# How do you implicitly differentiate 2=ytanx-xy ?

Jan 3, 2016

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

#### Explanation:

Find the derivative of each part separately.

$\frac{d}{\mathrm{dx}} \left(2\right) = 0$

Next, use the product rule. Remember that when implicitly differentiating, the chain rule will kick in and spit out $y '$ terms.

$\frac{d}{\mathrm{dx}} \left(y \tan x\right) = y ' \tan x + y {\sec}^{2} x$

Again, use the product rule.

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

Combine all the derivatives and solve for $y '$.

$0 = y ' \tan x + y {\sec}^{2} x - y - x y '$

$y - y {\sec}^{2} x = y ' \left(\tan x - x\right)$

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