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Help simplifying derivitive

  1. Nov 22, 2005 #1
    okay im supposed to prove the derivitive of: sqrt x/y + sqrt y/x
    is equal to y/x

    okay im stuck where i have finally isolated y' i dont know how to reduce it further im stuck with this at the moment (see attachment)
    can you please tell me the process you went through to simplify it?

    Attached Files:

  2. jcsd
  3. Nov 22, 2005 #2


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    Homework Helper

    It's better if you write out the equation is expanded form:

    [tex]\frac{\frac{1}{\sqrt{xy}} - \frac{\sqrt{y}}{x^\frac{3}{2}}}{\frac{\sqrt{x}}{y^\frac{3}{2}} - \frac{1}{\sqrt{xy}}}[/tex]

    Doing a couple of simplifying operations

    [tex]\frac{\frac{\sqrt{xy}}{xy} - \frac{\sqrt{y}}{x^\frac{3}{2}}}{\frac{\sqrt{x}}{y^\frac{3}{2}} - \frac{\sqrt{xy}}{xy}}[/tex]

    We try to match the denominators

    [tex]\frac{\frac{\sqrt{x}\sqrt{xy}}{x^\frac{3}{2}y} - \frac{y\sqrt{y}}{yx^\frac{3}{2}}}{\frac{x\sqrt{x}}{xy^\frac{3}{2}} - \frac{\sqrt{y}\sqrt{xy}}{xy^\frac{3}{2}}}[/tex]


  4. Nov 22, 2005 #3
    okay so how does this prove y' = y/x?
  5. Nov 22, 2005 #4


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    Gold Member

    [tex] \frac{\sqrt{xy^3}}{\sqrt{yx^3}}[/tex]
    [tex] = \sqrt{\frac{xy^3}{yx^3} [/tex]
    [tex] = \sqrt{\frac{y^2}{x^2}} = \frac{y}{x} [/tex]
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