# What is the derivative of  (x^3)-(xy)+(y^3)=1?

##### 1 Answer
Aug 16, 2017

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

#### Explanation:

We need to find the derivative with respect to $x$, $\mathrm{dy} / \mathrm{dx}$.

When we differentiate $y$, since it's a function that's not $x$, the chain rule will kick in and a $\mathrm{dy} / \mathrm{dx}$ term will arise thanks to the chain rule.

Also don't forget that differentiating $x y$ will use the product rule.

Differentiating:

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

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

$3 {x}^{2} - y - x \frac{\mathrm{dy}}{\mathrm{dx}} + 3 {y}^{2} \frac{\mathrm{dy}}{\mathrm{dx}} = 0$

Solve for $\frac{\mathrm{dy}}{\mathrm{dx}}$:

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

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