Implicit Differentiation: How Can I Solve for y' in e^(x/y)=x-y?

[JustinLiang]

Homework Statement

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?

 

[SammyS]

Homework Statement

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?

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) ?

 

[JustinLiang]

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) ?

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.

 

[SammyS]

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.

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. $\displaystyle \frac{d}{dx}\ln(x-y)=\frac{1-y'}{x-y}\,.$

Added in Edit:

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

 

[JustinLiang]

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. $\displaystyle \frac{d}{dx}\ln(x-y)=\frac{1-y'}{x-y}\,.$

Added in Edit:

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

Wow... I need some sleep haha. Thanks.