Single Variable Calculus, Chapter 3, 3.6, Section 3.6, Problem 9

Determine $\displaystyle \frac{dy}{dx}$ of $x^4(x+y) = y^2(3x-y)$ by Implicit Differentiation.

$\displaystyle \frac{d}{dx} \left[x^4 (x+y) \right] = \frac{d}{dx} \left[y^2 (3x-y) \right]$


$
\begin{equation}
\begin{aligned}
(x^4) \frac{d}{dx} (x + y) + (x+y) \frac{d}{dx} (x^4) & = (y^2) \frac{d}{dx} (3x-y) + (3x-y) \frac{d}{dx} (y^2)\\
\\
(x^4) \left[ \frac{d}{dx} (x) + \frac{d}{dx} (y) \right] + (x+y) \frac{d}{dx} (x^4) &= (y^2) \left[ 3 \frac{d}{dx} (x) - \frac{d}{dx} (y) \right] + (3x-y) \frac{d}{dx} (y^2)\\
\\
(x^4)\left( 1 + \frac{dy}{dx} \right) + (x+y)(4x^3) &= (y^2) \left[ (3)(1) - \frac{dy}{dx} \right] + (3x-y)(2y) \frac{dy}{dx}\\
\\
x^4 + x^4 \frac{dy}{dx} + 4x^4 + 4x^3y &= 3y^2 - y^2 \frac{dy}{dx} + 6xy \frac{dy}{dx} - 2y^2 \frac{dy}{dx}\\
\\
5x^4 + 4x^3y + x^4 \frac{dy}{dx} &= 3y^2 - 3y^2 \frac{dy}{dx} + 6xy \frac{dy}{dx}\\
\\
x^4 \frac{dy}{dx} + 3y^2 \frac{dy}{dx} - 6xy \frac{dy}{dx} &= 3y^2 - 5x^4 - 4x^3y
\end{aligned}
\end{equation}
$



$
\begin{equation}
\begin{aligned}
x^4y' + 3y^2y' - 6xyy' &= 3y^2 - 5x^4 - 4x^3y\\
\\
y'(x^4+3y^2 - 6xy) &= 3y^2 - 5x^4 - 4x^3y\\
\\
\frac{y'\cancel{(x^4+3y^2-6xy)}}{\cancel{x^4+3y^2-6xy}} &= \frac{3y^2-5x^4-4x^3y}{x^4+3y^2-6xy}\\
\\
y' &= \frac{3y^2 - 5x^4 - 4x^3y}{x^4 + 3y^2 - 6xy}
\end{aligned}
\end{equation}
$