# How do you implicitly differentiate 2=e^(xy)-xcosy ?

##### 1 Answer

$y ' = \frac{\cos y - y \cdot {e}^{x y}}{x \cdot \sin y + x \cdot {e}^{x y}}$

#### Explanation:

Start from the given equation

$2 = {e}^{x y} - x \cdot \cos y$

differentiate both sides of the equation with respect to x:

$\frac{d}{\mathrm{dx}} \left(2\right) = \frac{d}{\mathrm{dx}} \left({e}^{x y} - x \cdot \cos y\right)$

$0 = {e}^{x y} \cdot \left(x y ' + y \cdot \frac{\mathrm{dx}}{\mathrm{dx}}\right) - \left(x \left(- \sin y\right) \cdot y ' + \cos y \cdot \frac{\mathrm{dx}}{\mathrm{dx}}\right)$

$0 = {e}^{x y} \cdot \left(x y '\right) + y \cdot {e}^{x y} + x \cdot \sin y \cdot y ' - \cos y$

$0 = x y ' {e}^{x y} + y \cdot {e}^{x y} + x \cdot \sin y \cdot y ' - \cos y$

solving now for first derivative y'

$\cos y - y \cdot {e}^{x y} = \left(x \cdot {e}^{x y} + x \cdot \sin y\right) \cdot y '$

dividing both sides now by $\left(x \cdot {e}^{x y} + x \cdot \sin y\right)$

final answer becomes

$y ' = \frac{\cos y - y \cdot {e}^{x y}}{x \cdot \sin y + x \cdot {e}^{x y}}$