Question

Please explain me how to solve this problem?

Answer 1

\[\rm if~ x^xy^yz^z=C,show~that~at~x=y=z,~\\~\frac{\partial^2z}{\partial x~\partial y}=[xlogex]^{-1} \]

Answer 2

First i'll take log on both sides
so what i'll get is : \[\rm xlogx+ylogy+zlogz=logC\]

Answer 3

that is a good start

Answer 4

yep
but after that im struggling

Answer 5

how about the next step?
i need \(z_y\)

Answer 6

\[now~diff~w.r.ty~treating~x~as~constant~\\1+logy+(1+logz)\frac{\partial z}{\partial y}=0\\~\frac{\partial z}{\partial y}=-\frac{(1+logy)}{(1+\log z)}\]

Answer 7

or else if i diff the above equation again wrt x lets see what i get..

Answer 8

\[\rm So ~i~get~the~following~:\\ (1+logz)\frac{\partial^2 z}{\partial x~\partial y }+\frac{\partial z}{\partial y} \frac{\partial z}{\partial x}\frac{1}{z}=0\]

Answer 9

Looks good

Answer 10

substitute the values \(\dfrac{\partial z}{\partial y}\) and \(\dfrac{\partial z}{\partial x}\)

Answer 11

hey wait i think i got it!
\[\rm \frac{\partial z}{\partial y}= -\frac{(1+logy)}{(1+logz)} \\ \& \frac{\partial z}{\partial x}= -\frac{(1+logx)}{(1+logz)} \\~putting~x=y=z~\\~\\(1+logx)\frac{\partial^2 z}{\partial x \partial y }+ \frac{1}{x} \cdot \frac{-(1+logx)}{(1+logx)}~\cdot~\frac{-(1+logx)}{(1+logx)}=0~\\~\frac{\partial^2 z }{\partial x \partial y}=-\frac{1}{(x[1+logx])}\\~~~~~~~~~~=-\frac{1}{(xlog(ex))}\\~~~~~~~~~~=-[xlog(ex)]^{-1}\]

Answer 12

Yay!!
Thanks :)

Answer 13

You just wanted medals! Haha, just playing nice work!

Answer 14

-.-
i did not understand it
wanted help
@ganeshie8 helped me

Answer 15

worried about that negative sign :(
its -(xlogex)^{-1}

Answer 16

yes it has to be negative

Answer 17

ah
so i suppose then theres a printing mistake
Anyways Thanks @ganeshie8 :)