# Implicit Differentiation

1. Nov 5, 2011

### JustinLiang

1. The problem statement, all variables and given/known data
Find y' in
e^(x/y)=x-y

2. The attempt at a solution
I tried to differentiate it by changing it so that there would be a natural log (as seen in my attachment). However the end result is not the same as the answer key.

How the answer key did it was they used the chain rule on e^(x/y). So...

e^(x/y)*(x/y)'=1-y'

And then they solve for y'.

Why can I not change the e^(x/y) to the logarithmic notation?

#### Attached Files:

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Last edited: Nov 6, 2011
2. Nov 5, 2011

### SammyS

Staff Emeritus
Without seeing exactly what you did, there is no way to say whether or not your method is correct.

Is it any easier to find y' by differentiating (x/y) = ln(x-y) ?

3. Nov 6, 2011

### JustinLiang

Oops, I forgot to attach it. It is harder but I was just doing it for practice, it seems like I was unable to get the correct answer... Do you know why? I attached the photo.

4. Nov 6, 2011

### SammyS

Staff Emeritus
It does look difficult to get the answer into the same form once you get rid of the exponential. Taking the derivative then gets rid of the logarithm.

You do have a mistake in your work. $\displaystyle \frac{d}{dx}\ln(x-y)=\frac{1-y'}{x-y}\,.$

Actually, it's not that difficult to compare the results. Take the book answer and replace $\displaystyle e^{x/y}$ with $x-y\,.$