# Thread: Second Order Partial Derivatives

1. ## Second Order Partial Derivatives

Just need pointing in the right direction then hopefullly I'l understand these ....

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

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

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?

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

3. thanxxx

4. 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

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 $0$.