Question

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

Answer 1

what would you want to do?

Answer 2

its double der

Answer 3

x then y

Answer 4

ok, so you are finding

\[f_{xy}\]

Answer 5

yes

Answer 6

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

Answer 7

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

Answer 8

let me re-write the eqn first, it's tough to see it

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



\[{(xy)^{1/2} \over x^2y}\]

Answer 9

am I right?

Answer 10

already know that brain but thank you

Answer 11

ill write what i have first

Answer 12

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

Answer 13

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

Answer 14

quotient? i just did two product rules...

Answer 15

what you should notice first is that

the equation is equivalent to

\[\sqrt{1 \over yx^3}\]

Answer 16

this is a lot easier to find the derivatives

Answer 17

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

Answer 18

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?

Answer 19

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

Answer 20

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

Answer 21

to find f_x, I will write f as follows

\[{1 \over \sqrt y} * {1 \over x^{3/2}}\]

Answer 22

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

Answer 23

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

Answer 24

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

Answer 25

then f =

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

Answer 26

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

Answer 27

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

Answer 28

then f_x is

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

Answer 29

minus 1 even

Answer 30

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

Answer 31

well the first time with respect to x

Answer 32

now f_xy is

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

Answer 33

ok. Mini, what i got so far was:

Fx = \[(-2(\sqrt{xy}-1))\div(x^{3}y)\]

Fxy = \[(2\times(\sqrt{xy}-1))\div(y^{2}x ^{3})\]

Answer 34

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

Answer 35

then product x2

Answer 36

for respect to y

Answer 37

I am 100% sure with my answer

Answer 38

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

Answer 39

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

Answer 40

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

Answer 41

so its not fx*fy

Answer 42

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

Answer 43

no, you dont multiply both derivatives

Answer 44

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

Answer 45

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

Answer 46

f_xy means

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

Answer 47

yes

Answer 48

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

Answer 49

like y times the other part

Answer 50

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

Answer 51

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

Answer 52

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

Answer 53

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

Answer 54

rather respect to y

Answer 55

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

Answer 56

that question still holds

Answer 57

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

Answer 58

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

Answer 59

like: Fx * Fy

Answer 60

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

Answer 61

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

Answer 62

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

Answer 63

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

Answer 64

so let me give you an example

Answer 65

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

Answer 66

let's say

f(x,y) = \[3x^2y\]

Answer 67

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

Answer 68

6xy

Answer 69

then 6x

Answer 70

f_x would be

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

Answer 71

f_y would be

3x^2

Answer 72

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

Answer 73

well yeah i follow that, thats easy lol

Answer 74

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

Answer 75

If you would like me to answer your question

why (x^3)' and (x*x^2)' are not the same
they actually are

Answer 76

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

Answer 77

its my train of thought, thats all i can say

Answer 78

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

Answer 79

(x*x^2)' is

x'*x^2 + (x^2)'*x

=x^2+2x*x

=3x^2

Answer 80

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

Answer 81

thats my in terms of x

Answer 82

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 ?

Answer 83

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

Answer 84

yes!!!

Answer 85

MIni, you can use the product rule

Answer 86

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

Answer 87

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.

Answer 88

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

Answer 89

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

Answer 90

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

Answer 91

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

Answer 92

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

Answer 93

on both sides isnt it just two products?

Answer 94

isn't the main equation a fraction?

Answer 95

no, i changed them into multiplication

Answer 96

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

Answer 97

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

Answer 98

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