I saw a manipulation that goes like this:
$\displaystyle {{\partial E} \over {\partial A^2}} = {{\partial E} \over {2A \partial A}} $
Not familiar with this one...can anyone enlighten me?
Hmmm because $\displaystyle \partial (A^2)=2A \partial A$ ? it works like an usual differentiation : $\displaystyle \frac{d}{dx}(x^2)=2x \implies d(x^2)=2xdx$
Now, if we're talking about second differentiation, we should have had $\displaystyle \frac{\partial^{\color{red}2}E}{\partial A^2}$, if E is considered as a function of A.
Hello, chopet!
It would help if you copied it correctly!I saw a manipulation that goes like this: .$\displaystyle \frac{\partial E}{\partial \!A^2} \:= \:\frac{\partial E}{2A\, \partial\! A}$ . . . . I bet you didn't!
Not familiar with this one...can anyone enlighten me?
It actually says: .$\displaystyle \frac{\partial^2\!E}{\partial\! A^2} \;=\;\frac{\partial^2\!E}{\partial\! A\,\partial\! A}$
Let's say we have a multivariable function: .$\displaystyle E \;=\;f(A,B)$
. . $\displaystyle E$ is function of both $\displaystyle A$ and $\displaystyle B.$
Take the partial derivative with respect to $\displaystyle A\!:\;\;\frac{\partial E}{\partial\! A}$
Now take the partial derivative of that with respect to $\displaystyle A\!:\;\;\frac{\partial}{\partial\! A}\!\left(\frac{\partial E}{\partial\! A}\right)$
This can be written: .$\displaystyle \frac{\partial\!\cdot\partial E}{\partial\! A\cdot\partial\! A} \;=\;\boxed{\frac{\partial^2\!E}{\partial\! A\,\partial\! A}}\;\text{ or }\;\boxed{\frac{\partial^2\!E}{\partial\! A^2}}$