How to Simplify the Derivative of sqrt(x/y) + sqrt(y/x)?

[yourmom98]

okay I am supposed to prove the derivative of: sqrt x/y + sqrt y/x
is equal to y/x

okay I am stuck where i have finally isolated y' i don't know how to reduce it further I am stuck with this at the moment (see attachment)
can you please tell me the process you went through to simplify it?

 

[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

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

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

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

 

[yourmom98]

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

 

[dx]

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