fxy if f(x,y) = (xy)^(1/2)-1/(x^2*y)

- Mini

fxy if f(x,y) = (xy)^(1/2)-1/(x^2*y)

- Stacey Warren - Expert brainly.com

Hey! We 've verified this expert answer for you, click below to unlock the details :)

- chestercat

I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!

- yuki

what would you want to do?

- Mini

its double der

- Mini

x then y

Looking for something else?

Not the answer you are looking for? Search for more explanations.

## More answers

- yuki

ok, so you are finding
\[f_{xy}\]

- Mini

yes

- Mini

its just really long wanted to see someone else do it lol

- anonymous

derivate everything to respect x and then derivate that answer to respect y

- yuki

let me re-write the eqn first, it's tough to see it
(xy)^(1/2)-1/(x^2*y)
\[{(xy)^{1/2} \over x^2y}\]

- yuki

am I right?

- Mini

already know that brain but thank you

- Mini

ill write what i have first

- Mini

ill write what i have first\[.5(xy)^{-1/2}y+(x ^{2}y)^{-2}2xy\]

- anonymous

hmm.. it's going to be a quotien rule with two product rules. sorry but i'm kinda lazy to actualy write it

- Mini

quotient? i just did two product rules...

- yuki

what you should notice first is that
the equation is equivalent to
\[\sqrt{1 \over yx^3}\]

- yuki

this is a lot easier to find the derivatives

- Mini

i do chain rules and product, u dont have to use quotient

- yuki

Mini, 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?

- Mini

well product and quotient are the same thing though lol, its like comparing addition of negatives and subtraction

- anonymous

you are missing a -1 on the top part -> (xy)^(1/2)-1/(x^2*y):
\[(\sqrt{xy}-1)\div(x^{2}y)\]

- yuki

to find f_x, I will write f as follows
\[{1 \over \sqrt y} * {1 \over x^{3/2}}\]

- Mini

dont see what im missing i might of had a typo but my dx is + to the neg 2

- yuki

Brain, nice catch, I was going to solve the problem not knowing that lol

- anonymous

haha yea, sorry i was trying to type on the "Equation" thing it took me a while lol

- yuki

then f =
\[\sqrt{xy} - {1 \over x^2y}\]

- Mini

yuki 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

- Mini

i mean chain is pretty simple, bring to the front, copy, minus 2, der of inside

- yuki

then f_x is
\[{\sqrt{y} \over 2\sqrt{x}}+{2 \over yx^3 }\]

- Mini

minus 1 even

- Mini

yeah thats not what i got .5(xy)^(-1/2)y

- Mini

well the first time with respect to x

- yuki

now f_xy is
\[{1 \over 4\sqrt{xy}} - {2 \over x^3y^2}\]

- anonymous

ok. Mini, what i got so far was:
Fx = \[(-2(\sqrt{xy}-1))\div(x^{3}y)\]
Fxy = \[(2\times(\sqrt{xy}-1))\div(y^{2}x ^{3})\]

- Mini

\[.5(xy)^{-1/2}y+(x ^{2}y)^{-2}2xy\]

- Mini

then product x2

- Mini

for respect to y

- yuki

I am 100% sure with my answer

- yuki

I heard product rule, but you don't have to use it anywhere.
at least I didn't

- Mini

for respect to y\[-.25(xy)^{-3/2}xy+.5(xy)^{-1/2}-2(x ^{2}y)^{-3}2x ^{3}y+(x ^{2}y)^{-2}2x\]

- yuki

Mini, when you find the partial derivative of x, y is not considered a function of x, so you do not use the product rule

- Mini

so its not fx*fy

- yuki

so for the same reason we don't do chain rule as well

- anonymous

no, you dont multiply both derivatives

- Mini

but why is it not in this case? its with respect to y

- anonymous

you 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

- yuki

f_xy means
\[f_{xy} = d/dy(d/dx (f))\]

- anonymous

yes

- Mini

yes i know that much but if we are talking in terms of y, y isnt .25(xy)^(-3/2)y considered two?

- Mini

like y times the other part

- Mini

thats just like saying in terms of x(3-x)

- yuki

so 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

- Mini

its the same thing if instead you do x^3, as x times x^2, is it not?

- Mini

the 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

- Mini

rather respect to y

- Mini

its the same thing if instead you do x^3, as x times x^2, is it not?

- Mini

that question still holds

- anonymous

Mini, Are yu asking why can't you multiply fx and fy separatly?

- yuki

Mini, was my f(x,y) correct?
I'm wondering because you don't seem to have any interest in what I said.

- anonymous

like: Fx * Fy

- Mini

im asking why i cant just do \[.5(xy)^{-1/2}y+(x ^{2}y)^{-2}2xy\]

- Mini

and to the product on that, and i dont know yuki, im not worried about the answer im worried about the concept

- Mini

not trying to be a wingspan, plus i dont even have the answer haha

- yuki

Mini, 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

- yuki

so let me give you an example

- Mini

but why is that not the same as x*x^2 = x^3

- yuki

let's say
f(x,y) = \[3x^2y\]

- Mini

and separating those and doing one by product and one by chain

- Mini

6xy

- Mini

then 6x

- yuki

f_x would be
yes, 3y*(x^2)' = 6xy

- yuki

f_y would be
3x^2

- yuki

f_xy and f_yx must be the same, so let's check
f_xy = 6x(y)' = 6x
f_yx = 3(x^2)' = 6x
yay

- Mini

well yeah i follow that, thats easy lol

- anonymous

There's no need to do chain rule in this problem

- yuki

If you would like me to answer your question
why (x^3)' and (x*x^2)' are not the same
they actually are

- Mini

ah there might be no NEED, but can i do it that way

- Mini

its my train of thought, thats all i can say

- Mini

and i know they are the same, thats what im basing my principle on just doing the product rule on

- yuki

(x*x^2)' is
x'*x^2 + (x^2)'*x
=x^2+2x*x
=3x^2

- Mini

\[.5(xy)^{-1/2}y+(x ^{2}y)^{-2}2xy\]

- Mini

thats my in terms of x

- yuki

okay 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 ?

- Mini

my question still stands why i cant do the product rule on this

- Mini

yes!!!

- anonymous

MIni, you can use the product rule

- Mini

and that equation is now going to be derived in terms of y

- yuki

okay, 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.

- Mini

haha, thats what my professor said, i know he loves grading my papers but i see everything as addition and multiplication

- anonymous

hehe okay, yea, you can do product rule but yu still have to do another quotien rule

- anonymous

you will have to do two product rules inside a quotien rule

- yuki

mini, I want to know how you will consider f(x,y)

- Mini

where? once i do the product for \[.5(xy)^{-1/2}y+(x ^{2}y)^{-2}2xy\]

- Mini

on both sides isnt it just two products?

- anonymous

isn't the main equation a fraction?

- Mini

no, i changed them into multiplication

- anonymous

can you type the main equation and your "changed" equation?
please

- Mini

sqrt(xy)-1/(x^2*y)

- anonymous

yes that's the main, now what did yu change that for?

- Mini

changed to (xy)^(1/2) - (x^2y)^-1

- yuki

Mini, 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

- Mini

both 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

- yuki

Mini, if that is your claim
it doesn't look right in the first place

- Mini

and ignore the bad parathesis i just didtn want to put them all up, im pretty sure you followed it

- anonymous

ok, even thou you changed the equation for two multiplication instead a fraction. There is no chain rule in that equation

- Mini

(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?

- yuki

did you want to wright it as
\[x^{1/2}y^{1/2} -x^{-2}y^{-1}\]
?

- anonymous

no, 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)

- Mini

no, thats fine too its the same thing, u just multiplied the -1 out

- Mini

yes thats why y is still on the outside, its the constant...when u do the der of the inside

- anonymous

so it would be (1/2)(y)(xy)(-1/2) is that right

- Mini

yes
well to the negative half

- anonymous

(1/2)(y)(xy)^(-1/2) ops,

- Mini

it might be unorthodox like yuki said but thats how i did it

- anonymous

yea but, it would work. i don't see why it wouldn't

- Mini

like i get my \[.5(xy)^{-1/2}y+(x ^{2}y)^{-2}2xy\]

- Mini

and that was in terms of x

- Mini

now i just do products in terms of y

- yuki

okay 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.

- Mini

i dont know what question you asked yuki?

- Mini

if you mean if i wanted to write it as...i did answer you...

- Mini

damn yo you typing a lot, scaring me!

- anonymous

F = (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))

- Mini

oh pellet no wonder you were tying a long time, thought he chat froze haha

- anonymous

if yu can understand my Fxy.. that's what i got lol

- anonymous

haha yea.. this chat thing kinda sucks

- Mini

ill start from the begin then

- Mini

like of what you posted not the real begin

- anonymous

âˆ’.25(xy)âˆ’3/2xy+.5(xy)âˆ’1/2âˆ’2(x2y)âˆ’32x3y+(x2y)âˆ’22x
?

- Mini

should be xy on the first one

- Mini

at the end of mine the two last parts canceled out, and i was only left iwth the first 2

- Mini

xy on 2nd one***

- Mini

should be -.25(xy)^(-3/2)xy

- Mini

last ones i think looks good too, i just missed the 4 when i was simplifying, which doesnt cancel but just reduces

- Mini

ill just post it one product at a time

- Mini

then im takin a break and ill be back on prob in an hour! lol

- Mini

\[.5(xy)^{-1/2}y+(x ^{2}y)^{-2}2xy\]

- anonymous

i think i'm using the wrong equation
lol
is the main equation this : (xy)^(1/2)-1/(x^2*y) ?

- Mini

yes

- Mini

no u have the right one, and we are dead one together but i think u messed up the first product rule

- anonymous

[ (xy)^(1/2) ] - [ 1/(x^2*y)] ?

- Mini

when u derive .5(xy)^(-1/2)y

- Mini

but derive the first "function" not y

- Mini

thats where our answers are different

- Mini

u should get one fourth xy to the neg 3/2 then times x, then also times y from the orig

- Mini

right? haha its almost easier writing it out instead of using that dumb equation thing

- anonymous

so 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

- Mini

oh wow my brain just died after seeing that

- Mini

one at a time yo! one at a time!

- anonymous

haha
i just used a partial derivative calcualtor

- anonymous

http://www.calculator-grapher.com/derivative-calculator.html

- Mini

\[.5(xy)^{-1/2}y+(x ^{2}y)^{-2}2xy\]

- Mini

okay thats what we are doing in terms of y

- Mini

so im going to product the difference of them, 1 product then another

- Mini

or sum of them w/e

- Mini

the first product i get \[-.25(xy)^{-3/2}xy+.5(xy)^{-1/2}\]

- Mini

oh im using copy paste, makes it ALOT easier haha

- Mini

the second product is \[-2(x ^{2}y)^{-3}2x ^{3}y+(x ^{2}y)^{-2}2x\]

- anonymous

Fx = (.5)y(xy)^(.5) + 2xy/(x^2y)^2
Fxy = (.25)(y^2) / (sqrt(xy)) - 6/(x^4*y)

- anonymous

lol

- Mini

eh close enough

- anonymous

oh well.. i'm out lol good luck

- Mini

thanks man! have a good one

- anonymous

np y too

Looking for something else?

Not the answer you are looking for? Search for more explanations.