Single Variable Calculus, Chapter 3, 3.6, Section 3.6, Problem 11

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

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

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

\end{aligned}
\end{equation}
$



$
\begin{equation}
\begin{aligned}
2x^2yy'+2xy^2+xy' \cos y + \sin y &= 0\\
\\
2x^2yy' + xy' \cos y &= -2xy^2 - \sin y\\
\\
y'(2x^2y+ x \cos y) &= -2xy^2-\sin y\\
\\
\frac{y'\cancel{(2x^2+ x \cos y)}}{\cancel{2x^2y + x \cos y}} &= \frac{-2xy^2 - \sin y}{2x^2y + x \cos y}\\
\\
y' &= \frac{-2xy^2 - \sin y}{2x^2 y + x \cos y}

\end{aligned}
\end{equation}
$