# How do you implicitly differentiate -1=xy-tan(x-y) ?

##### 1 Answer
Feb 7, 2016

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

#### Explanation:

$- 1 = x y - \tan \left(x - y\right)$

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

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

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

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

Now we just have to "shift terms" to make $\frac{\mathrm{dy}}{\mathrm{dx}}$ the subject of the formula.

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

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