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Math Help - how to know here on what variable its a derivative of..(diff)

  1. #1
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    how to know here on what variable its a derivative of..(diff)

    (2x^2ylny-x)y'=y
    (2x^2ylny-x)dy=ydx
    then i divide both sides by dy
    (2x^2ylny-x)=yx'
    then i divide both sides by y
    (2x^2lny-\frac{x}{y})=x'
    x'+\frac{x}{y}=2x^2lny
    so i have here a bernuly foruma
    i divide both sides by x^2
    \frac{x'}{x^2}+\frac{1}{xy}=2lny
    z=x^{-1}
    z'=-1x^{-2}x'
    -z'+\frac{z}{y}=2ln y

    z is defined to be a function of x
    so z'=\frac{dz}{dx}
    why the book interprets z'=\frac{dz}{dy}
    ??
    Last edited by transgalactic; August 16th 2009 at 02:46 AM.
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  2. #2
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    z=\frac{1}{x} and x is a function of y, therefore z is also a function of y
    This is already what you have considered when you wrote z' = -x^{-2}x'
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  3. #3
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    z is linked to y not in a direct way .
    but z linked to x in a direct way
    z and x are more close to each other.

    i cant see a mathematical way of figuring it out
    its all intuition.and i my intuition is very bad
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  4. #4
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    When you write z=\frac{1}{x} with x function of y (in other words x = x(y)), you write a relation between two functions

    It is the same as writing z(y)=\frac{1}{x(y)}
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  5. #5
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    ok thanks
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