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

Oct 13, 2016

I got

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

Since $y$ is a function of $x$ (either $y$ in the equation), the derivative with respect to $x$ must include $\frac{\mathrm{dy}}{\mathrm{dx}}$ whenever the derivative of $y$ with respect to $x$ is taken.

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

$\frac{\mathrm{dy}}{\mathrm{dx}} = 2 {\left(x + y\right)}^{1} \cdot \stackrel{\text{Chain Rule}}{\overbrace{\frac{d}{\mathrm{dx}} \left[x + y\right]}}$

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

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

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

Now isolate the similar terms.

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

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

$\implies \textcolor{b l u e}{\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{2 x + 2 y}{1 - 2 x - 2 y}}$

Oct 13, 2016

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

#### Explanation:

Expand inside the parentheses:

$y = \left(x + y\right) \left(x + y\right)$

$y = {x}^{2} + x y + x y + {y}^{2}$

$y = {x}^{2} + 2 x y + {y}^{2}$

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

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

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

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

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

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

Hopefully this helps!