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

Nov 23, 2017

Differentiate each term, put each term back into the equation, and then solve for $\frac{\mathrm{dy}}{\mathrm{dx}}$

#### Explanation:

Differentiate: ${e}^{x y} + y = x - 1$

The first term:

Let $u = x y$, then $\frac{\mathrm{du}}{\mathrm{dx}} = \frac{d \left(x y\right)}{\mathrm{dx}}$

Use the product rule:

$\frac{d \left(x y\right)}{\mathrm{dx}} = \frac{d \left(x\right)}{\mathrm{dx}} y + x \frac{d \left(y\right)}{\mathrm{dx}}$

$\frac{\mathrm{du}}{\mathrm{dx}} = y + x \frac{\mathrm{dy}}{\mathrm{dx}}$

Use the chain rule:

$\frac{d \left({e}^{x y}\right)}{\mathrm{dx}} = \frac{d \left({e}^{u}\right)}{\mathrm{du}} \frac{\mathrm{du}}{\mathrm{dx}}$

$\frac{d \left({e}^{x y}\right)}{\mathrm{dx}} = {e}^{u} \frac{\mathrm{du}}{\mathrm{dx}}$

$\frac{d \left({e}^{x y}\right)}{\mathrm{dx}} = {e}^{u} \left(y + x \frac{\mathrm{dy}}{\mathrm{dx}}\right)$

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

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

The second term:

$\frac{d \left(y\right)}{\mathrm{dx}} = \frac{\mathrm{dy}}{\mathrm{dx}}$

The third term:

$\frac{d \left(x\right)}{\mathrm{dx}} = 1$

The fouth term:

$\frac{d \left(- 1\right)}{\mathrm{dx}} = 0$

Put the terms back into the equation:

$y {e}^{x y} + x {e}^{x y} \frac{\mathrm{dy}}{\mathrm{dx}} + \frac{\mathrm{dy}}{\mathrm{dx}} = 1$

Solve for $\frac{\mathrm{dy}}{\mathrm{dx}}$:

$x {e}^{x y} \frac{\mathrm{dy}}{\mathrm{dx}} + \frac{\mathrm{dy}}{\mathrm{dx}} = 1 - y {e}^{x y}$

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

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1 - y {e}^{x y}}{x {e}^{x y} + 1}$

Nov 23, 2017

See explanation

#### Explanation:

First, we differentiate both sides with respect to x:

$\frac{d}{\mathrm{dx}} {e}^{x y} + \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{d}{\mathrm{dx}} \left(x + 1\right)$
$\to \frac{d}{\mathrm{dx}} {e}^{x y} + y ' = 1$ (1)

For differentiating ${e}^{x y}$, recall that $\frac{d}{\mathrm{dx}} {e}^{u} = \frac{\mathrm{du}}{\mathrm{dx}} \cdot {e}^{u}$

With $u = x y , \frac{\mathrm{du}}{\mathrm{dx}} = \frac{d}{\mathrm{dx}} \left(x y\right)$

Using the chain rule...

$\frac{d}{\mathrm{dx}} \left(x y\right) = \frac{\mathrm{dx}}{\mathrm{dx}} \cdot y + x \cdot \frac{\mathrm{dy}}{\mathrm{dx}} = y + x y '$

Then back to (1), we get:
$\frac{d}{\mathrm{dx}} {e}^{x y} + y ' = 1 \to y {e}^{x y} + y ' x {e}^{x y} + y ' = 1$

We now move all terms without y' as a factor to the right hand side...

$\to y ' x {e}^{x y} + y ' = 1 - y {e}^{x y}$

Factor out y' from the left hand side...

$\to y ' \left(x {e}^{x y} + 1\right) = 1 - y {e}^{x y}$

And divide both sides by $\left(x {e}^{x y} + 1\right) . .$

y' = (1-ye^(xy))/(xe^(xy)+1