1. ## Please explain a simple differentiation

I saw a manipulation that goes like this:
${{\partial E} \over {\partial A^2}} = {{\partial E} \over {2A \partial A}}$

Not familiar with this one...can anyone enlighten me?

2. Originally Posted by chopet
I saw a manipulation that goes like this:
${{\partial E} \over {\partial A^2}} = {{\partial E} \over {2A \partial A}}$

Not familiar with this one...can anyone enlighten me?
Hmmm because $\partial (A^2)=2A \partial A$ ? it works like an usual differentiation : $\frac{d}{dx}(x^2)=2x \implies d(x^2)=2xdx$

Now, if we're talking about second differentiation, we should have had $\frac{\partial^{\color{red}2}E}{\partial A^2}$, if E is considered as a function of A.

3. Hello, chopet!
I saw a manipulation that goes like this: . $\frac{\partial E}{\partial \!A^2} \:= \:\frac{\partial E}{2A\, \partial\! A}$ . . . . I bet you didn't!

Not familiar with this one...can anyone enlighten me?
It would help if you copied it correctly!

It actually says: . $\frac{\partial^2\!E}{\partial\! A^2} \;=\;\frac{\partial^2\!E}{\partial\! A\,\partial\! A}$

Let's say we have a multivariable function: . $E \;=\;f(A,B)$
. . $E$ is function of both $A$ and $B.$

Take the partial derivative with respect to $A\!:\;\;\frac{\partial E}{\partial\! A}$

Now take the partial derivative of that with respect to $A\!:\;\;\frac{\partial}{\partial\! A}\!\left(\frac{\partial E}{\partial\! A}\right)$

This can be written: . $\frac{\partial\!\cdot\partial E}{\partial\! A\cdot\partial\! A} \;=\;\boxed{\frac{\partial^2\!E}{\partial\! A\,\partial\! A}}\;\text{ or }\;\boxed{\frac{\partial^2\!E}{\partial\! A^2}}$

4. ok. I've had the equation scanned from the economics textbook dealing with Asset Pricing:

5. Originally Posted by Moo
Hmmm because $\partial (A^2)=2A \partial A$ ? it works like an usual differentiation : $\frac{d}{dx}(x^2)=2x \implies d(x^2)=2xdx$
couldn't see this at first.
thanks!