# How do you differentiate e^y cos(x) = 6 + sin(xy)?

##### 1 Answer
Apr 17, 2018

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

#### Explanation:

Here,

${e}^{y} \cos x = 6 + \sin \left(x y\right)$

Diff.w.r.t. $x$,$\text{using "color(blue)"product and chain rule}$

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

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

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

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

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

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