Single Variable Calculus, Chapter 3, 3.6, Section 3.6, Problem 10

Determine $\displaystyle \frac{dy}{dx}$ of $y^5 + x^2y^3 = 1 + x^4 y$ by Implicit Differentiation.

$\displaystyle \frac{d}{dx}(y^5) + \frac{d}{dx}(x^2y^3) = \frac{d}{dx} (1) + \frac{d}{dx}(x^4y)$


$
\begin{equation}
\begin{aligned}
\frac{d}{dx} (y^5) + \left[ (x^2) \frac{d}{dx}(y^3) + (y^3) \frac{d}{dx} (x^2) \right] &= \frac{d}{dx} (1) \left[ (x^4) \frac{d}{dx} (y) + (y) \frac{d}{dx} (x^4) \right]\\
\\
(5y^4) \frac{dy}{dx} + (x^2)(3y^2) \frac{dy}{dx} + (y^3)(2x) &= 0 + (x^4) \frac{dy}{dx} + (y)(4x^3)

\end{aligned}
\end{equation}
$



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