# How do you implicitly differentiate  y-(3y-x)^2-1/y^2=x^3 + y^3- xy?

##### 1 Answer
Apr 1, 2016

By differentiating $x$ and $y$ like your usual differetiation practices.

#### Explanation:

$y - {\left(3 y - x\right)}^{2} - \frac{1}{y} ^ 2 = {x}^{3} + {y}^{3} - x y$

I will let $\frac{\mathrm{dy}}{\mathrm{dx}} = y '$

Now begin the implicit differentiation

$y ' - 2 \left(3 y - x\right) \left(3 y ' - 1\right) - \left(- 2\right) \frac{1}{y} ^ 3 y ' = 3 {x}^{2} + 3 {y}^{2} y ' - \left(y + x y '\right)$

I moved all the terms to one side and factored out $y '$

$y ' \left[7 - 18 y + \frac{2}{y} ^ 2 - 3 {y}^{2} + x\right] + 7 y - 2 x - 3 {x}^{2} = 0$

Trick is; after differentiating any form of $y$, add $y '$

Example: $\frac{d}{\mathrm{dx}} {y}^{2} = 2 y y '$

Hope this helps. Cheers