Question

find the second partial derivative of z = y cos x
So far I know that
dz/dx is -sinx
dz/dy, here im unsure, is it -y sin x??

Answer 1

@OOOPS

Answer 2

Nope, partial derivative w.r.t x, then y is constant,
\[\dfrac{dz}{dx}=-ysinx\]

Answer 3

\[\dfrac{dz}{dy}=cosx\]

Answer 4

got me??

Answer 5

yeah I think so, I will try for the second derative.

Answer 6

you mean \(\dfrac{d^2z}{dx^2}\)?? second partial derivative??

Answer 7

(dz/dx)^2 = -ycos x?
(dz/dy)^2 = -sin x?

Answer 8

yes

Answer 9

\[\dfrac{d^2z}{dx^2}= -ycosx\]
but \(\dfrac{d^2z}{dy^2}=0\) because to y, cos x is constant and (cosx)' =0

Answer 10

oh okey, in my book it says
-ycosx, -sin x and 0

Answer 11

One more thing, pay attention at NOTATION, it is \(\dfrac{d^2z}{dx^2}\), NOT \((\dfrac{dz}{dx})^2\) they are different!!!

Answer 12

You know why??

Answer 13

oh i write the d^2z/dx^2 in my book but here I write like the wrong way, no I dont know why? please tell me:)

Answer 14

\(\dfrac{dz}{dx}= -ysinx\rightarrow \dfrac{d^2z}{dx^2}= -ycosx\\while~~ (\dfrac{dz}{dx})^2 =(-ysinx)^2=y^2sin^2x\)
See??

Answer 15

oh, I didn't even realise that, thanx for making such equation.
Btw where in the US did u study?

Answer 16

yup

Answer 17

So where di U study in USA?

Answer 18

not did, do . I am currently student in the U.S

Answer 19

oh which university?

Answer 20

Temple university

Answer 21

Hmm, never heard of it, where in US is it?

Answer 22

http://temple.edu/

Answer 23

cool it looks like a good university:)

Answer 24

I don't know. I apply to that school because it is my neighbor.

Answer 25

well atleast the education system is good there?

Answer 26

yes,

Answer 27

o'well, thanx for the time and help:)

Answer 28

:)