Partial Derivatives of natural log

Click For Summary

Homework Help Overview

The discussion revolves around finding the partial derivatives of a function involving the exponential function, specifically the expression z=(e^(x+y))/(e^x+e^y). Participants are exploring their understanding of partial derivatives in the context of functions that include the exponential function.

Discussion Character

  • Exploratory, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants are attempting to clarify their understanding of partial derivatives, particularly in relation to expressions involving the exponential function. There is confusion regarding the correct application of differentiation rules and the interpretation of given derivatives.

Discussion Status

Some participants have provided guidance on the differentiation process, while others are still grappling with the concepts. There is an acknowledgment of misunderstandings in the initial posts, and attempts to clarify the problem have been made, but no consensus has been reached on the correct approach.

Contextual Notes

Participants are working under the constraints of homework rules, which may limit the amount of direct assistance they can receive. There is an emphasis on understanding the process rather than simply obtaining the correct answer.

nicolette2413
Messages
8
Reaction score
0
Hey all. I'm having some problems with the partial derivatives of e. I understand the basics such as exy2. where I'm getting confused is with the following

dz/dx=e(x+y)

and

dz/dx=1/ex+ey

Can anyone help me out with understanding these??
 
Physics news on Phys.org
nicolette2413 said:
Hey all. I'm having some problems with the partial derivatives of e. I understand the basics such as exy2. where I'm getting confused is with the following

dz/dx=e(x+y)

and

dz/dx=1/ex+ey
I don't understand what you are asking. You ask about finding the derivative but then give the derivative. Are asking for the partial derivatives of ex+y and 1/(ex+ ey or are you saying these are the derivatives and you want to find z?

Can anyone help me out with understanding these??
 
I'm sorry for being confusing. Yes I'm thrying to find the partial derivative of each item. I have no problems with partial derivatives that don't contain "e" but these I am having problems grasping.
 
Do you know what the derivative of ex is? The ordinary derivative, not partial derivative. It's the world's easiest derivative!

Do you know what the derivative of Cex is? ex+y is exey.

1/ex+ ex= e-x+ ex.

Those are both easier than [itex]e^{xy^2}[/itex]!
 
I guess I'm not asking this very well. I'm trying to solve the problem in parts but that's not helping me understand this any better. The full problem is...

z=(ex+y)/(ex+ey)

And I need to find the partial derivatives in respect to "x" and "y"
I am coming up with an answer that fully cancels out and winds up being zero when added back together.

Here's my work...

dz/dx=(ex+ey)(exey)-(exey)(ex+ey)/(ex+ey)2
dz/dy= the same as above.

I'm sure I'm missing something, but can't find where. Can you help please?
 
nicolette2413 said:
I guess I'm not asking this very well. I'm trying to solve the problem in parts but that's not helping me understand this any better. The full problem is...

z=(ex+y)/(ex+ey)
It is always a good idea to state the entire problem!

And I need to find the partial derivatives in respect to "x" and "y"
I am coming up with an answer that fully cancels out and winds up being zero when added back together.

Here's my work...

dz/dx=(ex+ey)(exey)-(exey)(ex+ey)/(ex+ey)2
dz/dy= the same as above.

I'm sure I'm missing something, but can't find where. Can you help please?
The derivative of exey with respect to x is just exey again and I think you have that in your answer. But the derivative of ex+ ey is just ex because ey is a constant with respect to x and its derivative is 0.

dz/dx=(ex+ey)(exey)
is correct
-(exey)(ex+ey)
this is wrong. You should have -(exey)(ex)
(the denominator was correct so I didn't mention it.)
 
Thank you, now that you pointed out where i was wrong, it makes much more sense!:biggrin:
 

Similar threads

Replies
5
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
Replies
5
Views
4K
  • · Replies 28 ·
Replies
28
Views
3K
Replies
4
Views
1K
  • · Replies 8 ·
Replies
8
Views
6K
  • · Replies 8 ·
Replies
8
Views
2K
Replies
2
Views
2K
  • · Replies 3 ·
Replies
3
Views
3K