Single Variable Calculus, Chapter 3, 3.6, Section 3.6, Problem 13

Determine $\displaystyle \frac{dy}{dx}$ of $4 \cos x \sin y = 1$ by Implicit Differentiation.

$\displaystyle \frac{d}{dx} (4 \cos x \sin y) = \frac{d}{dx} (1)$


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



$
\begin{equation}
\begin{aligned}
4y' \cos x \cos y - 4 \sin x \sin y &= 0\\
\\
4y' \cos x \cos y &= 4 \sin x \sin y\\
\\
\frac{\cancel{4}y' \cancel{\cos x \cos y}}{\cancel{4}\cancel{\cos x \cos y}} &= \frac{\cancel{4}\sin x \sin y}{\cancel{4} \cos x \cos y}\\
\\
y' & = \frac{\sin x \sin y}{\cos x \cos y}\\
\\
y' &= \left( \frac{\sin x}{\cos x}\right) \left( \frac{\sin y }{\cos y} \right)\\
\\
y' &= \tan x \tan y
\end{aligned}
\end{equation}
$