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

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


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