# How do you find the dy/dx for tan x + sec y - y = 0?

Jul 17, 2015

$\frac{\mathrm{dy}}{\mathrm{dx}} = {\sec}^{2} \frac{x}{1 - \sec y \tan y}$

#### Explanation:

Differentiate w.r.t x the terms,

${\sec}^{2} x + \sec y \tan y \frac{\mathrm{dy}}{\mathrm{dx}} - \frac{\mathrm{dy}}{\mathrm{dx}} = 0$

$\left(1 - \sec y \tan y\right) \frac{\mathrm{dy}}{\mathrm{dx}} = {\sec}^{2} x$

$\frac{\mathrm{dy}}{\mathrm{dx}} = {\sec}^{2} \frac{x}{1 - \sec y \tan y}$