# How do you differentiate tan x + sec y - y = 0?

Jul 4, 2016

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

#### Explanation:

$d \left(\tan x + \sec y - y = 0\right)$

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

$\implies {\sec}^{2} x \setminus \mathrm{dx} + \mathrm{dy} \setminus \left(\sec y \tan y - 1\right) = 0$

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

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