# What is the implicit derivative of 1=ytanx-y^2?

Dec 19, 2015

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

#### Explanation:

$\frac{d}{\mathrm{dx}} \left[1 = y \tan x - {y}^{2}\right]$

Remember that taking the derivative of any term with a $y$ when implicitly differentiating will put the chain rule into effect and split out a $\frac{\mathrm{dy}}{\mathrm{dx}}$ term. Also remember that finding $\frac{d}{\mathrm{dx}} \left[y \tan x\right]$ will require the product rule.

$0 = \tan x \frac{\mathrm{dy}}{\mathrm{dx}} + y \frac{d}{\mathrm{dx}} \left[\tan x\right] - 2 y \frac{\mathrm{dy}}{\mathrm{dx}}$

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

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

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

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