# How do you implicitly differentiate 2= e^(xy^2-x^3y)-y^2x^3+y ?

Apr 22, 2018

y' = -(y(3x^2ye^(x^3y)+3xe^(xy^2)-ye^(xy^2)))/(2x^3ye^(x^3y)-e^(x^3y)-2xye^(xy^2)+x^3e^(xy^2)

#### Explanation:

Given: $2 = {e}^{x {y}^{2} - {x}^{3} y} - {y}^{2} {x}^{3} + y$

Derivative rules:
$\left(k\right) ' = 0$
$\left({e}^{u}\right) ' = u ' {e}^{u}$
Product rule: $\left(u v\right) ' = u v ' + v u '$

Use the product rule twice for $e$:
Let $u = x {y}^{2} - {x}^{3} y$

$u ' = 2 x y y ' + {y}^{2} - {x}^{3} y ' - 3 {x}^{2} y$

$\left({e}^{u}\right) ' = \left(2 x y y ' + {y}^{2} - {x}^{3} y ' - 3 {x}^{2} y\right) {e}^{x {y}^{2} - {x}^{3} y}$

$\left(- {y}^{2} {x}^{3}\right) ' = - 3 {x}^{2} {y}^{2} - 2 {x}^{3} y y '$

Put it all together
$0 = \left(2 x y y ' + {y}^{2} - {x}^{3} y ' - 3 {x}^{2} y\right) {e}^{x {y}^{2} - {x}^{3} y} - 3 {x}^{2} {y}^{2} - 2 {x}^{3} y y ' + y '$

Distribute the $e$:

$0 = 2 x y y ' {e}^{x {y}^{2} - {x}^{3} y} + {y}^{2} {e}^{x {y}^{2} - {x}^{3} y} - {x}^{3} y ' {e}^{x {y}^{2} - {x}^{3} y} - 3 {x}^{2} y {e}^{x {y}^{2} - {x}^{3} y} - 3 {x}^{2} {y}^{2} - 2 {x}^{3} y y ' + y '$

Move all the $y '$ terms to the left side:
$- 2 x y y ' {e}^{x {y}^{2} - {x}^{3} y} + {x}^{3} y ' {e}^{x {y}^{2} - {x}^{3} y} + 2 {x}^{3} y y ' - y ' = {y}^{2} {e}^{x {y}^{2} - {x}^{3} y} - 3 {x}^{2} y {e}^{x {y}^{2} - {x}^{3} y} - 3 {x}^{2} {y}^{2}$

Factor the $y '$:
$y ' \left(- 2 x y {e}^{x {y}^{2} - {x}^{3} y} + {x}^{3} {e}^{x {y}^{2} - {x}^{3} y} + 2 {x}^{3} y - 1\right) = {y}^{2} {e}^{x {y}^{2} - {x}^{3} y} - 3 {x}^{2} y {e}^{x {y}^{2} - {x}^{3} y} - 3 {x}^{2} {y}^{2}$

Divide:

$y ' = \frac{{y}^{2} {e}^{x {y}^{2} - {x}^{3} y} - 3 {x}^{2} y {e}^{x {y}^{2} - {x}^{3} y} - 3 {x}^{2} {y}^{2}}{\left(- 2 x y {e}^{x {y}^{2} - {x}^{3} y} + {x}^{3} {e}^{x {y}^{2} - {x}^{3} y} + 2 {x}^{3} y - 1\right)}$

Factor out a negative from the numerator and rearrange:
$y ' = - \frac{3 {x}^{2} {y}^{2} + 3 {x}^{2} y {e}^{x {y}^{2} - {x}^{3} y} - {y}^{2} {e}^{x {y}^{2} - {x}^{3} y}}{2 {x}^{3} y - 1 - 2 x y {e}^{x {y}^{2} - {x}^{3} y} + {x}^{3} {e}^{x {y}^{2} - {x}^{3} y}}$

Make ${e}^{x {y}^{2} - {x}^{3} y} = {e}^{x {y}^{2}} {e}^{- {x}^{3} y}$

$y ' = - \frac{3 {x}^{2} {y}^{2} + 3 {x}^{2} y {e}^{x {y}^{2}} {e}^{- {x}^{3} y} - {y}^{2} {e}^{x {y}^{2}} {e}^{- {x}^{3} y}}{2 {x}^{3} y - 1 - 2 x y {e}^{x {y}^{2}} {e}^{- {x}^{3} y} + {x}^{3} {e}^{x {y}^{2}} {e}^{- {x}^{3} y}}$

Factor $y {e}^{- {x}^{3} y}$ from both numerator and ${e}^{- {x}^{3} y}$ from the denominator.
Realize that ${e}^{- {x}^{3} y} {e}^{{x}^{3} y} = {e}^{0} = 1$

$y ' = - \frac{y {e}^{- {x}^{3} y} \left(3 {x}^{2} y {e}^{{x}^{3} y} + 3 x {e}^{x {y}^{2}} - y {e}^{x {y}^{2}}\right)}{{e}^{- {x}^{3} y} \left(2 {x}^{3} y {e}^{{x}^{3} y} - {e}^{{x}^{3} y} - 2 x y {e}^{x {y}^{2}} + {x}^{3} {e}^{x {y}^{2}}\right)}$

Cancel common factors:
y' = -(y(3x^2ye^(x^3y)+3xe^(xy^2)-ye^(xy^2)))/(2x^3ye^(x^3y)-e^(x^3y)-2xye^(xy^2)+x^3e^(xy^2)