1. Implicit differentiation help

I need help with yet another implicit differentiation
e^x/y=x-y
I've tried

((1/y)(e^(x/y))((y-xy')/(y^2)))=1-y'

(e^(x/y))-1=(-y'y^2)/(y-xy')
What am I doing wrong? I don't see a way

2. Originally Posted by ppark
I need help with yet another implicit differentiation
e^x/y=x-y
I've tried

((1/y)(e^(x/y))((y-xy')/(y^2)))=1-y'

(e^(x/y))-1=(-y'y^2)/(y-xy')
What am I doing wrong? I don't see a way
I presume this is
$\displaystyle \displaystyle e^{x/y} = x + y$

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

$\displaystyle \displaystyle e^{x/y} \cdot \frac{1}{y} - e^{x/y} \cdot \frac{x}{y^2} \cdot y' = 1 + y'$

Can you take it from there?

-Dan

3. Originally Posted by topsquark
I presume this is
$\displaystyle \displaystyle e^{x/y} = x + y$

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

$\displaystyle \displaystyle e^{x/y} \cdot \frac{1}{y} - e^{x/y} \cdot \frac{x}{y^2} \cdot y' = 1 + y'$

Can you take it from there?

-Dan
Does the 1/y on e^(x/y) not need to come down when taking the derivative of e

EX: d/dx e^(x/y)= 1/y e^(x/y)

4. Originally Posted by ppark
Does the 1/y on e^(x/y) not need to come down when taking the derivative of e

EX: d/dx e^(x/y)= 1/y e^(x/y)
It's there. Try it using the chain rule and put u = x/y. Then
$\displaystyle \displaystyle \frac{d}{dx} e^{x/y} = \frac{d}{dx}e^u = e^u \cdot \frac{d}{dx}u$

$\displaystyle \displaystyle \frac{d}{dx}u = \frac{d}{dx}\left ( \frac{x}{y} \right ) = \frac{1}{y} - \frac{x}{y^2}y'$ <--- This is correct version of the term you wanted to "bring down."

So we get
$\displaystyle \displaystyle \frac{d}{dx}e^{x/y} = e^{x/y} \left ( \frac{1}{y} - \frac{x}{y^2}y' \right )$
as before.

-Dan