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Mini

  • 5 years ago

Integrate: xsin(pix^2)dx I'm going U sub with u equaling to pix^2 but I get stuck trying to compensate for the du = 2pixdx then dx = du/(2pix)

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  1. amistre64
    • 5 years ago
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    u = pi x^2 du = 2pi x dx dx = du/2pi x

  2. amistre64
    • 5 years ago
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    x sin(u) sin(u) ----- = ----- du 2pi x 2pi

  3. Mini
    • 5 years ago
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    so i can pull out 1/2pi in front

  4. amistre64
    • 5 years ago
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    yep; it aint gonna change any :)

  5. Mini
    • 5 years ago
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    if thats the case then the answer he has listed is wrong, the bounds are from 0 to 1 his answer listed is 1

  6. amistre64
    • 5 years ago
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    you need to have a 2pi up top to counter your integration

  7. Mini
    • 5 years ago
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    So when I set something dx = blah blah blah, i have to account for what im plugging in

  8. anonymous
    • 5 years ago
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    (the answer isn't 1)

  9. Mini
    • 5 years ago
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    for instance dx = 2xdu, i need to replace dx with the 2xdu but also put in a 1/2x?

  10. amistre64
    • 5 years ago
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    whatch.... D(-cos(2pix^2)) = 2pix sin(2pix^2) right?

  11. amistre64
    • 5 years ago
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    4pix lol...srry

  12. Mini
    • 5 years ago
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    lol newton his answers are rarely ever what they seem to be, its also a double integration though

  13. amistre64
    • 5 years ago
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    post the entire questionthen; its rather difficult to step in half way and try to determine where something messed up at

  14. amistre64
    • 5 years ago
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    4pi x sin(2pix^2) -------------- dx is doable right? 4pi

  15. Mini
    • 5 years ago
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    |0to1|cube root y to 1 (sin(pix^2))/x^2 dxdy

  16. Mini
    • 5 years ago
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    hard to type a double integral with this crappy window!

  17. amistre64
    • 5 years ago
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    your u was ok but we messed up the derivative of it lol u = 2pix^2 du = 4pi x dx dx = du/4pi x

  18. Mini
    • 5 years ago
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    uh why is u = 2pi...?

  19. Mini
    • 5 years ago
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    u = pix^2

  20. anonymous
    • 5 years ago
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    \[\iint \] \.[\iint \]

  21. amistre64
    • 5 years ago
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    0<y<1; and cbrt(y)<x<1 is your bounds right?

  22. amistre64
    • 5 years ago
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    you think changeing them would make it easier?

  23. Mini
    • 5 years ago
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    hold on phone lol

  24. amistre64
    • 5 years ago
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    {SS} [ sin(pix^2)/x^2] dy dx

  25. amistre64
    • 5 years ago
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    y=1 y=0 x=1 x=cbrt(y) can be changed to make it easier to start with a dy

  26. anonymous
    • 5 years ago
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    Amistre, brah, one day you're gonna have to give in and learn some LaTeX.

  27. amistre64
    • 5 years ago
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    lol.....nvr ;)

  28. amistre64
    • 5 years ago
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    heres our domain

    1 Attachment
  29. amistre64
    • 5 years ago
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    we can use this to switch to dydx

  30. Mini
    • 5 years ago
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    thats what i did, makes the first one extremely easy

  31. Mini
    • 5 years ago
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    u just ad a y t hen its all good

  32. amistre64
    • 5 years ago
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    0 < y < x^3 0 < x < 1 right?

  33. Mini
    • 5 years ago
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    ya then u do the first one, and its just the regular eq with a y added to it

  34. amistre64
    • 5 years ago
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    yes :)

  35. amistre64
    • 5 years ago
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    i gotta work thru it to see where se go tho :)

  36. Mini
    • 5 years ago
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    then u sub it all in and do the bounds and the x's cancel til u get xsin(pix^2)

  37. Mini
    • 5 years ago
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    thats why i skipped though, i was pretty sure my work til there was accurate

  38. Mini
    • 5 years ago
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    now ive been doing u = pix^2 du = 2pix

  39. amistre64
    • 5 years ago
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    sin(pix^2) -------- (x^3) = x sin(pix^2) x^2 lol.... its not that i dont trust you; i just dont trust me :)

  40. amistre64
    • 5 years ago
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    {S} x sin(pix^2) dx ; [0,1] right?

  41. Mini
    • 5 years ago
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    so then dx = du/2pix

  42. amistre64
    • 5 years ago
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    forget the u sub for the moment....

  43. Mini
    • 5 years ago
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    oh yeah kinda working ahead, ive done it up until here so many times haha

  44. Mini
    • 5 years ago
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    oh okay, well i tried by parts too, that pellet was NOT COOL

  45. amistre64
    • 5 years ago
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    if this was: 2pi x sin(pix^2) you could solve it right?

  46. Mini
    • 5 years ago
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    well yeah thats easy, because it cancels itself

  47. Mini
    • 5 years ago
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    its just in terms of say udu

  48. amistre64
    • 5 years ago
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    2pi [S] --- x sin(pix^2) dx then multiply by a useful form of 1 lol 2pi

  49. amistre64
    • 5 years ago
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    drag out the bottom 2pi and integrate it lol

  50. Mini
    • 5 years ago
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    but i still have the 1/2pi

  51. Mini
    • 5 years ago
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    no no see thats not my problem, he has the answer a 1 which is throwing me off

  52. amistre64
    • 5 years ago
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    ok...let me see what I gets :) i see the issue, but not the solution :)

  53. Mini
    • 5 years ago
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    follow me on my u sub i have u = pix^2 du = 2pixdx dx = du/2pix

  54. Mini
    • 5 years ago
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    when i completely substitute dx from the original problem to this du/2pix, do i have to account or add anything to that? or do i just literally plug it in the equation

  55. amistre64
    • 5 years ago
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    -cos(pi x^2) ---------- at 0 = -1/(2pi) 2pi -cos(pi x^2) ---------- at 1 = 1/(2pi) right? 2pi

  56. amistre64
    • 5 years ago
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    nah, just plug it in; its good...

  57. Mini
    • 5 years ago
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    no wait dont u have to change the bounds?

  58. amistre64
    • 5 years ago
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    1 1 --- + --- = pi .... so thats the answer 2pi 2pi

  59. Mini
    • 5 years ago
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    like say u OF 1 @ u = pix^2

  60. amistre64
    • 5 years ago
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    only if you wanna work with u and not pix^2

  61. amistre64
    • 5 years ago
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    same difference tho

  62. Mini
    • 5 years ago
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    so after i do the integration as long as i plug u back in before i use the bounds i dont have the change them correct?

  63. amistre64
    • 5 years ago
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    correct; i always mess that part up so I just re-stsute it back :)

  64. Mini
    • 5 years ago
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    yea i always just confuse myself and think too much into it

  65. amistre64
    • 5 years ago
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    but i get pi as an answer.... as long as it was all done right

  66. amistre64
    • 5 years ago
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    err....1/pi lol

  67. Mini
    • 5 years ago
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    yeah i just got that too, he still has 1 as the answer though, pisses me off

  68. Mini
    • 5 years ago
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    what goods a test review if you have the wrong answers listed!

  69. amistre64
    • 5 years ago
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    tell him to prove it; show him your work for verification :)

  70. Mini
    • 5 years ago
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    i wont see him until my exam on monday, cant really do that haha

  71. amistre64
    • 5 years ago
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    i know; right answers build confidence :)

  72. Mini
    • 5 years ago
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    well i know ive done it a dozen times and come up with the same thing, so im not worried about it, hes had wrong answers before

  73. anonymous
    • 5 years ago
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    Just use wolfram (or equivalent) to check answer.

  74. Mini
    • 5 years ago
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    thats the ONLY thing that was confusing me is because it didnt match my answers so i thought i did my u sub wrong

  75. amistre64
    • 5 years ago
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    your u sub was fine; a little more work than I would have put into it; but fine

  76. Mini
    • 5 years ago
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    i used wolfram earlier and got the same general answer with an infinite bounds

  77. Mini
    • 5 years ago
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    which is essentially the same thing

  78. amistre64
    • 5 years ago
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    i just read up on double ints last night; they remind me of nested functions in programing

  79. Mini
    • 5 years ago
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    i wont worry about it now though however...

  80. Mini
    • 5 years ago
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    if you arent busy i have a long intensive problem for u lol

  81. amistre64
    • 5 years ago
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    give me a shot at it; i gotta warm up for calc1 on monday :)

  82. Mini
    • 5 years ago
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    lets see

  83. Mini
    • 5 years ago
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    fxy if f(x,y) = sqrt(xy) - 1/(((x^2)y)

  84. Mini
    • 5 years ago
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    ill just type it as i would say it to clarify, squareroot of both x and y, minus

  85. Mini
    • 5 years ago
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    one over x squared times y

  86. amistre64
    • 5 years ago
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    \[\frac{\sqrt{xy}-1}{x^2y}\]

  87. Mini
    • 5 years ago
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    no no

  88. Mini
    • 5 years ago
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    the x^2y

  89. Mini
    • 5 years ago
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    is ONLY under the 1, nothing else

  90. amistre64
    • 5 years ago
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    \[\sqrt{xy} -\frac{1}{x^2y}\]

  91. Mini
    • 5 years ago
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    i was working on a dry erase board so i dont have the work anymore but i ended up doing i think 4 product rules lol

  92. Mini
    • 5 years ago
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    yeah thats it

  93. amistre64
    • 5 years ago
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    x^(-2) y^(-1)

  94. amistre64
    • 5 years ago
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    what are we to do with this function?

  95. amistre64
    • 5 years ago
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    x^(1/2) y^(1/2) - x^(-2) y^(-1)

  96. Mini
    • 5 years ago
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    u find der with respect to x then y

  97. Mini
    • 5 years ago
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    but when you find the der with respect to y, u use what you found with respect to x

  98. amistre64
    • 5 years ago
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    dF -\[\frac{dF}{dy} = \frac{\sqrt{x}}{2 \sqrt{y}} + \frac{\ln(y)}{x^2}\] dx

  99. amistre64
    • 5 years ago
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    \[\frac{dF}{dx} = \frac{\sqrt{y}}{2\sqrt{x}} + \frac{2}{xy}\]

  100. Mini
    • 5 years ago
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    no

  101. Mini
    • 5 years ago
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    u do dx first, then when u get that der

  102. amistre64
    • 5 years ago
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    prolly dint understand the questionthen lol

  103. Mini
    • 5 years ago
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    u do d/dy to tHAT one

  104. amistre64
    • 5 years ago
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    ohh... is my dx right lol

  105. Mini
    • 5 years ago
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    yeah its screwy because we thought that too, but it seemed to easy and then once u get the two different der what do u do

  106. Mini
    • 5 years ago
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    hold on lemme work it again

  107. Mini
    • 5 years ago
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    \[1/2(xy)^{-1/2}y+(x ^{2}y)^{-2}2xy\]

  108. Mini
    • 5 years ago
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    thats the first one

  109. amistre64
    • 5 years ago
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    \[D[(dF/dx)]/dy = \frac{1}{4\sqrt{xy}} - \frac{2x}{(xy)^2}\]

  110. Mini
    • 5 years ago
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    change everything to chain its easier to do, cause our answers arent matching up

  111. amistre64
    • 5 years ago
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    that last is messed

  112. amistre64
    • 5 years ago
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    working in a dim light and no paper :)

  113. amistre64
    • 5 years ago
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    the first term is good, or shoul dbe

  114. amistre64
    • 5 years ago
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    but +2/x * 1/y = 2ln(y)/x

  115. Mini
    • 5 years ago
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    dont even worry about ln, just makes it worse than it already is, but i dont think your first is right

  116. Mini
    • 5 years ago
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    (xy)^1/2 in terms of X

  117. amistre64
    • 5 years ago
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    Dx(sqrt(xy)) = sqrt(x)/2sqrt(y)

  118. amistre64
    • 5 years ago
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    x and y are transposed, but that should be it

  119. amistre64
    • 5 years ago
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    Csqrt(x) = C/2sqrt(x) right?

  120. Mini
    • 5 years ago
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    one half goes to the front then u have (xy) then u subtract one and get (xy)^(-1/2) all divided by 2

  121. Mini
    • 5 years ago
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    then chain the inside and get y

  122. amistre64
    • 5 years ago
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    the y term is constant with respect to x.... right?

  123. Mini
    • 5 years ago
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    yes thats why when u chain the inside u only get y, not x

  124. Mini
    • 5 years ago
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    its like 9x so to speak

  125. amistre64
    • 5 years ago
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    Csqrt(x) - C/2sqrt(x)

  126. amistre64
    • 5 years ago
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    sqrt(y)/C -> 1/C2sqrt(y) ends up 1/4sqrt(xy)

  127. Mini
    • 5 years ago
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    dont see where u are getting this other half from, you only get a half from the chain rule because its a sqrt

  128. amistre64
    • 5 years ago
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    sqrt(xy) = sqrt(x)* sqrt(y)

  129. Mini
    • 5 years ago
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    yes, but i dont see why you would do that in the first place

  130. amistre64
    • 5 years ago
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    to work them seperately; do dx first then dy and come to the proper solution..

  131. amistre64
    • 5 years ago
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    that is what you said to do right?

  132. Mini
    • 5 years ago
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    no that doesnt make any sense

  133. Mini
    • 5 years ago
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    no you have to do with respect to x to the WHOLE equation first

  134. Mini
    • 5 years ago
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    once you find with respect to x u hvae to apply with respect to y after that

  135. amistre64
    • 5 years ago
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    oh, like implicit then?

  136. Mini
    • 5 years ago
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    no lol

  137. amistre64
    • 5 years ago
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    then what you want is to hold y constant and dx first right?

  138. Mini
    • 5 years ago
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    yes i thought we already said that earlier, y is just y, it does nothing now

  139. amistre64
    • 5 years ago
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    so i split the sqrt to work with only the x

  140. amistre64
    • 5 years ago
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    sqrt(y) ------ 2sqrt(x)

  141. Mini
    • 5 years ago
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    no

  142. Mini
    • 5 years ago
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    nevermind lol

  143. amistre64
    • 5 years ago
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    you say no...but tell me why not?

  144. Mini
    • 5 years ago
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    okay we have the orignal problem blah blah

  145. Mini
    • 5 years ago
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    we take the derivative of that with ONLY respect to x

  146. Mini
    • 5 years ago
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    once we get that function we take the derivative of that with respect to y

  147. amistre64
    • 5 years ago
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    and since it is a sum we can do each term seperately...lets work the first term.

  148. amistre64
    • 5 years ago
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    sqrt(xy) = sqrt(x) * sqrt(y) by definition that is appropriate

  149. Mini
    • 5 years ago
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    u just chain rule so .5(xy)^(-1/2) times y

  150. Mini
    • 5 years ago
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    no quit doin that!

  151. Mini
    • 5 years ago
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    just look at sqrt(xy) as (xy)^(1/2)

  152. amistre64
    • 5 years ago
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    ok

  153. Mini
    • 5 years ago
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    now instead of splitting stuff up, u just made the whole thing a chain rule which is reaaaaaaaaally easy

  154. Mini
    • 5 years ago
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    you kno what chain rule is right?

  155. amistre64
    • 5 years ago
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    sqrt(Cx) ->> C/2sqrt(Cx) right?

  156. Mini
    • 5 years ago
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    yes

  157. Mini
    • 5 years ago
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    because its just (Cx)^(1/2)

  158. Mini
    • 5 years ago
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    do you kno what the chain rule is though?

  159. amistre64
    • 5 years ago
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    so that translates to: y ------- correct? 2sqrt(xy)

  160. Mini
    • 5 years ago
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    yes!

  161. amistre64
    • 5 years ago
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    lol...no, i just do double integrals for a hobby ;)

  162. Mini
    • 5 years ago
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    well with this problem its easier if you just put everything to a negative exponent

  163. Mini
    • 5 years ago
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    its verrrrrrrry long, and that makes it faster, especially when u are trying to do it in terms of x or y

  164. amistre64
    • 5 years ago
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    easier is a subjective term;

  165. Mini
    • 5 years ago
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    well bringing something to the front, minus 1, then der of the inside is pretty easy, dont think thats subjective

  166. amistre64
    • 5 years ago
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    so we got: y ------ and should dy that now right? 2sqrt(xy)

  167. Mini
    • 5 years ago
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    no u still have to do dx to the last part, that was only the first part lol

  168. amistre64
    • 5 years ago
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    the last term can wait; its part of another section of the addition

  169. Mini
    • 5 years ago
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    the eq is \[(xy)^{1/2}-1/(x ^{2}y)\]

  170. amistre64
    • 5 years ago
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    right; so the first term is derivable n its own right without the second

  171. Mini
    • 5 years ago
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    you can, you are going to have to end up doing the second, and for thought it seems simpler to just do all the dx first

  172. Mini
    • 5 years ago
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    now you just product the 1st term get y, then product the second term

  173. amistre64
    • 5 years ago
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    the length of this posting is messing up my system :) ill let you have at it alone ;)

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