# Question #60b11

Nov 16, 2017

When $y = \sqrt{\frac{x}{m}} + \sqrt{\frac{m}{x}}$, we observe that $2 x y \frac{\mathrm{dy}}{\mathrm{dx}} - \frac{x}{m} + \frac{m}{x} = 0$.

#### Explanation:

$y = \sqrt{\frac{x}{m}} + \sqrt{\frac{m}{x}}$

Then I assume you're asking, what's the expression $2 x y \frac{\mathrm{dy}}{\mathrm{dx}} - \frac{x}{m} + \frac{m}{x}$ equivalent to?

Well, we need to find $\frac{\mathrm{dy}}{\mathrm{dx}}$. First let's rewrite $y$.

$y = {m}^{- \frac{1}{2}} {x}^{\frac{1}{2}} + {m}^{\frac{1}{2}} {x}^{- \frac{1}{2}}$

So now we can differentiate with the power rule:

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1}{2} {m}^{- \frac{1}{2}} {x}^{- \frac{1}{2}} - \frac{1}{2} {m}^{\frac{1}{2}} {x}^{- \frac{3}{2}} = \frac{1}{2 \sqrt{m x}} - \frac{\sqrt{m}}{2 x \sqrt{x}}$

So then we want to know about:

$2 x y \frac{\mathrm{dy}}{\mathrm{dx}} - \frac{x}{m} + \frac{m}{x}$

$= 2 x \left(\sqrt{\frac{x}{m}} + \sqrt{\frac{m}{x}}\right) \left(\frac{1}{2 \sqrt{m x}} - \frac{\sqrt{m}}{2 x \sqrt{x}}\right) - \frac{x}{m} + \frac{m}{x}$

FOIL the two binomials:

$= 2 x \left(\frac{\sqrt{x}}{\sqrt{m}} \frac{1}{2 \sqrt{m} \sqrt{x}} - \frac{\sqrt{x}}{\sqrt{m}} \frac{\sqrt{m}}{2 x \sqrt{x}} + \frac{\sqrt{m}}{\sqrt{x}} \frac{1}{2 \sqrt{m} \sqrt{x}} - \frac{\sqrt{m}}{\sqrt{x}} \frac{\sqrt{m}}{2 x \sqrt{x}}\right) - \frac{x}{m} + \frac{m}{x}$

And simplify:

$= 2 x \left(\frac{1}{2 m} - \frac{1}{2 x} + \frac{1}{2 x} - \frac{m}{2 {x}^{2}}\right) - \frac{x}{m} + \frac{m}{x}$

Distribute:

$= \left(\frac{x}{m} - \frac{m}{x}\right) - \frac{x}{m} + \frac{m}{x}$

$= 0$

Cool!