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

Sep 3, 2015

#### Explanation:

${e}^{y} \sin x = x + x y$

Differentiating with respect to x, differentiate terms in y with respect to y, and multiply by $\frac{\mathrm{dy}}{\mathrm{dx}}$, and differentiate terms in x as per normal.

LHS: ${e}^{y} \left(\frac{\mathrm{dy}}{\mathrm{dx}}\right) \sin x$ + ${e}^{y} \cos x$ (using product rule)
RHS: $1 + y + x \left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)$ (using product rule as well)

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

Make $\frac{\mathrm{dy}}{\mathrm{dx}}$ the subject.

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

I hope that helped!