xyz)xz \frac{\partial z}{\partial y}\ = 0+2+3\frac{\partial z}{\partial y}\ *

    [snip]
    If z is an implicit function of x and y then the partial derivative of xyz with respect to y is xz + xy \frac{\partial z}{\partial y}.

    Therefore the left hand side of the line marked * should be \cos (xyz) \left(xz + xy \frac{\partial z}{\partial y} \right).
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