No, EVS, the question was stated that way, and it is 3x(partial of z wrt y at constant x) = 2y(partial of z wrt x at constant y), just to make sure i didn't type anything wrong in the file i attached.

But yeah, am I allowed to imply that z = x^2 x y^3 ?

Reply 10

DeanK2

It seems from the document that all you have been asked to show is the scanned equation holds true for a function z you are free to choose.

It looks like the statemnt if z = (x^2)(y^3) is written by you.

Why don't you try a different function for z?

Reply 11

EierVonSatan

21

Taking a random example (as the relation should hold for all 'nice' functions) f(u) = u² - 1 and let u = x^2y^3

knowing that x and y are independant (i.e. z = f(x, y) =/= 0)

Partial (dz/dx) = 4x^3y^6 and partial (dz/dy) = 6x^4y^5

so 2y(4x^3y^6) =/= 3x(6x^4y^5) according to the relation

but 3x(4x^3y^6)= 2y(6x^4y^5) ?

Reply 12

shengoc

OP

12

EierVonSatan

Taking a random example (as the relation should hold for all 'nice' functions) f(u) = u² - 1 and let u = x^2y^3

knowing that x and y are independant (i.e. z = f(x, y) =/= 0)

Partial (dz/dx) = 4x^3y^6 and partial (dz/dy) = 6x^4y^5

so 2y(4x^3y^6) =/= 3x(6x^4y^5) according to the relation

but 3x(4x^3y^6)= 2y(6x^4y^5) ?

I see your point, but i was told that what we are asked to show in this question can be done. Hmm, thanks anyway, for all your help.

Reply 13

shengoc

OP

12

EierVonSatan

Taking a random example (as the relation should hold for all 'nice' functions) f(u) = u² - 1 and let u = x^2y^3

knowing that x and y are independant (i.e. z = f(x, y) =/= 0)

Partial (dz/dx) = 4x^3y^6 and partial (dz/dy) = 6x^4y^5

so 2y(4x^3y^6) =/= 3x(6x^4y^5) according to the relation

but 3x(4x^3y^6)= 2y(6x^4y^5) ?

Yes, i managed to show what the question ask to show. Apparently, a chain rule works like you mentioned above. If I let z=f(u), u = x^2 y^3

therefore dz/dx = dz/du x du/dx

i did the same for dz/dy,

then make dz/du as subject of formula and equate the two, then i managed to show it. Does this make sense to you?

p/s - I just realised there may be errors in the question, i got 3x(partial z wrt x) in the end instead of what the question stated. hmm, i will leave it that way then, thanks a lot, EVS

Reply 14

EierVonSatan

21

shengoc

p/s - I just realised there may be errors in the question, i got 3x(partial z wrt x) in the end instead of what the question stated. hmm, i will leave it that way then, thanks a lot, EVS

You do realise that this is what I was asking you from the beginning? :rolleyes:

Reply 15

cpchem

14

shengoc

Yes, i managed to show what the question ask to show. Apparently, a chain rule works like you mentioned above. If I let z=f(u), u = x^2 y^3

therefore dz/dx = dz/du x du/dx

i did the same for dz/dy,

then make dz/du as subject of formula and equate the two, then i managed to show it. Does this make sense to you?

p/s - I just realised there may be errors in the question, i got 3x(partial z wrt x) in the end instead of what the question stated. hmm, i will leave it that way then, thanks a lot, EVS

There are bound to be errors in the question - it's Grout.

Reply 16

shengoc

OP

12

EierVonSatan

You do realise that this is what I was asking you from the beginning? :rolleyes:

no, actually i don't; i didn't have a clue how to do that until the chain rule comes up and only then, i realise that, my apologies for that, EVS.

Reply 17

shengoc

OP

12

cpchem

There are bound to be errors in the question - it's Grout.

Thanks, yeah, i checked all my workings again, and it seemed to work out logically, if the question was set otherwise. Anyway, thanks for letting me know.