Help simplifying derivitive

1. Nov 22, 2005

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?

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2. Nov 22, 2005

mezarashi

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

$$\frac{\frac{1}{\sqrt{xy}} - \frac{\sqrt{y}}{x^\frac{3}{2}}}{\frac{\sqrt{x}}{y^\frac{3}{2}} - \frac{1}{\sqrt{xy}}}$$

Doing a couple of simplifying operations

$$\frac{\frac{\sqrt{xy}}{xy} - \frac{\sqrt{y}}{x^\frac{3}{2}}}{\frac{\sqrt{x}}{y^\frac{3}{2}} - \frac{\sqrt{xy}}{xy}}$$

We try to match the denominators

$$\frac{\frac{\sqrt{x}\sqrt{xy}}{x^\frac{3}{2}y} - \frac{y\sqrt{y}}{yx^\frac{3}{2}}}{\frac{x\sqrt{x}}{xy^\frac{3}{2}} - \frac{\sqrt{y}\sqrt{xy}}{xy^\frac{3}{2}}}$$

$$\frac{\frac{x-y}{\sqrt{y}x^\frac{3}{2}}}{\frac{x-y}{\sqrt{x}y^\frac{3}{2}}}$$

.
$$\frac{\sqrt{xy^3}}{\sqrt{yx^3}}$$

3. Nov 22, 2005

yourmom98

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

4. Nov 22, 2005

dx

$$\frac{\sqrt{xy^3}}{\sqrt{yx^3}}$$
$$= \sqrt{\frac{xy^3}{yx^3}$$
$$= \sqrt{\frac{y^2}{x^2}} = \frac{y}{x}$$