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Math Help - Partial Derivatives

  1. #1
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    Partial Derivatives

    Could someone show me where to go at ??? please

    dz/dx is representing partial z with respect to x.

    Find the value of dz/dx at the point (1,1,1) if the equation

    xy+(z^3)x-2yz=0

    defines z as a function of the two independent variables x and y and the partial derivative.

    so we need to differentiate with respect to x?

    d/dx(xz) + d/dx((z^3)x)-d/dx(2yz) = 0

    x dz/dx + ??? - 2y dz/dx = 0
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  2. #2
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by ur5pointos2slo View Post
    Could someone show me where to go at ??? please

    dz/dx is representing partial z with respect to x.

    Find the value of dz/dx at the point (1,1,1) if the equation

    xy+(z^3)x-2yz=0

    defines z as a function of the two independent variables x and y and the partial derivative.

    so we need to differentiate with respect to x?

    d/dx(xz) + d/dx((z^3)x)-d/dx(2yz) = 0

    x dz/dx + ??? - 2y dz/dx = 0
    Yes, you partially differentiate both sides of the equation with respect to x.

    \frac{\partial}{\partial x}\left[xy+z^3x-2yz\right]=\frac{\partial}{\partial x}0\implies y+\frac{\partial}{\partial x}\left(z^3x\right)-2\frac{\partial}{\partial x}\left[yz\right]=0

    Now since z=z\!\left(x,y\right), we use the product rule to evaluate \frac{\partial}{\partial x}\left[z^3x\right]:

    \frac{\partial}{\partial x}\left[z^3x\right]=z^3+2z^2x\frac{\partial z}{\partial x}

    In the case of \frac{\partial}{\partial x}\left[yz\right], we just have y\frac{\partial z}{\partial x} since y is a constant and z depends on x.

    Thus, it follows that we have y+z^3+2z^2x\frac{\partial z}{\partial x}-2y\frac{\partial z}{\partial x}=0\implies y+z^3=\left(2y-2z^2x\right)\frac{\partial z}{\partial x}\implies \frac{\partial z}{\partial x}=\frac{y+z^3}{2y-2z^2x}

    Does this make sense?
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  3. #3
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    That makes sense. Thank you.
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