# Second Order Partial Derivatives

• July 2nd 2010, 03:48 AM
DwightHoward
Second Order Partial Derivatives
Just need pointing in the right direction then hopefullly I'l understand these (Headbang)....

Find 1st and 2nd Order Partial Derivatives:

z = 3x² - 2y⁴

First Order
= 2x
= -8y³

Now I know there should be 4 second order Derivatives and i remember how to work the first two out which are:

= 2
= -24y²

However I become stuck on working out the last two which I believe should be the same.

I know what they are they are the Zxy and Zyx and I know they are the same but could someone explain how i calculate these thanks.

�I know what theI kk
• July 2nd 2010, 03:54 AM
HallsofIvy
Quote:

Originally Posted by DwightHoward
Just need pointing in the right direction then hopefullly I'l understand these (Headbang)....

Find 1st and 2nd Order Partial Derivatives:

z = 3x² - 2y⁴

First Order
= 2x
= -8y³

Better to write explicitely that $z_x=6x$ (NOT 2x) and $z_y= -8y^3$.
$z_{xx}= (z_x)_x= 6$ (NOT 2) and $z_{yy}= (z_y)_y= -24y^2$/

$z_{xy}= (z_x)_y$. What is the derivative of 6x with respect to y?
$z_{yx}= (z_y)_x$. What is the derivative of [itex]-8y^3[itex] with respect to x?

Quote:

Now I know there should be 4 second order Derivatives and i remember how to work the first two out which are:

= 2
= -24y²

However I become stuck on working out the last two which I believe should be the same.

I know what they are they are the Zxy and Zyx and I know they are the same but could someone explain how i calculate these thanks.

�I know what theI kk
• July 2nd 2010, 04:04 AM
DwightHoward
thanxxx
• July 2nd 2010, 04:50 AM
drumist
Quote:

Originally Posted by DwightHoward
Firstly just realised myself i forgot to times by 3 but thats no problem.

Secondly am I rite to assume 6x cannot be differentiated with respect to y and is therefore 0?

If not would you please explain how it is calculated

Thanks for your help

Yes, your answer is correct. However, the correct wording would be to say that $6x$ does not depend on $y$, and therefore $z_{xy}=0$. Just for clarification: $6x$ can be differentiated with respect to $y$, but that differentiation gives a result of $0$.