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

1. Yuki

what would you want to do?

2. Mini

its double der

3. Mini

x then y

4. Yuki

ok, so you are finding $f_{xy}$

5. Mini

yes

6. Mini

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

7. anonymous

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

8. Yuki

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

9. Yuki

am I right?

10. Mini

already know that brain but thank you

11. Mini

ill write what i have first

12. Mini

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

13. anonymous

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

14. Mini

quotient? i just did two product rules...

15. Yuki

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

16. Yuki

this is a lot easier to find the derivatives

17. Mini

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

18. 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?

19. Mini

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

20. anonymous

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

21. Yuki

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

22. Mini

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

23. Yuki

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

24. anonymous

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

25. Yuki

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

26. 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

27. Mini

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

28. Yuki

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

29. Mini

minus 1 even

30. Mini

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

31. Mini

well the first time with respect to x

32. Yuki

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

33. anonymous

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

34. Mini

$.5(xy)^{-1/2}y+(x ^{2}y)^{-2}2xy$

35. Mini

then product x2

36. Mini

for respect to y

37. Yuki

I am 100% sure with my answer

38. Yuki

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

39. Mini

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

40. 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

41. Mini

so its not fx*fy

42. Yuki

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

43. anonymous

no, you dont multiply both derivatives

44. Mini

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

45. 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

46. Yuki

f_xy means $f_{xy} = d/dy(d/dx (f))$

47. anonymous

yes

48. Mini

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

49. Mini

like y times the other part

50. Mini

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

51. 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

52. Mini

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

53. 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

54. Mini

rather respect to y

55. Mini

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

56. Mini

that question still holds

57. anonymous

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

58. Yuki

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

59. anonymous

like: Fx * Fy

60. Mini

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

61. Mini

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

62. Mini

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

63. 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

64. Yuki

so let me give you an example

65. Mini

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

66. Yuki

let's say f(x,y) = $3x^2y$

67. Mini

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

68. Mini

6xy

69. Mini

then 6x

70. Yuki

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

71. Yuki

f_y would be 3x^2

72. 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

73. Mini

well yeah i follow that, thats easy lol

74. anonymous

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

75. Yuki

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

76. Mini

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

77. Mini

its my train of thought, thats all i can say

78. Mini

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

79. Yuki

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

80. Mini

$.5(xy)^{-1/2}y+(x ^{2}y)^{-2}2xy$

81. Mini

thats my in terms of x

82. 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 ?

83. Mini

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

84. Mini

yes!!!

85. anonymous

MIni, you can use the product rule

86. Mini

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

87. 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.

88. Mini

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

89. anonymous

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

90. anonymous

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

91. Yuki

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

92. Mini

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

93. Mini

on both sides isnt it just two products?

94. anonymous

isn't the main equation a fraction?

95. Mini

no, i changed them into multiplication

96. anonymous

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

97. Mini

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

98. anonymous

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

99. Mini

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

100. 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

101. 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

102. Yuki

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

103. Mini

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

104. anonymous

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

105. 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?

106. Yuki

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

107. 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)

108. Mini

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

109. Mini

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

110. anonymous

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

111. Mini

yes well to the negative half

112. anonymous

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

113. Mini

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

114. anonymous

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

115. Mini

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

116. Mini

and that was in terms of x

117. Mini

now i just do products in terms of y

118. 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.

119. Mini

i dont know what question you asked yuki?

120. Mini

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

121. Mini

damn yo you typing a lot, scaring me!

122. 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))

123. Mini

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

124. anonymous

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

125. anonymous

haha yea.. this chat thing kinda sucks

126. Mini

ill start from the begin then

127. Mini

like of what you posted not the real begin

128. anonymous

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

129. Mini

should be xy on the first one

130. Mini

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

131. Mini

xy on 2nd one***

132. Mini

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

133. Mini

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

134. Mini

ill just post it one product at a time

135. Mini

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

136. Mini

$.5(xy)^{-1/2}y+(x ^{2}y)^{-2}2xy$

137. anonymous

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

138. Mini

yes

139. Mini

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

140. anonymous

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

141. Mini

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

142. Mini

but derive the first "function" not y

143. Mini

thats where our answers are different

144. Mini

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

145. Mini

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

146. 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

147. Mini

oh wow my brain just died after seeing that

148. Mini

one at a time yo! one at a time!

149. anonymous

haha i just used a partial derivative calcualtor

150. anonymous
151. Mini

$.5(xy)^{-1/2}y+(x ^{2}y)^{-2}2xy$

152. Mini

okay thats what we are doing in terms of y

153. Mini

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

154. Mini

or sum of them w/e

155. Mini

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

156. Mini

oh im using copy paste, makes it ALOT easier haha

157. Mini

the second product is $-2(x ^{2}y)^{-3}2x ^{3}y+(x ^{2}y)^{-2}2x$

158. anonymous

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

159. anonymous

lol

160. Mini

eh close enough

161. anonymous

oh well.. i'm out lol good luck

162. Mini

thanks man! have a good one

163. anonymous

np y too