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Math Help - Partial differentation

  1. #1
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    Partial differentation

    Hi everyone
    Need help with this question,lifesaver guaranteed.

    Find dz/dx & dz/dy for
    4ln(4xyz)+sin(xz2)=0
    This is the working but i'm not sure if it is corect,need help.
    F_x=4*\frac {4yz}{4xyz}+z^2cosxz^2=\frac{4}{x}+z^2cosxz^2

    F_z=4*\frac{4xy}{4xyz}+2zxcosxz^2=\frac{4}{z}+2zxc  osxz^2
    F_y=4*4xz


    Thank you for all your help & support.
    Last edited by anderson; March 6th 2010 at 11:32 AM.
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  2. #2
    Senior Member
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    So, our starting equation is

    4 \ln (4xyz) + \sin (xz^2) = 0

    When computing \frac{\partial z}{\partial x}, we are considering z to be a function of x, i.e., z(x). So, any time we have a term containing z multiplied with a term containing x, we must use the product rule.

    So now we differentiate:

    \frac{\partial}{\partial x}\left[4 \ln (4xyz) + \sin (xz^2)\right] = \frac{\partial}{\partial x}(0)

    4 \frac{\frac{\partial}{\partial x}(4xyz)}{4xyz} + \cos (xz^2) \frac{\partial}{\partial x} (xz^2) = 0

    Looking at these two partials:

    \frac{\partial}{\partial x}(4xyz) = 4y\frac{\partial}{\partial x}(xz) = 4y\left(x \frac{\partial z}{\partial x} + z\right)

    and

    \frac{\partial}{\partial x}(xz^2) = 2xz \frac{\partial z}{\partial x} + z^2

    Thus:

    4 \frac{4y(x \frac{\partial z}{\partial x} + z)}{4xyz} + \cos (xz^2) \left(2xz \frac{\partial z}{\partial x} + z^2\right) = 0

    At this stage, you can simplify and solve for \frac{\partial z}{\partial x}.

    I'll leave the other partial for you to solve using the same technique.
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  3. #3
    MHF Contributor

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    Did you notice that ln(xyz)= ln(x)+ ln(y)+ ln(z)?

    That's why \frac{\partial ln(xyz)}{\partial x}= \frac{yz}{xyz}= \frac{1}{x}= \frac{\partial ln(x)}{\partial x}
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