Partial Derivatives of natural log

In summary, you are trying to find the partial derivatives of e with respect to x and y. The derivative of ex+y is just exey again and the derivative of ex is just ex because ey is a constant with respect to x.
  • #1
nicolette2413
8
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
  • #2
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??
 
  • #3
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.
 
  • #4
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]!
 
  • #5
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?
 
  • #6
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.)
 
  • #7
Thank you, now that you pointed out where i was wrong, it makes much more sense!:biggrin:
 

FAQ: Partial Derivatives of natural log

1. What is a partial derivative of natural log?

A partial derivative of natural log is a mathematical concept used in multivariate calculus to measure the rate of change of a function with respect to one of its variables while holding other variables constant. It is denoted by ∂ln(f)/∂x, where f is the function and x is the variable of interest.

2. How is a partial derivative of natural log calculated?

To calculate a partial derivative of natural log, you need to apply the chain rule. First, take the derivative of the natural log function, which is 1/x. Then, multiply it by the derivative of the inner function with respect to the variable of interest. Finally, simplify the expression if possible.

3. What is the significance of partial derivatives of natural log in real life?

Partial derivatives of natural log have various applications in real life, such as in economics, physics, and engineering. They are used to optimize functions and find the maximum or minimum values, which can be useful in decision-making processes and problem-solving.

4. Can partial derivatives of natural log be negative?

Yes, partial derivatives of natural log can be negative. This indicates a decreasing rate of change of the function with respect to the variable of interest. Similarly, a positive partial derivative indicates an increasing rate of change.

5. Are there any limitations to using partial derivatives of natural log?

One limitation of using partial derivatives of natural log is that they only measure the rate of change of a function in one direction. They do not provide information about the overall behavior of the function. Additionally, they can only be used for differentiable functions.

Back
Top