# How do you find dy/dx given x=y+2sqrty?

Jan 23, 2017

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

#### Explanation:

We have:

$x = y + 2 \sqrt{y}$

Completing the square at the second member:

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

we can extract the square root, and as $1 + \sqrt{y} > 0$ we keep only the positive value of the root at the first member:

$\sqrt{x + 1} = 1 + \sqrt{y}$

$\sqrt{y} = \sqrt{x + 1} - 1$

$y = {\left(\sqrt{x + 1} - 1\right)}^{2} = \left(x + 1\right) - 2 \sqrt{x + 1} + 1 = x - 2 \sqrt{x + 1} + 2$

we have thus made $y \left(x\right)$ explicit, so that:

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

Jan 23, 2017

$x \ge - 1 \mathmr{and} y \ge 0$. See the Socratic graph.
$y ' = \frac{\sqrt{y}}{\sqrt{y} + 1}$.

#### Explanation:

graph{y+sqrty-x=0 [-10, 10, -5, 5]}

For real x, $y \ge 0$

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

$y ' = \frac{1}{\frac{\mathrm{dx}}{\mathrm{dy}}} = \frac{\sqrt{y}}{\sqrt{y} + 1}$.

Of course, inversely,

$y = \sqrt{1 + x} - 1 \ge 0$, giving $x \ge 0$