4
$\begingroup$

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.

$\endgroup$
6
$\begingroup$

$( \partial u / \partial x )^2$ is the derivative squared, while $\partial ^2 u / \partial x ^2$ is the second derivative (that's the notation for 2nd derivative).

$\endgroup$
5
$\begingroup$

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 $$ \partial^2 u / \partial x^2 = \partial / \partial x (2xy) = 2y. $$

$\endgroup$
1
$\begingroup$

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 $x$.

$\endgroup$
1
$\begingroup$

It's not.

Let $f = \partial u / \partial x $.

Then, $( \partial u / \partial x )^2 = f^2 $ (assuming $f$ returns a number) and $\partial f / \partial x = \partial ^2 u / \partial x ^2$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.