What does d^2u/dx^2 notation mean in partial derivatives?

[mr_coffee]

Hello everyone I have no idea how to start this problem because I'm confused on the notation, what does it mean?
here is a picture:
http://img291.imageshack.us/img291/1177/lastscan2lc.jpg
I know how to take partial derivatives, but the d^2 part is confusing and the dx^2? what the!

 

[FrogPad]

The "d^2/dt^2" part means it is a second order derivative. It basically means how many times you take that derivative.

So when you have: $$\frac{d^1 x^2}{dx^1}$$ then this means... take the derivative of x^2 one time. So you get 2x.

Now if you you had $$\frac{d^2 x^2}{dx^2}$$ then this means you take the derivative twice. So in TI-89 syntax you would have:

d(d(x^2,x),x)

which equals 2.

 

[TD]

It may be a but confusing at first, but you'll get used to it. Note that the "square" is at the 'd'-sign in the numerator and above the x (or any other variable) in the denominator. Of course, we still mean the variable x, and not x². In the nominator, it still has to be clear that we're differentiating f, and not f².

So (I'm using normal derivatives here, not partials, but the notation is similar)
$$\frac{{d^2 f}}{{dx^2 }} = \frac{d}{{dx}}\left( {\frac{{df}}{{dx}}} \right)$$

But watch out, not one of the following:
$$\frac{{df^2 }}<br /> {{dx^2 }},\frac{{d^2 f}}<br /> {{d^2 x}},\frac{{df^2 }}<br /> {{d^2 x}}$$