# Second Order Partial Derivatives

• Jul 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
• Jul 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 \$\displaystyle z_x=6x\$ (NOT 2x) and \$\displaystyle z_y= -8y^3\$.
\$\displaystyle z_{xx}= (z_x)_x= 6\$ (NOT 2) and \$\displaystyle z_{yy}= (z_y)_y= -24y^2\$/

\$\displaystyle z_{xy}= (z_x)_y\$. What is the derivative of 6x with respect to y?
\$\displaystyle 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
• Jul 2nd 2010, 04:04 AM
DwightHoward
thanxxx
• Jul 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