# How do you find dy/dx by implicit differentiation given sqrtx+sqrty=x+y?

Feb 6, 2018

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

#### Explanation:

We have ${x}^{\frac{1}{2}} + {y}^{\frac{1}{2}} = x + y$

First we take $\frac{d}{\mathrm{dx}}$ of each term:
$\frac{d}{\mathrm{dx}} \left[{x}^{\frac{1}{2}}\right] + \frac{d}{\mathrm{dx}} \left[{y}^{\frac{1}{2}}\right] = \frac{d}{\mathrm{dx}} \left[x\right] + \frac{d}{\mathrm{dx}} \left[y\right]$

${x}^{- \frac{1}{2}} / 2 + \frac{d}{\mathrm{dx}} \left[{y}^{\frac{1}{2}}\right] = 1 + \frac{d}{\mathrm{dx}} \left[y\right]$

Using the chain rule: $\frac{d}{\mathrm{dx}} = \frac{\mathrm{dy}}{\mathrm{dx}} \cdot \frac{d}{\mathrm{dy}}$

${x}^{- \frac{1}{2}} / 2 + \frac{\mathrm{dy}}{\mathrm{dx}} \frac{d}{\mathrm{dy}} \left[{y}^{\frac{1}{2}}\right] = 1 + \frac{\mathrm{dy}}{\mathrm{dx}} \frac{d}{\mathrm{dy}} \left[y\right]$

${x}^{- \frac{1}{2}} / 2 + \frac{\mathrm{dy}}{\mathrm{dx}} {y}^{- \frac{1}{2}} / 2 = 1 + \frac{\mathrm{dy}}{\mathrm{dx}} 1$

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

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

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

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

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

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

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

Feb 6, 2018

$\frac{\frac{1}{2 \sqrt{x}} - 1}{1 - \frac{1}{2 \sqrt{y}}}$

#### Explanation:

Differentiate the two sides:

$\frac{\mathrm{ds} q r t x}{\mathrm{dx}} + \frac{\mathrm{ds} q r t y}{\mathrm{dx}} = \frac{\mathrm{dx}}{\mathrm{dx}} + \frac{\mathrm{dy}}{\mathrm{dx}}$

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

Leave $\frac{\mathrm{dy}}{\mathrm{dx}}$ alone:

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

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

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