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anonymous
 5 years ago
fxy if f(x,y) = (xy)^(1/2)1/(x^2*y)
anonymous
 5 years ago
fxy if f(x,y) = (xy)^(1/2)1/(x^2*y)

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0what would you want to do?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ok, so you are finding \[f_{xy}\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0its just really long wanted to see someone else do it lol

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0derivate everything to respect x and then derivate that answer to respect y

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0let me rewrite the eqn first, it's tough to see it (xy)^(1/2)1/(x^2*y) \[{(xy)^{1/2} \over x^2y}\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0already know that brain but thank you

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ill write what i have first

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ill write what i have first\[.5(xy)^{1/2}y+(x ^{2}y)^{2}2xy\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0hmm.. it's going to be a quotien rule with two product rules. sorry but i'm kinda lazy to actualy write it

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0quotient? i just did two product rules...

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0what you should notice first is that the equation is equivalent to \[\sqrt{1 \over yx^3}\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0this is a lot easier to find the derivatives

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i do chain rules and product, u dont have to use quotient

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Mini, just to let you know the quotient rule makes these problems far more faster, although I know that it is a pain. so I will just do this one with quotient rule and you try to confirm it with product rule, ok?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0well product and quotient are the same thing though lol, its like comparing addition of negatives and subtraction

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0you are missing a 1 on the top part > (xy)^(1/2)1/(x^2*y): \[(\sqrt{xy}1)\div(x^{2}y)\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0to find f_x, I will write f as follows \[{1 \over \sqrt y} * {1 \over x^{3/2}}\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0dont see what im missing i might of had a typo but my dx is + to the neg 2

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Brain, nice catch, I was going to solve the problem not knowing that lol

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0haha yea, sorry i was trying to type on the "Equation" thing it took me a while lol

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0then f = \[\sqrt{xy}  {1 \over x^2y}\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0yuki i honestly have no idea what you are doing, im changing everything to exponents then doing chain and products, it might be "slower" but its faster for myself

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i mean chain is pretty simple, bring to the front, copy, minus 2, der of inside

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0then f_x is \[{\sqrt{y} \over 2\sqrt{x}}+{2 \over yx^3 }\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0yeah thats not what i got .5(xy)^(1/2)y

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0well the first time with respect to x

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0now f_xy is \[{1 \over 4\sqrt{xy}}  {2 \over x^3y^2}\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ok. Mini, what i got so far was: Fx = \[(2(\sqrt{xy}1))\div(x^{3}y)\] Fxy = \[(2\times(\sqrt{xy}1))\div(y^{2}x ^{3})\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[.5(xy)^{1/2}y+(x ^{2}y)^{2}2xy\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I am 100% sure with my answer

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I heard product rule, but you don't have to use it anywhere. at least I didn't

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0for respect to y\[.25(xy)^{3/2}xy+.5(xy)^{1/2}2(x ^{2}y)^{3}2x ^{3}y+(x ^{2}y)^{2}2x\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Mini, when you find the partial derivative of x, y is not considered a function of x, so you do not use the product rule

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so for the same reason we don't do chain rule as well

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0no, you dont multiply both derivatives

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0but why is it not in this case? its with respect to y

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0you take the derivate to respect x first to get Fx, then you take the derivate to respect y for the new functions whch will turn into Fxy

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0f_xy means \[f_{xy} = d/dy(d/dx (f))\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0yes i know that much but if we are talking in terms of y, y isnt .25(xy)^(3/2)y considered two?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0like y times the other part

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0thats just like saying in terms of x(3x)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so you take the derivative of f in terms of x while y is constant and derivative of f_x in terms of y while x is constant

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0its the same thing if instead you do x^3, as x times x^2, is it not?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0the last part is in terms of y, which applies to both parts so im still missing why i cant do the product once i did fx

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0its the same thing if instead you do x^3, as x times x^2, is it not?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0that question still holds

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Mini, Are yu asking why can't you multiply fx and fy separatly?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Mini, was my f(x,y) correct? I'm wondering because you don't seem to have any interest in what I said.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0im asking why i cant just do \[.5(xy)^{1/2}y+(x ^{2}y)^{2}2xy\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0and to the product on that, and i dont know yuki, im not worried about the answer im worried about the concept

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0not trying to be a wingspan, plus i dont even have the answer haha

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Mini, because y is not a function of x. You do not do product rule with partial derivatives because in partial derivatives, other variables are constant

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so let me give you an example

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0but why is that not the same as x*x^2 = x^3

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0let's say f(x,y) = \[3x^2y\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0and separating those and doing one by product and one by chain

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0f_x would be yes, 3y*(x^2)' = 6xy

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0f_xy and f_yx must be the same, so let's check f_xy = 6x(y)' = 6x f_yx = 3(x^2)' = 6x yay

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0well yeah i follow that, thats easy lol

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0There's no need to do chain rule in this problem

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0If you would like me to answer your question why (x^3)' and (x*x^2)' are not the same they actually are

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ah there might be no NEED, but can i do it that way

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0its my train of thought, thats all i can say

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0and i know they are the same, thats what im basing my principle on just doing the product rule on

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0(x*x^2)' is x'*x^2 + (x^2)'*x =x^2+2x*x =3x^2

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[.5(xy)^{1/2}y+(x ^{2}y)^{2}2xy\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0thats my in terms of x

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0okay now I think we are on the same page, so you are trying to use the product rule for this problem anyways, am I right ?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0my question still stands why i cant do the product rule on this

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0MIni, you can use the product rule

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0and that equation is now going to be derived in terms of y

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0okay, gotcha. just to let you know that is very unorthodox, and I haven't tried it yet, so give me some time to work it out first.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0haha, thats what my professor said, i know he loves grading my papers but i see everything as addition and multiplication

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0hehe okay, yea, you can do product rule but yu still have to do another quotien rule

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0you will have to do two product rules inside a quotien rule

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0mini, I want to know how you will consider f(x,y)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0where? once i do the product for \[.5(xy)^{1/2}y+(x ^{2}y)^{2}2xy\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0on both sides isnt it just two products?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0isn't the main equation a fraction?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0no, i changed them into multiplication

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0can you type the main equation and your "changed" equation? please

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0yes that's the main, now what did yu change that for?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0changed to (xy)^(1/2)  (x^2y)^1

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Mini, this is your f(x,y) \[\sqrt{xy}{1 \over x^2y}\] what do you want to do from here ? without that I cannot tell what you are doing

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0both of you that before u jus typed is what i changed it to, and i changed it just to do the chain because its easier for myself

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Mini, if that is your claim it doesn't look right in the first place

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0and ignore the bad parathesis i just didtn want to put them all up, im pretty sure you followed it

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ok, even thou you changed the equation for two multiplication instead a fraction. There is no chain rule in that equation

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0(xy)^(1/2) isnt a chain rule lol? one half goes tot he front, leave xy, minus 1, then put y outside because its in terms of x?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0did you want to wright it as \[x^{1/2}y^{1/2} x^{2}y^{1}\] ?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0no, you are taking the derivative to respec x which makes y into a constant. it's same as taking the derivative of (x3)^(1/2)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0no, thats fine too its the same thing, u just multiplied the 1 out

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0yes thats why y is still on the outside, its the constant...when u do the der of the inside

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so it would be (1/2)(y)(xy)(1/2) is that right

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0yes well to the negative half

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0(1/2)(y)(xy)^(1/2) ops,

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0it might be unorthodox like yuki said but thats how i did it

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0yea but, it would work. i don't see why it wouldn't

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0like i get my \[.5(xy)^{1/2}y+(x ^{2}y)^{2}2xy\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0and that was in terms of x

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0now i just do products in terms of y

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0okay Mini, I cannot help you unless you answer me so... just let me know if you are ready to focus. I'm pretty sure there are others who can help.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i dont know what question you asked yuki?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0if you mean if i wanted to write it as...i did answer you...

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0damn yo you typing a lot, scaring me!

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0F = (xy)^(1/2)  (x^2y)^1 Fx = .5y(xy)^(1/2) + 2xy(x^2y)^2) Fxy = .5(xy)^(1/2) + .5y(.5(xy)^(3/2) + 2x(x^2 y)^(2) + 2xy(2x^2(x^2y)^(2))

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0oh pellet no wonder you were tying a long time, thought he chat froze haha

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0if yu can understand my Fxy.. that's what i got lol

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0haha yea.. this chat thing kinda sucks

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ill start from the begin then

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0like of what you posted not the real begin

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0−.25(xy)−3/2xy+.5(xy)−1/2−2(x2y)−32x3y+(x2y)−22x ?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0should be xy on the first one

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0at the end of mine the two last parts canceled out, and i was only left iwth the first 2

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0should be .25(xy)^(3/2)xy

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0last ones i think looks good too, i just missed the 4 when i was simplifying, which doesnt cancel but just reduces

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ill just post it one product at a time

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0then im takin a break and ill be back on prob in an hour! lol

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[.5(xy)^{1/2}y+(x ^{2}y)^{2}2xy\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i think i'm using the wrong equation lol is the main equation this : (xy)^(1/2)1/(x^2*y) ?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0no u have the right one, and we are dead one together but i think u messed up the first product rule

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0[ (xy)^(1/2) ]  [ 1/(x^2*y)] ?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0when u derive .5(xy)^(1/2)y

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0but derive the first "function" not y

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0thats where our answers are different

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0u should get one fourth xy to the neg 3/2 then times x, then also times y from the orig

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0right? haha its almost easier writing it out instead of using that dumb equation thing

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so your Fxy is: (.5) * (((.5)  1) * (x * y) ^ (((.5)  1)  1) * x * y + (x * y) ^ ((.5)  1)) + 2 * x * ((x ^ 2 * y) ^ 2  y * 2 * (x ^ 2 * y) * x ^ 2) / (x ^ 2 * y) ^ 2 ^ 2

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0oh wow my brain just died after seeing that

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0one at a time yo! one at a time!

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0haha i just used a partial derivative calcualtor

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0http://www.calculatorgrapher.com/derivativecalculator.html

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[.5(xy)^{1/2}y+(x ^{2}y)^{2}2xy\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0okay thats what we are doing in terms of y

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so im going to product the difference of them, 1 product then another

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0the first product i get \[.25(xy)^{3/2}xy+.5(xy)^{1/2}\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0oh im using copy paste, makes it ALOT easier haha

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0the second product is \[2(x ^{2}y)^{3}2x ^{3}y+(x ^{2}y)^{2}2x\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Fx = (.5)y(xy)^(.5) + 2xy/(x^2y)^2 Fxy = (.25)(y^2) / (sqrt(xy))  6/(x^4*y)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0oh well.. i'm out lol good luck

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0thanks man! have a good one
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