Help simplifying derivitive

yourmom98

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 Images

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mezarashi

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]

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

.
[tex]\frac{\sqrt{xy^3}}{\sqrt{yx^3}}[/tex]

yourmom98

okay so how does this prove y' = y/x?

dx

[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]