Single Variable Calculus, Chapter 3, 3.6, Section 3.6, Problem 8

Determine $\displaystyle \frac{dy}{dx}$ of $2x^3 + x^2y - xy^3 = 2$ by Implicit Differentiation.

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


$
\begin{equation}
\begin{aligned}
2 \frac{d}{dx} (x^3) + \left[ (x^2) \frac{d}{dx} (y) + (y) \frac{d}{dx} (x^2) \right] - \left[ (x) \frac{d}{dx} (y^3) + (y^3) \frac{d}{dx} (x) \right] &= \frac{d}{dx} (2)\\
\\
(2)(3x^2) + \left[ x^2 \frac{dy}{dx} + (y) (2x) \right] - \left[ (x) (3y^2) \frac{dy}{dx} + (y^3)(1) \right] &= 0\\
\\
6x^2 + x^2y' + 2xy - (3xy^2y'+y^3) &= 0\\
\\
6x^2 + x^2 y' + 2xy - 3xy^2y' - y^3 &= 0
\end{aligned}
\end{equation}
$



$
\begin{equation}
\begin{aligned}
x^2y' - 3xy^2y' &= y^3 - 6x^2 - 2xy\\
\\
y'(x^2-3xy^2) &= y^3 - 6x^2 - 2xy\\
\\
\frac{y'\cancel{(x^2-3xy^2)}}{\cancel{x^2-3xy^2}} &= \frac{y^3-6x^2-2xy}{x^2-3xy^2}\\
\\
y' &= \frac{y^3-6x^2-2xy}{x^2-3xy^2}
\end{aligned}
\end{equation}
$