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

Jun 28, 2017

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

#### Explanation:

Write the equation as:

$x {e}^{y} = 1 + \sin \left(x y\right)$

and differentiate both sides with respect to $x$.

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

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

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

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

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

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

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

and substituting $x {e}^{y}$ from the original equation:

$y ' = \frac{y \cos \left(x y\right) - {e}^{y}}{1 + \sin \left(x y\right) - x \cos \left(x y\right)}$