# What is the first derivative of (y)*(sqrt(x)) - (x)*(sqrt(y)) = 16?

##### 1 Answer
May 16, 2017

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{y \left(2 \sqrt{x} - 1\right)}{x \left(2 \sqrt{y} - 1\right)}$

#### Explanation:

Differentiate both sides of the equation with respect to $x$:

$\frac{d}{\mathrm{dx}} \left(y \sqrt{x} - x \sqrt{y}\right) = 0$

$\frac{d}{\mathrm{dx}} \left(y \sqrt{x}\right) = \frac{d}{\mathrm{dx}} \left(x \sqrt{y}\right)$

using the product rule:

$\frac{y}{2 \sqrt{x}} + \frac{\mathrm{dy}}{\mathrm{dx}} \sqrt{x} = \frac{x}{2 \sqrt{y}} \frac{\mathrm{dy}}{\mathrm{dx}} + \sqrt{y}$

solving for $\frac{\mathrm{dy}}{\mathrm{dx}}$:

$\frac{\mathrm{dy}}{\mathrm{dx}} \left(\sqrt{x} - \frac{x}{2 \sqrt{y}}\right) = \sqrt{y} - \frac{y}{2 \sqrt{x}}$

$\frac{\mathrm{dy}}{\mathrm{dx}} \left(\frac{2 \sqrt{x} \sqrt{y} - x}{2 \sqrt{y}}\right) = \frac{2 \sqrt{x} \sqrt{y} - y}{2 \sqrt{x}}$

$\frac{\mathrm{dy}}{\mathrm{dx}} = \left(\frac{2 \sqrt{x} \sqrt{y} - y}{2 \sqrt{x}}\right) \left(\frac{2 \sqrt{y}}{2 \sqrt{x} \sqrt{y} - x}\right)$

$\frac{\mathrm{dy}}{\mathrm{dx}} = \left(\frac{2 y \sqrt{x} - y}{2 x \sqrt{y} - x}\right)$

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{y \left(2 \sqrt{x} - 1\right)}{x \left(2 \sqrt{y} - 1\right)}$