# How do you differentiate (e^y)(cos(x))=1+sin(xy)?

May 11, 2018

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

#### Explanation:

Given that, ${e}^{y} \cos x = 1 + \sin \left(x y\right)$.

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

Applying the Product Rule, and, the Chain Rule, we have,

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

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

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

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

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

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

as desired!