Question

I have no idea how to do this sum :/
Please help!

Answer 1

\[\rm If~\frac{x^2}{a+u}+\frac{y^2}{b+u}=1~\\~\\ ~~\\~u^2_x+u^2_y=2(xu_x+yu_y)\]

Answer 2

And those are partials?

Answer 3

yes
\(\rm u_x=\Large \frac{\partial u}{\partial x}\)

Answer 4

Gotcha

Answer 5

So what exactly do you want to do xD

Answer 6

i have to prove that result

Answer 7

Though so, ok did you try to take the partial derivative of your expression haha?

Answer 8

im not understanding what to do ;~;

Answer 9

What happens if you take partial derivatives for both x and y for \[\frac{ x^2 }{ a+u }+\frac{ y^2 }{ b+u }=1\]

Answer 10

hmm..?

Answer 11

@ganeshie8 please explain me

Answer 12

\[\rm \frac{x^2}{a+u}+\frac{y^2}{b+u}=1\]

taking differential gives
\[\rm \frac{2x}{a+u}dx+\frac{2y}{b+u}dy - \left(\dfrac{x^2}{(a+u)^2}+\dfrac{y^2}{(b+u)^2}\right)dz=0\]

Answer 13

wait what did you do?

Answer 14

is it total differentiation?
then i dont know about it

Answer 15

okay, then you will need to find the partials separately and plug them into the given equaiton i guess

Answer 16

\[\frac{ x^2 }{ a+u }+\frac{ y^2 }{ b+u }=1\]

treat \(y\) as constant and find \(u_x\)

Answer 17

so it will be :
wait im confused
;~;

Answer 18

its going to be a painful algebra
i hope there is some other clever way to work this

Answer 19

;~;

Answer 20

@zepdrix @mukushla @Kainui @bibby @baru

Answer 21

wait what's u?

Answer 22

that is the question
nothing more is provided
i suppose f(x,y,u)

Answer 23

u is a function of x n y

Answer 24

How many questions is there on that review/or paper you're doing I think I know that and I can help you! Through em

Answer 25

hmm..?
@jerry101 i did not understand you
could you please explain me

Answer 26

nvm lol

Answer 27

:/

Answer 28

not every question is a sum.... its only used for addition.... rather call it a question....

Answer 29

so how would i prove it?

Answer 30

take partial derivatives separately... take the left hand side and proceed for right hand side..

Answer 31

Thats what m struggling with :/

Answer 32

If we compute the first partial derivatives of the starting equation, we get, after a simplification, this condition:
\[\Large \frac{{x{u_y}}}{{a + u}} = \frac{{y{u_x}}}{{b + u}}\]