Just a quick question about partial derivatives

  • Thread starter vorcil
  • Start date
  • #1
395
0
Not a homework question, but It will help me none the less,

In my book it says

[tex] \frac{d}{dt} \int_{-\infty}^{\infty} |\Psi(x,t)|^2 dx [/tex]

is equivalent to

[tex] \int_{-\infty}^{\infty} \frac{\partial}{\partial t}|\Psi(x,t)|^2 dx [/tex]

I understand how It becomes a partial derivative, since I'm differentiating psi with respect to x and t

But what I don't understand Is how you can just move [tex] \frac{d}{dt} [/tex]
into an integral, why can you do that? I don't understand the rules behind it,

for example
(i'm just making this up)
[tex] \frac{d}{dt} \int_0^2 2xt dx [/tex]
I'm first going to integrate the inside function then differentiate it with respect to t
before I try the other way of moving dt inside the integral

[tex] \int_0^2 2xt dx = \left \frac{2x^2t}{2} \right|_0^2 [/tex]
the equation becomes
[tex] \left x^2t\right|_0^2 [/tex]
putting in 2, I get 2^2 *t
= 4t
differentiating that,
[tex] \frac{d}{dt} 4t = 4 [/tex]

now i'll try the other way by moving in the partial derivative

[tex] \int_{-\infty}^{\infty} \frac{\partial}{\partial t} 2xt dx [/tex]
I've only done one math paper before, and 5 physics papers,
so I don't know how to solve partial derivatives in the conventional sense,
but what I have picked up is, that a physicists first line of attack to any partial derivative, is separation of variables,
so uh, i'm not too sure how i do it in this context, since I can't separate x,t the way I can separate [tex] \Psi , \textrm{ which I can rewrite as } \psi(x)\phi(x) [/tex]
 

Answers and Replies

  • #2
395
0
[tex] \int_{-\infty}^{\infty} \frac{\partial}{\partial t} 2xt dx, = 2x [/tex]

that's from differentiating 2xt, treating x as a constant, d/dt 2t = 2, so the t disappears

[tex] \int_0^2 2x dx = \int_0^2 \frac{2x^2}{2} dx = \left x^2\right|_0^2 [/tex]

so 2^2 = 4

so i've just proved for the equation 2xy

[tex] \frac{d}{dy} \int_a^b F(x,y) dx \textrm{ is equivalent to } \int_a^b \frac{\partial}{\partial y} F(x,y) dx [/tex]

meh

Is this true for all equations?
 
  • #3
D H
Staff Emeritus
Science Advisor
Insights Author
15,415
687
But what I don't understand Is how you can just move [tex] \frac{d}{dt} [/tex]
into an integral, why can you do that? I don't understand the rules behind it,
This is an application of the general concept of "differentiation under the integral sign" or the Leibiz Integral Rule. http://mathworld.wolfram.com/LeibnizIntegralRule.html.
 
  • #4
D H
Staff Emeritus
Science Advisor
Insights Author
15,415
687
[tex] \frac{d}{dy} \int_a^b F(x,y) dx \textrm{ is equivalent to } \int_a^b \frac{\partial}{\partial y} F(x,y) dx [/tex]

meh

Is this true for all equations?

If a and b are constants, yes. This is essentially the fundamental theorem of calculus.

If either a or b is a function of y (or if both are a function of y) then the above is not true. You need to use the more generic form of the Liebniz Integral Rule to compute derivative.
 

Related Threads on Just a quick question about partial derivatives

  • Last Post
Replies
1
Views
588
Replies
2
Views
1K
Replies
2
Views
1K
Replies
9
Views
735
Replies
3
Views
3K
  • Last Post
Replies
1
Views
1K
Replies
14
Views
1K
  • Last Post
Replies
1
Views
2K
  • Last Post
Replies
6
Views
2K
  • Last Post
Replies
3
Views
803
Top