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Math Help - Partial Derivatives

  1. #1
    Junior Member
    Joined
    May 2012
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    Philadelphia
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    Partial Derivatives

    \begin{array}\\\text{Given:}\\
    z=sin(2u)cos(3v)\\
    \text{find}\frac{\partial{z}}{\partial{u}}\\
    \\
    \\
    \text{The answer is the following:}\\
    \\
    [sin(2u)(0)]+[cos(3v)(cos(2u)(2)]\\
    \\
    \text{however, i'm not sure why the answer doesn't simply regard the cos(3v) as a constant, making the answer:}\\
    \\
    [cos(3v)][cos(2u)(2)],\\
    \\
    \text{as in the problem:}\\


    f(x,y,z)=x^{2}y^{5}+xz^{2}\;\;\;\text{where}\;\;f_ {x}=y^{5}2x+z^{2}(1).\\
    \text{Here, }y^{5}\text{and }z^{2}\text{are treated as if they were constants.}\\
    \\
    \text{I suspect the chain rule has something to do with it, but i just don't know. Could someone help to clarify this?}








    \end{array}
    " alt="\begin{array}\\\text{Given:}\\
    z=sin(2u)cos(3v)\\
    \text{find}\frac{\partial{z}}{\partial{u}}\\
    \\
    \\
    \text{The answer is the following:}\\
    \\
    [sin(2u)(0)]+[cos(3v)(cos(2u)(2)]\\
    \\
    \text{however, i'm not sure why the answer doesn't simply regard the cos(3v) as a constant, making the answer:}\\
    \\
    [cos(3v)][cos(2u)(2)],\\
    \\
    \text{as in the problem:}\\


    f(x,y,z)=x^{2}y^{5}+xz^{2}\;\;\;\text{where}\;\;f_ {x}=y^{5}2x+z^{2}(1).\\
    \text{Here, }y^{5}\text{and }z^{2}\text{are treated as if they were constants.}\\
    \\
    \text{I suspect the chain rule has something to do with it, but i just don't know. Could someone help to clarify this?}








    \end{array}
    " />
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  2. #2
    Junior Member
    Joined
    May 2012
    From
    Philadelphia
    Posts
    28

    Re: Partial Derivatives

    Sorry everybody. I tried to post in latex, but obviously I messed up.
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