Derivative Problem

  • Thread starter Lancelot59
  • Start date
  • #1
643
1
I'm unsure of my work when completing this problem:

[tex]y=e^{-x^{2}} \int^{x}_{0} e^{t^{2}} dt + c_{1}e^{-x^{2}}[/tex]

I applied the product rule to the left bit.

[tex]\frac{dy}{dx}=e^{-x^{2}} e^{x^{2}} + e^{-x^{2}}(-2x)\int^{x}_{0} e^{t^{2}} dt + (-2x)c_{1}e^{-x^{2}}[/tex]

I'm fairly certain I did this wrong.
 

Answers and Replies

  • #2
391
1
I believe that:
[tex] \int^{x}_{0} e^{t^{2}} dt [/tex]
Is a constant with respect to x, so you don't need to use the product rule, just treat it like any other constant
 
  • #3
gb7nash
Homework Helper
805
1
I believe that:
[tex] \int^{x}_{0} e^{t^{2}} dt [/tex]
Is a constant with respect to x, so you don't need to use the product rule, just treat it like any other constant

No, this is a function with respect to x, so the product rule is valid here.
 
  • #4
643
1
So I derived it correctly?
 
  • #5
Ray Vickson
Science Advisor
Homework Helper
Dearly Missed
10,706
1,722
I'm certain you did it right. Why do you think otherwise?

RGV
 
  • #6
643
1
I'm certain you did it right. Why do you think otherwise?

RGV

It's from a problem where I need to verify that it is a solution to:

[tex]y'+2xy=1[/tex]

I was concerned that the integral still containing the "t" variable wouldn't cancel, however upon re-inspection I think it should.
 

Related Threads on Derivative Problem

  • Last Post
Replies
12
Views
2K
  • Last Post
Replies
3
Views
1K
  • Last Post
Replies
2
Views
1K
  • Last Post
Replies
3
Views
936
  • Last Post
Replies
1
Views
751
  • Last Post
Replies
5
Views
2K
  • Last Post
Replies
10
Views
676
  • Last Post
Replies
2
Views
1K
  • Last Post
Replies
1
Views
1K
  • Last Post
Replies
4
Views
1K
Top