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Math Help - Partial Derrivatives Assignment problem

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    Partial Derrivatives Assignment problem

    Hi,
    I am an engineering student undertaking a subject called "mathematics for scientist and engineers". I was hopeing someone here could help me better understand this question as I really have no idea on how to even begin to solve it.
    Thank You
    Elbarto
    [IMG]file:///C:/DOCUME%7E1/Barto/LOCALS%7E1/Temp/moz-screenshot.jpg[/IMG]
    Attached Thumbnails Attached Thumbnails Partial Derrivatives Assignment problem-q3-partial-diff.jpg  
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  2. #2
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    Quote Originally Posted by elbarto View Post
    Hi,
    I am an engineering student undertaking a subject called "mathematics for scientist and engineers". I was hopeing someone here could help me better understand this question as I really have no idea on how to even begin to solve it.
    Thank You
    Elbarto
    [IMG]file:///C:/DOCUME%7E1/Barto/LOCALS%7E1/Temp/moz-screenshot.jpg[/IMG]
    z = f(x^2 + y^2)

    When we take the partial derivative of z with respect to y we take an "ordinary" derivative, but assume that y is constant. Thus:
    (del)z/(del)x = f'(x^2 + y^2) * (2x)
    where (del)z/(del)x represents the partial derivative and f'(u) represents the derivative of f(u) with respect to its argument. This is nothing more than the chain rule where we have set y equal to a constant.

    Similarly:
    (del)z/(del)y = f'(x^2 + y^2) * (2y)

    So:
    y * (del)z/(del) - x * (del)z/(del)y

    = y* f'(x^2 + y^2) * (2x) - x * f'(x^2 + y^2) * 2y

    = 2xy* f'(x^2 + y^2) - 2xy * f'(x^2 + y^2) = 0
    as advertised.

    -Dan
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    Hi Dan,
    Thank you very much for your fast reply. Makes so much more sense after someone explains it. I now understand what they mean when they say f is a differentiable function of one variable. I cant thank you enough.
    Regards Elbarto
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