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Thread: Is this formula correct?

  1. #1
    Feb 2009

    Is this formula correct?

    hi everyone

    need help to verify this formula for partial derivative, is this correct?thank you in advance for all help & support.

    $\displaystyle \frac{\partial z}{\partial x}= - \frac {F_x}{F_z}$
    $\displaystyle \frac{\partial z}{\partial y}= - \frac {F_y}{F_z}$

    Find $\displaystyle \frac{\partial z}{\partial x}$ and $\displaystyle \frac{\partial z}{\partial y}$ for

    a) xyz - cos(x + y + z)=0
    $\displaystyle F_x = yz + sin(x + y + z)$,
    $\displaystyle F_z=xy+sin(x+y+z)$
    $\displaystyle F_y=xz+sin(x+y+z)$

    $\displaystyle \frac{\partial z}{\partial x}= - \frac {F_x}{F_z}$
    $\displaystyle =-\frac{yz + sin(x + y + z)}{xy+sin(x+y+z)}$
    $\displaystyle \frac{\partial z}{\partial y}= - \frac {F_y}{F_z}$
    $\displaystyle -\frac{xz+sin(x+y+z)}{xz+sin(x+y+z)}$

    b) $\displaystyle 4ln(4xyz) + sin(xz^2))=0 $
    $\displaystyle F_x=\frac{4}{x}+z^2cos(xz^2)$
    $\displaystyle F_y=\frac{4}{y}$
    $\displaystyle F_z=\frac{4}{x}+2xz^2cos(xz^2)$
    $\displaystyle \frac{\partial z}{\partial x}=-\frac{\frac{4}{x}+z^2cos(xz^2)}{\frac{4}{x}+2xz^2c os(xz^2)}$
    $\displaystyle \frac{\partial z}{\partial y}=\frac{\frac{4}{y}}{\frac{4}{x}+2xz^2cos(xz^2)}$

    really appreciate if someone can help verify & confirm,thank you in advance for all your kind help & support,really appreciate.

    Thank you & regards
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  2. #2
    MHF Contributor
    Oct 2008
    All good up until Fz in b).
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