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Ika_Kim

ive tried to prove it many times but , i still didnt get it.... if dy/dx = -y/x+2y prove that d''y/dx''=8/(x+2y)^3

  • one year ago
  • one year ago

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  1. tkhunny
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    Do you mean \(\dfrac{-y}{x+2y}\)? That is NOT what you have written. Folks in calculus should know about the Order of Operations. Have you considered the Quotient Rule for differentiation?

    • one year ago
  2. precal
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    use the quotient rule

    • one year ago
  3. Ika_Kim
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    really ? sorry i dont know how to write it in the right way...yes i use that quotient rule , but then i got this\[\frac{ -\frac{ dy }{ dx }.x+2y-(2y.\frac{ dy }{ dx }) }{ (x+2y)^2 }\] then what should i do ?

    • one year ago
  4. tkhunny
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    You should try that again. The whole numerator is a mess.

    • one year ago
  5. Ika_Kim
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    ok..wait

    • one year ago
  6. Ika_Kim
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    someby pls help me..i dont get it..... :{

    • one year ago
  7. tkhunny
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    Numerator: (x+2y)(-y') - (-y)(1+2y') Look at it very carefully.

    • one year ago
  8. Ika_Kim
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    oh yeahh my numerator is really a mess..then what should i do ?

    • one year ago
  9. Ika_Kim
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    what should i do ~~~~~ omgg

    • one year ago
  10. hartnn
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    even i tried this more than once, and i don't get numerator = 8, is anything else given ?

    • one year ago
  11. Ika_Kim
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    the original question .

    • one year ago
  12. Ika_Kim
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    idk what im doing -.-

    • one year ago
  13. Ika_Kim
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    first question

    • one year ago
  14. hartnn
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    ohhh...so, xy+y^2 =4.... then its easy, did you get that numerator first ? next thing will be to simplify it.

    • one year ago
  15. hartnn
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    got this : (x+2y)(-y') - (-y)(1+2y') first ?

    • one year ago
  16. Ika_Kim
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    yes i got it just now .haha

    • one year ago
  17. Ika_Kim
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    then ?

    • one year ago
  18. hartnn
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    now use, y'=-y/(x+2y)

    • one year ago
  19. Ika_Kim
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    yes im using that..then quotient rule ?

    • one year ago
  20. hartnn
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    (x+2y)(-y') - (-y)(1+2y') is the result of numerator AFTER applying quotient rule, now you have to simplify

    • one year ago
  21. tkhunny
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    Just for the record, it is ALWAYS more beneficial to show the ORIGINAL problem statement.

    • one year ago
  22. hartnn
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    yeah, that would have saved me few minutes...

    • one year ago
  23. Ika_Kim
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    to be honest im stucking again

    • one year ago
  24. hartnn
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    what u got after substituting y'=-y/(x+2y) in that numerator ?

    • one year ago
  25. Ika_Kim
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    \[2y-\frac{ 2y^2 }{ x+2y }\] what about this ? for that numerator

    • one year ago
  26. Ika_Kim
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    \[\frac{ 2xy+2y }{ (x+2y)^3 }\]

    • one year ago
  27. Ika_Kim
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    this ???????

    • one year ago
  28. Ika_Kim
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    then subtitute (0,2) into that equation ?

    • one year ago
  29. hartnn
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    its actually this : \(\huge \frac{ 2xy+2y^2 }{ (x+2y)^3 }=\frac{ 2(xy+y^2) }{ (x+2y)^3 }\) now use xy+y^2 =4 and you are done

    • one year ago
  30. hartnn
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    \(2y-\frac{ 2y^2 }{ x+2y }\) was correct

    • one year ago
  31. Ika_Kim
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    \[\frac{ 2xy+2y^2}{(x+2y)^3}\]

    • one year ago
  32. Ika_Kim
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    this ????am i right ??

    • one year ago
  33. hartnn
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    yeah, factoring out 2 will give you 2 (xy+y^2) and xy+y^2 as per the question is 4

    • one year ago
  34. Ika_Kim
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    ohhhh........i see ur answer........and i totally got it ! thanks a lot @hartnn !!:) really appreciate ^_^

    • one year ago
  35. hartnn
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    welcome ^_^

    • one year ago
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