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Math Help - partial derivative question

  1. #1
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    partial derivative question

    I need some clarification on how to get the \frac{\partial z}{\partial x} and \frac{\partial z}{\partial y} on:

    x^2 +y^2+z^2 = 3xyz

    I know that the solution is \frac{\partial z}{\partial x} = \frac{3yz-2x}{2z-3xy} and \frac{\partial z}{\partial y} = \frac{3xz-2y}{2z-3xy} but I have no idea how they got to it.
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  2. #2
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    Let's do \frac{\partial{z}}{\partial{x}}

    We can use implicit differentiation letting y be a constant.

    x^{2}+y^{2}+z^{2}=3xyz

    2x+2z\frac{dz}{dx}=3xy\frac{dz}{dx}+3yz

    2z\frac{dz}{dx}-3xy\frac{dz}{dx}=3yz-2x

    \frac{dz}{dx}(2z-3xy)=3yz-2x

    \frac{dz}{dx}=\frac{3yz-2x}{2z-3xy}
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  3. #3
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    Quote Originally Posted by lllll View Post
    I need some clarification on how to get the \frac{\partial z}{\partial x} and \frac{\partial z}{\partial y} on:

    x^2 +y^2+z^2 = 3xyz

    I know that the solution is \frac{\partial z}{\partial x} = \frac{3yz-2x}{2z-3xy} and \frac{\partial z}{\partial y} = \frac{3xz-2y}{2z-3xy} but I have no idea how they got to it.
    Use implicit differentiation (z van be treated as an implicit function of x and y):

    Differentiate both sides partially with resepct to x:

    2x + 2z \frac{\partial z}{\partial x} = 3yz + 3xy \frac{\partial z}{\partial x}.

    The chain rule has been used on the left hand side to differentiate z^2 and the product rule has been used on the right to differentiate the product.

    Now re-arrange the result to make \frac{\partial z}{\partial x} the subject.

    Same idea when getting the partial derivative with respect to y.

    Note that you can re-arrange the given rule to make z the subject and then take the partial derivative with respect to x in the usual way.
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