# What is the implicit derivative of 1=y/x-x-tany?

Jan 20, 2016

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

#### Explanation:

Using implicit differentiation:

$1 = \frac{y}{x} - x - \tan y$

$\implies \frac{d}{\mathrm{dx}} 1 = \frac{d}{\mathrm{dx}} \left(\frac{y}{x} - x - \tan y\right)$

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

$\implies \frac{x \left(\frac{d}{\mathrm{dx}} y\right) - y \left(\frac{d}{\mathrm{dx}} x\right)}{x} ^ 2 - 1 - {\sec}^{2} y \left(\frac{d}{\mathrm{dx}} y\right) = 0$

$\implies \frac{1}{x} \frac{\mathrm{dy}}{\mathrm{dx}} - \frac{y}{x} ^ 2 - 1 - {\sec}^{2} y \frac{\mathrm{dy}}{\mathrm{dx}} = 0$

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

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

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