# How do you find dy/dx by implicit differentiation of x^3-xy+y^2=4?

Nov 28, 2016

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{3 {x}^{2} + y}{x - 3 y}$

#### Explanation:

${x}^{3} - x y + {y}^{2} = 4$

We differentiate everything wrt $x$:

$\frac{d}{\mathrm{dx}} \left({x}^{3}\right) - \frac{d}{\mathrm{dx}} \left(x y\right) + \frac{d}{\mathrm{dx}} \left({y}^{2}\right) = \frac{d}{\mathrm{dx}} \left(4\right)$

We can just deal with the ${x}^{3}$ and the constant term;

$3 {x}^{2} - \frac{d}{\mathrm{dx}} \left(x y\right) + \frac{d}{\mathrm{dx}} \left({y}^{2}\right) = 0$

For the second term we apply the product rule;

 3x^2 - { (x)(d/dx(y)) +(d/dx(x))(y } + d/dx(y^2) = 0
$\therefore 3 {x}^{2} - \left\{x \frac{\mathrm{dy}}{\mathrm{dx}} + \left(1\right) \left(y\right)\right\} + \frac{d}{\mathrm{dx}} \left({y}^{2}\right) = 0$
$\therefore 3 {x}^{2} - x \frac{\mathrm{dy}}{\mathrm{dx}} + y + \frac{d}{\mathrm{dx}} \left({y}^{2}\right) = 0$

And for the last term we use the chain rule so that we can differentiate wrt $y$ (this is the "Implicit" part of the differentiation

$3 {x}^{2} - x \frac{\mathrm{dy}}{\mathrm{dx}} + y + \frac{\mathrm{dy}}{\mathrm{dx}} \frac{d}{\mathrm{dy}} \left({y}^{2}\right) = 0$
$\therefore 3 {x}^{2} - x \frac{\mathrm{dy}}{\mathrm{dx}} + y + \frac{\mathrm{dy}}{\mathrm{dx}} 2 y = 0$
$\therefore x \frac{\mathrm{dy}}{\mathrm{dx}} - 2 y \frac{\mathrm{dy}}{\mathrm{dx}} = 3 {x}^{2} + y$
$\therefore \left(x - 3 y\right) \frac{\mathrm{dy}}{\mathrm{dx}} = 3 {x}^{2} + y$
$\therefore \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{3 {x}^{2} + y}{x - 3 y}$