Help simplifying derivitive

  • Thread starter yourmom98
  • Start date
  • #1
42
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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?
 

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Answers and Replies

  • #2
mezarashi
Homework Helper
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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]
 
  • #3
42
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okay so how does this prove y' = y/x?
 
  • #4
dx
Homework Helper
Gold Member
2,011
18
[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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