# How do you implicitly differentiate x^3 + y^3 = 9xy?

##### 1 Answer
Jul 9, 2015

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

#### Explanation:

When deriving implicitly you must remember to trea $y$ as a function of $x$ so that, for example, if you have:
${y}^{2}$ derived gives you: $2 y \cdot \frac{\mathrm{dy}}{\mathrm{dx}}$
In your case you have:
$3 {x}^{2} + 3 {y}^{2} \frac{\mathrm{dy}}{\mathrm{dx}} = 9 y + 9 x \frac{\mathrm{dy}}{\mathrm{dx}}$ where on the right I used the Product Rule:

Collecting $\frac{\mathrm{dy}}{\mathrm{dx}}$ you get:

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{3 {x}^{2} - 9 y}{9 x - 3 {y}^{2}} = \frac{\cancel{3} \left({x}^{2} - 3 y\right)}{\cancel{3} \left(3 x - {y}^{2}\right)} = \frac{{x}^{2} - 3 y}{3 x - {y}^{2}}$