Single Variable Calculus, Chapter 3, 3.6, Section 3.6, Problem 20

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

$\displaystyle \frac{d}{dx} (\sin x) + \frac{d}{dx} (\cos y) = \frac{d}{dx} (\sin x \cos y) $


$
\begin{equation}
\begin{aligned}
\frac{d}{dx} (\sin x) + \frac{d}{dx} (\cos y) &= (\sin x) \frac{d}{dx} (\cos y) + (\cos y) \frac{d}{dx} (\sin x)\\
\\
\cos x + (-\sin y) \frac{dy}{dx} &= (\sin x)(-\sin y) \frac{dy}{dx} + (\cos y)(\cos x)

\end{aligned}
\end{equation}
$



$
\begin{equation}
\begin{aligned}
\cos x - y' \sin y &= y' \sin x \sin y + \cos x \cos y\\
\\
y'(\sin x \sin y - \sin y) & = \cos x \cos y - \cos x\\
\\
\frac{y'\cancel{(\sin x \sin y - \sin y)}}{\cancel{\sin x \sin y - \sin y}} &= \frac{\cos x \cos y - \cos x}{\sin x \sin y - \sin y}\\
\\
y' &= \frac{\cos x \cos y - \cos x}{\sin x \sin y - \sin y}
\end{aligned}
\end{equation}
$