# How do you use implicit differentiation to find dy/dx given sqrtx+sqrt(yx^2)=x?

May 26, 2017

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

#### Explanation:

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

$\frac{d}{\mathrm{dx}} \left(\sqrt{x} + \sqrt{y {x}^{2}}\right) = \frac{d}{\mathrm{dx}} \left(x\right)$

$\frac{d}{\mathrm{dx}} \left(\sqrt{x}\right) + \frac{d}{\mathrm{dx}} \left(\sqrt{y {x}^{2}}\right) = 1$

$\frac{1}{2 \sqrt{x}} + \frac{1}{2 \sqrt{y {x}^{2}}} \frac{d}{\mathrm{dx}} \left(y {x}^{2}\right) = 1$

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

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

Note that both $x > 0$ and $y > 0$ if the equation is in the real domain, otherwise $\sqrt{x}$ and $\sqrt{y {x}^{2}}$ would not be defined,
so $\frac{x}{\sqrt{{x}^{2}}} = 1$ and $\frac{y}{\sqrt{y}} = \sqrt{y}$

$\frac{1}{2 \sqrt{x}} + \sqrt{y} + \frac{x}{2 \sqrt{y}} \frac{\mathrm{dy}}{\mathrm{dx}} = 1$

$\frac{x}{2 \sqrt{y}} \frac{\mathrm{dy}}{\mathrm{dx}} = 1 - \frac{1}{2 \sqrt{x}} - \sqrt{y}$

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

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