Partial Differentiation

Main Question or Discussion Point

f(x,y)=2e^(x^2y), then fx(0,1) = ?

fx = 4xye^(x^2y)
=4(0)(1)e^(0)(1)
=e^0
=1?

I'm told that I'm getting this question wrong but don't know how. Can anybody please help show me what I'm doing wrong on this one?
 

Answers and Replies

Gib Z
Homework Helper
3,344
4
You are doing partial differentiation with respect to x, so you can treat y as a constant.

It's just simply using the chain rule and [itex](e^x)' = e^x[/itex].

If you do that correctly you should get up to [itex] 2 \cdot ( 2y \cdot x^{2y-1} ) \exp (x^{2y})[/itex].

The x term is a nice fat zero, giving the final result of 0.
 
Crap! I see now. Thanks Gib.
 
HallsofIvy
Science Advisor
Homework Helper
41,738
897
f(x,y)=2e^(x^2y), then fx(0,1) = ?

fx = 4xye^(x^2y)
=4(0)(1)e^(0)(1)
=e^0
=1?

I'm told that I'm getting this question wrong but don't know how. Can anybody please help show me what I'm doing wrong on this one?
Is that
[tex]2e^{x^{2y}}[/tex]
as Gib Z assumed or
[tex]2e^{x^2y}[/tex]
which is what I would assume?
If it is the latter then
[tex]f_x= 4xye^{x^2y}[/tex]
as you have. Then [itex]f_x(0,1)= 4(0)(1)e^{(0)(1)}= 0[/itex], not 1, because of the "0" multiplying the exponential.
 
Gib Z
Homework Helper
3,344
4
Damn i just realised that :( Looking at his working it looks like you're right Halls =] Sorry guys.
 
It was the latter. But I realized what I was doing wrong from Gib's reply. Thanks guys!
 

Related Threads for: Partial Differentiation

  • Last Post
Replies
6
Views
645
  • Last Post
Replies
5
Views
3K
  • Last Post
Replies
15
Views
28K
  • Last Post
Replies
10
Views
751
Replies
1
Views
1K
Replies
3
Views
3K
  • Last Post
Replies
7
Views
754
Top