# How do you find the derivative of tan x + sec y - y = 0?

Jan 23, 2017

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

#### Explanation:

we do this by implicit differentiation

$\frac{d}{\mathrm{dx}} \left(\tan x + \sec y = 0\right)$

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

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