# What is the implicit derivative of 1= xye^y-xcos(xy) ?

Jul 16, 2016

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

#### Explanation:

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

To do these derivatives, remember $\frac{d}{\mathrm{dx}} \left(y\right) = \frac{\mathrm{dy}}{\mathrm{dx}}$ and hammer your product and chain rules.

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

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

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