Please explain a simple differentiation

**chopet** · August 22nd 2008, 09:07 AM

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?

**Moo** · August 22nd 2008, 09:41 AM

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.

**Soroban** · August 22nd 2008, 10:52 AM

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

**chopet** · August 23rd 2008, 07:45 AM

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

**chopet** · August 23rd 2008, 07:49 AM

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!

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