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

##### 1 Answer
Aug 6, 2017

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

#### Explanation:

$\text{differentiate "color(blue)"implicitly with respect to x}$

$\text{differentiate "ytan(xy)" using the "color(blue)"product rule}$

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

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

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

$\text{differentiate "-xy" using the "color(blue)"product rule}$

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

$\text{putting the right side all together}$

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

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

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