Mini
  • Mini
fxy if f(x,y) = (xy)^(1/2)-1/(x^2*y)
Mathematics
jamiebookeater
  • jamiebookeater
See more answers at brainly.com
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this
and thousands of other questions

yuki
  • yuki
what would you want to do?
Mini
  • Mini
its double der
Mini
  • Mini
x then y

Looking for something else?

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

More answers

yuki
  • yuki
ok, so you are finding \[f_{xy}\]
Mini
  • Mini
yes
Mini
  • Mini
its just really long wanted to see someone else do it lol
anonymous
  • anonymous
derivate everything to respect x and then derivate that answer to respect y
yuki
  • 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
  • yuki
am I right?
Mini
  • Mini
already know that brain but thank you
Mini
  • Mini
ill write what i have first
Mini
  • Mini
ill write what i have first\[.5(xy)^{-1/2}y+(x ^{2}y)^{-2}2xy\]
anonymous
  • 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
  • Mini
quotient? i just did two product rules...
yuki
  • yuki
what you should notice first is that the equation is equivalent to \[\sqrt{1 \over yx^3}\]
yuki
  • yuki
this is a lot easier to find the derivatives
Mini
  • Mini
i do chain rules and product, u dont have to use quotient
yuki
  • 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
  • Mini
well product and quotient are the same thing though lol, its like comparing addition of negatives and subtraction
anonymous
  • 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
  • yuki
to find f_x, I will write f as follows \[{1 \over \sqrt y} * {1 \over x^{3/2}}\]
Mini
  • Mini
dont see what im missing i might of had a typo but my dx is + to the neg 2
yuki
  • yuki
Brain, nice catch, I was going to solve the problem not knowing that lol
anonymous
  • anonymous
haha yea, sorry i was trying to type on the "Equation" thing it took me a while lol
yuki
  • yuki
then f = \[\sqrt{xy} - {1 \over x^2y}\]
Mini
  • 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
  • Mini
i mean chain is pretty simple, bring to the front, copy, minus 2, der of inside
yuki
  • yuki
then f_x is \[{\sqrt{y} \over 2\sqrt{x}}+{2 \over yx^3 }\]
Mini
  • Mini
minus 1 even
Mini
  • Mini
yeah thats not what i got .5(xy)^(-1/2)y
Mini
  • Mini
well the first time with respect to x
yuki
  • yuki
now f_xy is \[{1 \over 4\sqrt{xy}} - {2 \over x^3y^2}\]
anonymous
  • 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
  • Mini
\[.5(xy)^{-1/2}y+(x ^{2}y)^{-2}2xy\]
Mini
  • Mini
then product x2
Mini
  • Mini
for respect to y
yuki
  • yuki
I am 100% sure with my answer
yuki
  • yuki
I heard product rule, but you don't have to use it anywhere. at least I didn't
Mini
  • 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
  • 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
  • Mini
so its not fx*fy
yuki
  • yuki
so for the same reason we don't do chain rule as well
anonymous
  • anonymous
no, you dont multiply both derivatives
Mini
  • Mini
but why is it not in this case? its with respect to y
anonymous
  • 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
  • yuki
f_xy means \[f_{xy} = d/dy(d/dx (f))\]
anonymous
  • anonymous
yes
Mini
  • 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
  • Mini
like y times the other part
Mini
  • Mini
thats just like saying in terms of x(3-x)
yuki
  • 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
  • Mini
its the same thing if instead you do x^3, as x times x^2, is it not?
Mini
  • 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
  • Mini
rather respect to y
Mini
  • Mini
its the same thing if instead you do x^3, as x times x^2, is it not?
Mini
  • Mini
that question still holds
anonymous
  • anonymous
Mini, Are yu asking why can't you multiply fx and fy separatly?
yuki
  • 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
  • anonymous
like: Fx * Fy
Mini
  • Mini
im asking why i cant just do \[.5(xy)^{-1/2}y+(x ^{2}y)^{-2}2xy\]
Mini
  • Mini
and to the product on that, and i dont know yuki, im not worried about the answer im worried about the concept
Mini
  • Mini
not trying to be a wingspan, plus i dont even have the answer haha
yuki
  • 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
  • yuki
so let me give you an example
Mini
  • Mini
but why is that not the same as x*x^2 = x^3
yuki
  • yuki
let's say f(x,y) = \[3x^2y\]
Mini
  • Mini
and separating those and doing one by product and one by chain
Mini
  • Mini
6xy
Mini
  • Mini
then 6x
yuki
  • yuki
f_x would be yes, 3y*(x^2)' = 6xy
yuki
  • yuki
f_y would be 3x^2
yuki
  • 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
  • Mini
well yeah i follow that, thats easy lol
anonymous
  • anonymous
There's no need to do chain rule in this problem
yuki
  • 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
  • Mini
ah there might be no NEED, but can i do it that way
Mini
  • Mini
its my train of thought, thats all i can say
Mini
  • Mini
and i know they are the same, thats what im basing my principle on just doing the product rule on
yuki
  • yuki
(x*x^2)' is x'*x^2 + (x^2)'*x =x^2+2x*x =3x^2
Mini
  • Mini
\[.5(xy)^{-1/2}y+(x ^{2}y)^{-2}2xy\]
Mini
  • Mini
thats my in terms of x
yuki
  • 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
  • Mini
my question still stands why i cant do the product rule on this
Mini
  • Mini
yes!!!
anonymous
  • anonymous
MIni, you can use the product rule
Mini
  • Mini
and that equation is now going to be derived in terms of y
yuki
  • 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
  • Mini
haha, thats what my professor said, i know he loves grading my papers but i see everything as addition and multiplication
anonymous
  • anonymous
hehe okay, yea, you can do product rule but yu still have to do another quotien rule
anonymous
  • anonymous
you will have to do two product rules inside a quotien rule
yuki
  • yuki
mini, I want to know how you will consider f(x,y)
Mini
  • Mini
where? once i do the product for \[.5(xy)^{-1/2}y+(x ^{2}y)^{-2}2xy\]
Mini
  • Mini
on both sides isnt it just two products?
anonymous
  • anonymous
isn't the main equation a fraction?
Mini
  • Mini
no, i changed them into multiplication
anonymous
  • anonymous
can you type the main equation and your "changed" equation? please
Mini
  • Mini
sqrt(xy)-1/(x^2*y)
anonymous
  • anonymous
yes that's the main, now what did yu change that for?
Mini
  • Mini
changed to (xy)^(1/2) - (x^2y)^-1
yuki
  • 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
  • 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
  • yuki
Mini, if that is your claim it doesn't look right in the first place
Mini
  • Mini
and ignore the bad parathesis i just didtn want to put them all up, im pretty sure you followed it
anonymous
  • anonymous
ok, even thou you changed the equation for two multiplication instead a fraction. There is no chain rule in that equation
Mini
  • 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
  • yuki
did you want to wright it as \[x^{1/2}y^{1/2} -x^{-2}y^{-1}\] ?
anonymous
  • 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
  • Mini
no, thats fine too its the same thing, u just multiplied the -1 out
Mini
  • Mini
yes thats why y is still on the outside, its the constant...when u do the der of the inside
anonymous
  • anonymous
so it would be (1/2)(y)(xy)(-1/2) is that right
Mini
  • Mini
yes well to the negative half
anonymous
  • anonymous
(1/2)(y)(xy)^(-1/2) ops,
Mini
  • Mini
it might be unorthodox like yuki said but thats how i did it
anonymous
  • anonymous
yea but, it would work. i don't see why it wouldn't
Mini
  • Mini
like i get my \[.5(xy)^{-1/2}y+(x ^{2}y)^{-2}2xy\]
Mini
  • Mini
and that was in terms of x
Mini
  • Mini
now i just do products in terms of y
yuki
  • 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
  • Mini
i dont know what question you asked yuki?
Mini
  • Mini
if you mean if i wanted to write it as...i did answer you...
Mini
  • Mini
damn yo you typing a lot, scaring me!
anonymous
  • 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
  • Mini
oh pellet no wonder you were tying a long time, thought he chat froze haha
anonymous
  • anonymous
if yu can understand my Fxy.. that's what i got lol
anonymous
  • anonymous
haha yea.. this chat thing kinda sucks
Mini
  • Mini
ill start from the begin then
Mini
  • Mini
like of what you posted not the real begin
anonymous
  • anonymous
−.25(xy)−3/2xy+.5(xy)−1/2−2(x2y)−32x3y+(x2y)−22x ?
Mini
  • Mini
should be xy on the first one
Mini
  • Mini
at the end of mine the two last parts canceled out, and i was only left iwth the first 2
Mini
  • Mini
xy on 2nd one***
Mini
  • Mini
should be -.25(xy)^(-3/2)xy
Mini
  • Mini
last ones i think looks good too, i just missed the 4 when i was simplifying, which doesnt cancel but just reduces
Mini
  • Mini
ill just post it one product at a time
Mini
  • Mini
then im takin a break and ill be back on prob in an hour! lol
Mini
  • Mini
\[.5(xy)^{-1/2}y+(x ^{2}y)^{-2}2xy\]
anonymous
  • anonymous
i think i'm using the wrong equation lol is the main equation this : (xy)^(1/2)-1/(x^2*y) ?
Mini
  • Mini
yes
Mini
  • Mini
no u have the right one, and we are dead one together but i think u messed up the first product rule
anonymous
  • anonymous
[ (xy)^(1/2) ] - [ 1/(x^2*y)] ?
Mini
  • Mini
when u derive .5(xy)^(-1/2)y
Mini
  • Mini
but derive the first "function" not y
Mini
  • Mini
thats where our answers are different
Mini
  • Mini
u should get one fourth xy to the neg 3/2 then times x, then also times y from the orig
Mini
  • Mini
right? haha its almost easier writing it out instead of using that dumb equation thing
anonymous
  • 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
  • Mini
oh wow my brain just died after seeing that
Mini
  • Mini
one at a time yo! one at a time!
anonymous
  • anonymous
haha i just used a partial derivative calcualtor
anonymous
  • anonymous
http://www.calculator-grapher.com/derivative-calculator.html
Mini
  • Mini
\[.5(xy)^{-1/2}y+(x ^{2}y)^{-2}2xy\]
Mini
  • Mini
okay thats what we are doing in terms of y
Mini
  • Mini
so im going to product the difference of them, 1 product then another
Mini
  • Mini
or sum of them w/e
Mini
  • Mini
the first product i get \[-.25(xy)^{-3/2}xy+.5(xy)^{-1/2}\]
Mini
  • Mini
oh im using copy paste, makes it ALOT easier haha
Mini
  • Mini
the second product is \[-2(x ^{2}y)^{-3}2x ^{3}y+(x ^{2}y)^{-2}2x\]
anonymous
  • anonymous
Fx = (.5)y(xy)^(.5) + 2xy/(x^2y)^2 Fxy = (.25)(y^2) / (sqrt(xy)) - 6/(x^4*y)
anonymous
  • anonymous
lol
Mini
  • Mini
eh close enough
anonymous
  • anonymous
oh well.. i'm out lol good luck
Mini
  • Mini
thanks man! have a good one
anonymous
  • anonymous
np y too

Looking for something else?

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