Single Variable Calculus, Chapter 3, Review Exercises, Section Review Exercises, Problem 36

Find $y'$ of $y = x \tan y = y - 1$


$
\begin{equation}
\begin{aligned}
\frac{d}{dx} (x \tan y) &= \frac{d}{dx} (y) - \frac{d}{dx} (1)\\
\\
(x) \frac{d}{dx} (\tan y) + (\tan y) \frac{d}{dx} (x) &= \frac{dy}{dx} - 0 \\
\\
(x) (\sec^2y)\frac{dy}{dx} + (\tan y)(1) &= \frac{dy}{dx}\\
\\
x \sec^2 y \frac{dy}{dx} + \tan y &= \frac{dy}{dx}\\
\\
xy' \sec^2 y - y' + \tan y &= y'\\
\\
xy' \sec^2y-y' &= - \tan y\\
\\
y' \left( x \sec^2 y - 1 \right) &= -\tan y\\
\\
\frac{y' \cancel{\left( x \sec^2 y - 1 \right)} }{\cancel{ x \sec^2 y - 1 }} &= \frac{-\tan y}{ x \sec^2 y - 1 }\\
\\
y' &= \frac{-\tan y}{ x \sec^2 y - 1 } \qquad \text{or} \qquad y' = \frac{\tan y}{1- x \sec^2 y}
\end{aligned}
\end{equation}
$