# How do you find the derivative of x+xy=y^2?

May 3, 2016

I assume that you want the derivative of $y$ with respect to $x$. I get $\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1 + y}{2 y - x}$

#### Explanation:

There is no such thing as the derivative of an equation. We can find the derivative of either variable with respect to a variable. I assume that you want to find $\frac{\mathrm{dy}}{\mathrm{dx}}$.

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

We assume that $y$ is some function of $x$ that we haven't found. Think of it as "some stuff in parentheses".

${\underbrace{x}}_{\text{term 1"+underbrace(x("some stuff"))_"term 2" = underbrace(("some stuff")^2)_"term 3}}$

We have 3 terms and we will differentiate term-by-term.

In order to differentiate $x \left(\text{some stuff}\right)$, we'll need the product rule and for ${\left(\text{some stuff}\right)}^{2}$ we'll need the power rule and the chain rule.

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

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

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

There are 4 terms now. Two of them include a factor of $\frac{\mathrm{dy}}{\mathrm{dx}}$ and the other two do not. Solve for $\frac{\mathrm{dy}}{\mathrm{dx}}$.

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

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

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