# How do you implicitly differentiate 2=e^(x-y)-x/cosy ?

Mar 5, 2017

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

#### Explanation:

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

$0 = \left(1 - \frac{\mathrm{dy}}{\mathrm{dx}}\right) {e}^{x - y} - \frac{\cos y + x \sin y \left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)}{{\cos}^{2} y}$

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

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

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

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

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

This fraction could be manipulated more to make terms cancel (eg multiplying by the conjugate of the denominator, etc.), but I have isolated $\frac{\mathrm{dy}}{\mathrm{dx}}$.