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Ika_Kim

  • one year ago

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

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  1. tkhunny
    • one year ago
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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?

  2. precal
    • one year ago
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    use the quotient rule

  3. Ika_Kim
    • one year ago
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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 ?

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

  5. Ika_Kim
    • one year ago
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    ok..wait

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

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

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

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

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

  11. Ika_Kim
    • one year ago
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    the original question .

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

  13. Ika_Kim
    • one year ago
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    first question

  14. hartnn
    • one year ago
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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.

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

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

  17. Ika_Kim
    • one year ago
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    then ?

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

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

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

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

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

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

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

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

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

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

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

  29. hartnn
    • one year ago
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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

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

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

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

  33. hartnn
    • one year ago
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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

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

  35. hartnn
    • one year ago
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    welcome ^_^

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