# Math Help - Quick question on an implicit differentiation problem

1. ## Quick question on an implicit differentiation problem

Find dy/dx for $x^3y^3 - y = x$

Here's what I did;

$3x^2 3y^2 y' - y' = 1$
$y'(3x^2 3y^2 - 1) = 1$
$y' = 1 / (3x^2 3y^2 - 1)$

Could someone point out to me where I went wrong? The given answer is $1 - 3x^2 y^3 / (3x^3 y^2 - 1)$

2. Originally Posted by Archduke01
Find dy/dx for $x^3y^3 - y = x$

Here's what I did;

$3x^2 3y^2 y' - y' = 1$
$y'(3x^2 3y^2 - 1) = 1$
$y' = 1 / (3x^2 3y^2 - 1)$

Could someone point out to me where I went wrong? The given answer is $1 - 3x^2 y^3 / (3x^3 y^2 - 1)$
You have to apply the product rule to the first term.

$\frac{d}{dx}(x^3y^3)=3x^3y^2y'+3x^2y^3$

Does this make sense?

$\frac{d}{dx}(x^3y^3)=3x^3y^2y'+3x^2y^3$