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

##### 1 Answer
Dec 13, 2015

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

#### Explanation:

$\frac{d}{\mathrm{dx}} \left[1 = \frac{x}{y} - x \tan y\right]$

Find each derivative.

$\frac{d}{\mathrm{dx}} \left[1\right] = 0$

$\frac{d}{\mathrm{dx}} \left[\frac{x}{y}\right] = \frac{y \frac{d}{\mathrm{dx}} \left[x\right] - x \frac{d}{\mathrm{dx}} \left[y\right]}{y} ^ 2 = \frac{y - x \frac{\mathrm{dy}}{\mathrm{dx}}}{y} ^ 2$

$\frac{d}{\mathrm{dx}} \left[x \tan y\right] = \tan y \frac{d}{\mathrm{dx}} \left[x\right] + x \frac{d}{\mathrm{dx}} \left[\tan y\right] = \tan y + x {\sec}^{2} y \frac{\mathrm{dy}}{\mathrm{dx}}$

Plug the derivatives back in.

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

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

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

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

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