Single Variable Calculus, Chapter 3, 3.6, Section 3.6, Problem 19

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

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

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



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