# How do you find dy/ dx for 4 + xy = y^2?

Aug 30, 2015

Use implicit differentiation with the product and power rules.

#### Explanation:

Because of the topic under which this was posted (and because solving for $y$ then differentiating is tedious),

let's differentiate implicitly.

$4 + x y = {y}^{2}$

Remember that $y$ is some unknown function of $x$. It may help to think of it as "some stuff in parentheses".

$\frac{d}{\mathrm{dx}} \left(4 + x y\right) = \frac{d}{\mathrm{dx}} \left({y}^{2}\right)$

$\frac{d}{\mathrm{dx}} \left(4\right) + {\underbrace{\frac{d}{\mathrm{dx}} \left(x y\right)}}_{\text{product and chain}}$ $= {\underbrace{\frac{d}{\mathrm{dx}} \left({y}^{2}\right)}}_{\text{power and chain}}$

$0 + \left[\underbrace{\left(1\right) y + x \frac{d}{\mathrm{dx}} \left(y\right)}\right] = \underbrace{2 {y}^{1} \frac{d}{\mathrm{dx}} \left(y\right)}$

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

Solve algebraically for

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

Note
The product rule can be written in various orders. I've used $\frac{d}{\mathrm{dx}} \left(u v\right) = u ' v + u v '$