How do you implicitly differentiate -3=(xy)/(1-e^y)?

Jun 6, 2016

There is more than one answer to this question, depending on which method you use, and an interesting exercise would be to show that they are mathematically the same expression.

Explanation:

Given that $- 3 = \frac{x y}{1 - {e}^{y}}$, I assume the question asks for $\frac{\mathrm{dy}}{\mathrm{dx}}$ to be expressed in terms of $x$ and $y$, and that we are differentiating with respect to $x$. This works in principle if we differentiate with respect to $y$.

There are two methods to solve this.

Method 1: Brute force

$- 3 = \frac{x y}{1 - {e}^{y}}$

Differentiate with respect to $x$ on both sides:

$0 = \frac{\left(y + x \frac{\mathrm{dy}}{\mathrm{dx}}\right) \left(1 - {e}^{y}\right) - \left(x y\right) \left(- {e}^{y} \frac{\mathrm{dy}}{\mathrm{dx}}\right)}{1 - {e}^{y}} ^ 2$
$0 = \frac{y - y {e}^{y} + x \frac{\mathrm{dy}}{\mathrm{dx}} - x {e}^{y} \frac{\mathrm{dy}}{\mathrm{dx}} + x y {e}^{y} \frac{\mathrm{dy}}{\mathrm{dx}}}{1 - {e}^{y}} ^ 2$
$0 = \frac{y - y {e}^{y} + \frac{\mathrm{dy}}{\mathrm{dx}} \left(x - x {e}^{y} + x y {e}^{y}\right)}{1 - {e}^{y}} ^ 2$
$0 = y - y {e}^{y} + \frac{\mathrm{dy}}{\mathrm{dx}} \left(x - x {e}^{y} + x y {e}^{y}\right)$
$x \frac{\mathrm{dy}}{\mathrm{dx}} \left(1 - {e}^{y} + y {e}^{y}\right) = y \left({e}^{y} - 1\right)$
$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{y \left({e}^{y} - 1\right)}{x \left(1 - {e}^{y} + y {e}^{y}\right)}$ ... 

Method 2: Simplify before differentiating

$- 3 = \frac{x y}{1 - {e}^{y}}$
$3 \left({e}^{y} - 1\right) = x y$

Differentiate with respect to $x$ on both sides:
$3 {e}^{y} \frac{\mathrm{dy}}{\mathrm{dx}} = y + x \frac{\mathrm{dy}}{\mathrm{dx}}$
$3 {e}^{y} \frac{\mathrm{dy}}{\mathrm{dx}} - x \frac{\mathrm{dy}}{\mathrm{dx}} = y$
$\frac{\mathrm{dy}}{\mathrm{dx}} \left(3 {e}^{y} - x\right) = y$
$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{y}{3 {e}^{y} - x}$ ... 

For additional practice, show that $x = \frac{3 \left({e}^{y} - 1\right)}{y}$ and use this substitution to simplify both results,  and , into the expression $\frac{\mathrm{dy}}{\mathrm{dx}} = {y}^{2} / \left(3 \left(1 - {e}^{y} + y {e}^{y}\right)\right)$