# What is the derivative of e^y cos x = 3 + sin(xy)?

Mar 28, 2017

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

#### Explanation:

$\text{differentiate " e^ycosx" using "color(blue)"product rule}$

$\text{Given " f(x)=g(x)h(x)" then }$

$f ' \left(x\right) = g \left(x\right) h ' \left(x\right) + h \left(x\right) g ' \left(x\right)$

$\text{differentiate " sin(xy)" using "color(blue)" chain rule}$

$\text{Given " f(x)=g(h(x))" then}$

$f ' \left(x\right) = g ' \left(h \left(x\right)\right) . h ' \left(x\right)$

$\text{differentiate "color(blue)"implicitly with respect to x}$

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

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

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

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

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