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

Oct 12, 2016

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

#### Explanation:

1=e^y−xcos(xy)
rArr(d1)/dx=d/dx(e^y−xcos(xy))
$\Rightarrow 0 = \frac{{\mathrm{de}}^{y}}{\mathrm{dx}} - \frac{d \left(x \cos \left(x y\right)\right)}{\mathrm{dx}}$
$\Rightarrow 0 = \left(\frac{\mathrm{dy}}{\mathrm{dx}}\right) {e}^{y} - \left(\left(\frac{\mathrm{dx}}{\mathrm{dx}}\right) \cos x y + x \frac{\mathrm{dc} o s x y}{\mathrm{dx}}\right)$
$\Rightarrow 0 = \left(\frac{\mathrm{dy}}{\mathrm{dx}}\right) {e}^{y} - \left(\cos x y + x \frac{\mathrm{dx} y}{\mathrm{dx}} \left(- \sin x y\right)\right)$
$\Rightarrow 0 = \left(\frac{\mathrm{dy}}{\mathrm{dx}}\right) {e}^{y} - \left(\cos x y + x \left(\left(y + x \frac{\mathrm{dy}}{\mathrm{dx}}\right) \left(- \sin x y\right)\right)\right)$
$\Rightarrow 0 = \left(\frac{\mathrm{dy}}{\mathrm{dx}}\right) {e}^{y} - \left(\cos x y + x \left(- y \sin x y - x \frac{\mathrm{dy}}{\mathrm{dx}} \left(\sin x y\right)\right)\right)$
$\Rightarrow 0 = \left(\frac{\mathrm{dy}}{\mathrm{dx}}\right) {e}^{y} - \left(\cos x y - x y \sin x y - {x}^{2} \frac{\mathrm{dy}}{\mathrm{dx}} \left(\sin x y\right)\right)$
$\Rightarrow 0 = \left(\frac{\mathrm{dy}}{\mathrm{dx}}\right) {e}^{y} - \cos x y + x y \sin x y + {x}^{2} \frac{\mathrm{dy}}{\mathrm{dx}} \left(\sin x y\right)$
$\Rightarrow 0 = \left(\frac{\mathrm{dy}}{\mathrm{dx}}\right) {e}^{y} + {x}^{2} \frac{\mathrm{dy}}{\mathrm{dx}} \left(\sin x y\right) - \cos x y + x y \sin x y$
$\Rightarrow 0 = \left(\frac{\mathrm{dy}}{\mathrm{dx}}\right) \left({e}^{y} + {x}^{2} \left(\sin x y\right)\right) - \cos x y + x y \sin x y$
$\Rightarrow \cos x y - x y \sin x y = \left(\frac{\mathrm{dy}}{\mathrm{dx}}\right) \left({e}^{y} + {x}^{2} \left(\sin x y\right)\right)$

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