# How can I find the derivatives of the implicitly set function y=y(x), set with the equation xe^y + ye^x - e^(xy) = 0? the answer I have is y' = ye^xy -e^y -ye^x/xe^y + e^x -xe^xy

Mar 27, 2018

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

#### Explanation:

Given: $x {e}^{y} + y {e}^{x} - {e}^{x y} = 0$

Differentiate each term with respect to x:

$\frac{d \left(x {e}^{y}\right)}{\mathrm{dx}} + \frac{d \left(y {e}^{x}\right)}{\mathrm{dx}} - \frac{d \left({e}^{x y}\right)}{\mathrm{dx}} = \frac{d \left(0\right)}{\mathrm{dx}} \text{ [1]}$

The first term requires the use of the product rule:

$\frac{d \left(u v\right)}{\mathrm{dx}} = \frac{\mathrm{du}}{\mathrm{dx}} v + u \frac{\mathrm{dv}}{\mathrm{dx}}$

Where $u = x \mathmr{and} v = {e}^{y}$, then $\frac{\mathrm{du}}{\mathrm{dx}} = 1 \mathmr{and} \frac{\mathrm{dv}}{\mathrm{dx}} = \frac{d \left({e}^{y}\right)}{\mathrm{dx}}$

We need the chain rule for $\frac{d \left({e}^{y}\right)}{\mathrm{dx}}$:

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

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

Substituting into the product rule:

$\frac{d \left(x {e}^{y}\right)}{\mathrm{dx}} = {e}^{y} + x {e}^{y} \frac{\mathrm{dy}}{\mathrm{dx}} \text{ [2]}$

Substitute equation [2] into equation [1]:

${e}^{y} + x {e}^{y} \frac{\mathrm{dy}}{\mathrm{dx}} + \frac{d \left(y {e}^{x}\right)}{\mathrm{dx}} - \frac{d \left({e}^{x y}\right)}{\mathrm{dx}} = \frac{d \left(0\right)}{\mathrm{dx}} \text{ [1.1]}$

The next term, also, requires the use of the product rule:

$\frac{d \left(u v\right)}{\mathrm{dx}} = \frac{\mathrm{du}}{\mathrm{dx}} v + u \frac{\mathrm{dv}}{\mathrm{dx}}$

Where $u = y , \mathmr{and} v = {e}^{x}$, then $\frac{\mathrm{du}}{\mathrm{dx}} = \frac{\mathrm{dy}}{\mathrm{dx}} \mathmr{and}$(dv)/dx = e^x#

Substituting into the product rule:

$\frac{d \left(y {e}^{x}\right)}{\mathrm{dx}} = \frac{\mathrm{dy}}{\mathrm{dx}} {e}^{x} + y {e}^{x} \text{ [3]}$

Substitute equation [3] into equation [1.1]:

${e}^{y} + x {e}^{y} \frac{\mathrm{dy}}{\mathrm{dx}} + \frac{\mathrm{dy}}{\mathrm{dx}} {e}^{x} + y {e}^{x} - \frac{d \left({e}^{x y}\right)}{\mathrm{dx}} = \frac{d \left(0\right)}{\mathrm{dx}} \text{ [1.2]}$

If we let $u = x y$, then we can use the chain rule for the next term:

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

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

But we shall need the product rule to compute $\frac{\mathrm{du}}{\mathrm{dx}}$

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

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

Reverse the substitution for u:

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

Use the distributive property:

$\frac{d \left({e}^{x y}\right)}{\mathrm{dx}} = y {e}^{x y} + x {e}^{x y} \frac{\mathrm{dy}}{\mathrm{dx}} \text{ [4]}$

Substitute equation [4] into equation [1.2] (remember to distribute the leading -1):

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

The derivative of 0 is 0:

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

Move all of the terms that do NOT contain $\frac{\mathrm{dy}}{\mathrm{dx}}$ to the right:

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

Factor out $\frac{\mathrm{dy}}{\mathrm{dx}}$ on the left:

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

Divide both sides by $\left(x {e}^{y} + {e}^{x} - x {e}^{x y}\right)$:

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

Mar 27, 2018

$y ' = - \frac{1}{x}$

#### Explanation:

notice that $y {e}^{x}$ and ${e}^{x} y$ are same.Therefore $y {e}^{x} - {e}^{x} y = 0$

this will give you $x {e}^{y} = 0$

applying product rule $\left(f . g\right) ' = f . g ' + f ' . g$

so $y ' x {e}^{y} + {e}^{y} = 0$
rearrange the equation,
$y ' x {e}^{y} = - {e}^{y}$
$y ' = - {e}^{y} / \left(x {e}^{y}\right)$
y'=$- \frac{1}{x}$

alternatively,
$\ln \left(x {e}^{y}\right) = 0$
$\ln x + \ln {e}^{y} = 0$
$\ln x + y = 0$
$y = - \ln x$
$y ' = - \frac{1}{x}$