# How do you implicitly differentiate (x+y)^2-2xy=x+220?

Dec 12, 2015

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

#### Explanation:

Take the derivative with respect to $x$. Remember that any $y$-term that's differentiated will spit out a $\frac{\mathrm{dy}}{\mathrm{dx}}$ term thanks to the chain rule.

$\frac{d}{\mathrm{dx}} \left[{\left(x + y\right)}^{2} - 2 x y = x + 220\right]$

Find each part:

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

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

$\frac{d}{\mathrm{dx}} \left[x\right] = 1$

$\frac{d}{\mathrm{dx}} \left[220\right] = 0$

We get:

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

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

$\frac{\mathrm{dy}}{\mathrm{dx}} \left(4 x + 2 y\right) = 1 - 2 x$

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