# Just a quick question about partial derivatives

1. Apr 29, 2010

### vorcil

Not a homework question, but It will help me none the less,

In my book it says

$$\frac{d}{dt} \int_{-\infty}^{\infty} |\Psi(x,t)|^2 dx$$

is equivalent to

$$\int_{-\infty}^{\infty} \frac{\partial}{\partial t}|\Psi(x,t)|^2 dx$$

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 $$\frac{d}{dt}$$
into an integral, why can you do that? I don't understand the rules behind it,

for example
(i'm just making this up)
$$\frac{d}{dt} \int_0^2 2xt dx$$
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

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

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

$$\int_{-\infty}^{\infty} \frac{\partial}{\partial t} 2xt dx$$
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 $$\Psi , \textrm{ which I can rewrite as } \psi(x)\phi(x)$$

2. Apr 29, 2010

### vorcil

$$\int_{-\infty}^{\infty} \frac{\partial}{\partial t} 2xt dx, = 2x$$

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

$$\int_0^2 2x dx = \int_0^2 \frac{2x^2}{2} dx = \left x^2\right|_0^2$$

so 2^2 = 4

so i've just proved for the equation 2xy

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

meh

Is this true for all equations?

3. Apr 29, 2010

### D H

Staff Emeritus
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. Apr 29, 2010

### D H

Staff Emeritus
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.