# What is the implicit derivative of 1= e^(xy) ?

Mar 6, 2016

$\frac{\mathrm{dy}}{\mathrm{dx}} = - \frac{y}{x}$

#### Explanation:

When we differentiate we have to use the chain rule in conjunction with the product rule.

The left had side is a constant $1$ so its derivative with respect to $x$ is $0$

For the right hand side we use the chain rule and the product rule.

${e}^{x y} \left[y + x \frac{\mathrm{dy}}{\mathrm{dx}}\right]$

So together we have

$0 = {e}^{x y} \left[y + x \frac{\mathrm{dy}}{\mathrm{dx}}\right]$

Distribute ${e}^{x y}$

$0 = y {e}^{x y} + x {e}^{x y} \frac{\mathrm{dy}}{\mathrm{dx}}$

Isolate term with $\frac{\mathrm{dy}}{\mathrm{dx}}$

$\frac{\mathrm{dy}}{\mathrm{dx}} x {e}^{x y} = - y {e}^{x y}$

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{- y {e}^{x y}}{x {e}^{x y}}$

$\frac{\mathrm{dy}}{\mathrm{dx}} = - \frac{y}{x}$

Mar 6, 2016

I like the question and the answer can be written $\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{y}{x}$, but . . .

#### Explanation:

$1 = {e}^{x y}$ implies that $x y = 0$ which in turn implies that either $x = 0$ or $y = 0$.

The graph of this equation is the pair of axes.

Here is the graph of $1 = {e}^{x y}$ using Socratic's graphing utility:

graph{1=e^(xy) [-10, 10, -5, 5]}