Is it equal to $\partial ^2 u / \partial x ^2$?
I am trying to figure that out. I don't think it's the case, but I am just trying to make sure.
4
$\begingroup$
$\endgroup$
6
$\begingroup$
$\endgroup$
$( \partial u / \partial x )^2$ is the derivative squared, while $\partial ^2 u / \partial x ^2$ is the second derivative (that's the notation for 2nd derivative).
5
$\begingroup$
$\endgroup$
As others have mentioned, they are not the same. $$ ( \partial u / \partial x )^2 = ( \partial u / \partial x )( \partial u / \partial x ) $$ while $$ \partial^2 u / \partial x^2 = \partial / \partial x(\partial u / \partial x). $$ To show an example. If $u = f(x,y) = x^2y$, then $$ ( \partial u / \partial x )^2 = (2xy)^2 $$ and $$ \partial^2 u / \partial x^2 = \partial / \partial x (2xy) = 2y. $$
1
$\begingroup$
$\endgroup$
No, it is just the number $\partial u/\partial x$ multiplied by itself.
There's no simpler way of writing that, as long as you don't know how $u$ is a function of $x$.
answered Nov 11 '12 at 19:07
1
$\begingroup$
$\endgroup$
It's not.
Let $f = \partial u / \partial x $.
Then, $( \partial u / \partial x )^2 = f^2 $ (assuming $f$ returns a number) and $\partial f / \partial x = \partial ^2 u / \partial x ^2$.
