# How do you implicitly differentiate # x cos(y) + y cos(x) = 1 ?

Nov 19, 2015

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

#### Explanation:

I'm assuming we're differentiating with respect to $x$.

$\frac{d}{\mathrm{dx}} \left[x \cos \left(y\right) + y \cos \left(x\right) = 1\right]$

Lots of use of the Product Rule and Chain Rule:
$\cos \left(y\right) - x \sin \left(y\right) \left[\frac{\mathrm{dy}}{\mathrm{dx}}\right] + \left[\frac{\mathrm{dy}}{\mathrm{dx}}\right] \cos \left(x\right) - y \sin \left(x\right) = 0$
$\left[\frac{\mathrm{dy}}{\mathrm{dx}}\right] \cos \left(x\right) - x \sin \left(y\right) \left[\frac{\mathrm{dy}}{\mathrm{dx}}\right] = y \sin \left(x\right) - \cos \left(y\right)$
$\frac{\mathrm{dy}}{\mathrm{dx}} \left(\cos \left(x\right) - x \sin \left(y\right)\right) = y \sin \left(x\right) - \cos \left(y\right)$
$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{y \sin \left(x\right) - \cos \left(y\right)}{\cos \left(x\right) - x \sin \left(y\right)}$