Question

second partial derivative of z=xe^(-2y)

Answer 1

so i got e^(-2y) for Zx correct?

Answer 2

With respect to what?

Answer 3

trying to find Zx and Zy

Answer 4

does it mean that it's respect to Z?

Answer 5

Hmm your Zx looks correct.
The x drops to 1, and the constant sticks around.

Answer 6

With respect to x on the first one :o\[\Large Z_x \quad=\quad \frac{\partial Z}{\partial x}\]just in case there was any confusion on that

Answer 7

So you're finding \(Z_{xx},Z_{yy},Z_{xy},Z_{yx}\)

Answer 8

yes

Answer 9

so i got Zy=-2xe^(-2y) is this correct?

Answer 10

yes

Answer 11

ok now

Answer 12

earlier i got e^(-2y) for Zx right? now i don't know how to get Zxx ?

Answer 13

So Zx left us with something that is `constant` with respect to x correct?
So let's think about it like this,\[\Large Z_x\quad=\quad C\]\[\Large Z_{xx}\quad=\quad ?\]Derivative of a constant? :)

Answer 14

0?

Answer 15

good good good

Answer 16

oh so Zxx is just 0?

Answer 17

yes*

Answer 18

that was a weird explanation lol, so i erased it XD

Answer 19

lol ok now

Answer 20

Zy=-2xe^(-2y)

Answer 21

Zyy=?

Answer 22

i got 4xe^-2y times -2?

Answer 23

should i not multiply that -2? ( from the ^-2y )

Answer 24

4xe^-2y sounds correct. Not sure why you're getting yet ANOTHER factor of -2 though 0-O

You wrote 4xe^-2y*-2

that last -2 shouldn't be there, maybe I misunderstood you though.

Answer 25

oh no you are right haha thank you for your help ;)