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anonymous

  • 5 years ago

If x^6. y^2=(x+y)^8, then y''(double derivative) is equal to

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  1. watchmath
    • 5 years ago
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    I don't like this .... Go ahead amistre64 :D

  2. anonymous
    • 5 years ago
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    lol

  3. amistre64
    • 5 years ago
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    6x^5.y^2 + x^6.2y.y' = 7(x+y)^7 . y'

  4. mattfeury
    • 5 years ago
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    (for medal #600. no pressure amistre :). )

  5. anonymous
    • 5 years ago
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    need an 8 in front of that (x+y)^(7) term

  6. amistre64
    • 5 years ago
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    6x^5.y^2 = 7(x+y)^7 . y' - x^6.2y.y' 6x^5.y^2 = (7(x+y)^7 - x^6.2y) y' 6x^5.y^2 ----------------- = y' ; whew!! first is down lol (7(x+y)^7 - x^6.2y)

  7. amistre64
    • 5 years ago
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    DOH!!

  8. anonymous
    • 5 years ago
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    . is dot product here?

  9. amistre64
    • 5 years ago
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    6x^5.y^2 ----------------- = y' ; there lol (8(x+y)^7 - x^6.2y)

  10. amistre64
    • 5 years ago
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    yeah '.' is lazt multiplication

  11. amistre64
    • 5 years ago
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    y' = (6x^5) (y^2) (7(x+y)^7 - x^6.2y)^-1 to make life easier right?

  12. watchmath
    • 5 years ago
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    Go Amistre65 for the 600 medal!! many people are watching :).

  13. amistre64
    • 5 years ago
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    y'' = f'gh + fg'h + fgh' y'' = (6x^5) (y^2) (8(x+y)^7 - x^6.2y)^-1 <-- and i fixed the typo

  14. anonymous
    • 5 years ago
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    its 64 watchmath...lol

  15. amistre64
    • 5 years ago
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    itll be 65 after this; i think im aging lol

  16. watchmath
    • 5 years ago
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    after you give him one more medal his username will change to 65 :D

  17. anonymous
    • 5 years ago
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    :-D

  18. amistre64
    • 5 years ago
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    y'' = (30x^4) (y^2) (8(x+y)^7 - x^6.2y)^-1 + (6x^5) (2y)y' (8(x+y)^7 - x^6.2y)^-1 -(6x^5) (y^2) (8(x+y)^7 - x^6.2y )^-2 (innards)

  19. anonymous
    • 5 years ago
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    just so you know amistre, the answer is zero

  20. watchmath
    • 5 years ago
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    Oh, you shouldn't let him know right now. Tell him later after he finish :).

  21. amistre64
    • 5 years ago
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    innards = 56(x+y)^6.y' - [(x^6 . 2.y') + (6x^5.2y)] --------------------------------------- y'' = (30x^4) (y^2) (8(x+y)^7 - x^6.2y)^-1 + (6x^5) (2y)y' (8(x+y)^7 - x^6.2y)^-1 -(6x^5) (y^2) (8(x+y)^7 - x^6.2y )^-2 [*] *[56(x+y)^6.y' - (x^6. 2.y') - (6x^5.2y) ] = 0 lol

  22. amistre64
    • 5 years ago
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    fill in all the y' bits and go to town ;)

  23. amistre64
    • 5 years ago
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    thats all im doin for it by the way lol

  24. amistre64
    • 5 years ago
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    and its prolly chock full of typos :_)

  25. shadowfiend
    • 5 years ago
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    BOOOOOM! 600 MEDALER!

  26. mattfeury
    • 5 years ago
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    *pops champagne*

  27. amistre64
    • 5 years ago
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    Mormons dont drink champagne....

  28. mattfeury
    • 5 years ago
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    *pops sparkling cider*

  29. amistre64
    • 5 years ago
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    lol

  30. shadowfiend
    • 5 years ago
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    But everyone enjoys the nice popping sound of a cork!

  31. anonymous
    • 5 years ago
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    oh he opened it for himself

  32. anonymous
    • 5 years ago
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    one red punch for amistre and some sugar cookies

  33. amistre64
    • 5 years ago
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    and orange jello with carrots....

  34. shadowfiend
    • 5 years ago
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    400 fans, 600 medals. You add up to 1000 now!

  35. amistre64
    • 5 years ago
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    statistically; im an enigma ;)

  36. watchmath
    • 5 years ago
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    There should be a shorter way to do this. Anybody knows?

  37. anonymous
    • 5 years ago
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    \[x ^{m}.y ^{n}=(x+y)^{m+n}\] so,y'=y/x and y''=0 this is how its solved in my book. does anyone understand what they did in here?

  38. watchmath
    • 5 years ago
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    Now I like the problem :D.

  39. anonymous
    • 5 years ago
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    did you get it watchmath?

  40. watchmath
    • 5 years ago
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    No, not yet.

  41. anonymous
    • 5 years ago
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    ok...thats everything the solution says.

  42. amistre64
    • 5 years ago
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    that would mean theres a rule of exponents for adding unlike bases somewhere right?

  43. amistre64
    • 5 years ago
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    multiplying unlike bases...

  44. watchmath
    • 5 years ago
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    I got \(y'=-y/x\) :)

  45. watchmath
    • 5 years ago
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    Is it y/x or -y/x ?

  46. anonymous
    • 5 years ago
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    x ^{m}.y ^{n}=(x+y)^{m+n} this relation is not correct.. \[2^{2}.3^{2}=4*9=36\] RHS:\[(x+y)^{m+n}=(2+3)^{2+2}=5^{4}=625\]

  47. watchmath
    • 5 years ago
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    Take natural log of both sides we have \[m\ln x+n\ln y=(m+n)\ln(x+y)\] Then \(\frac{m}{x}+\frac{ny'}{y}=\frac{(m+n)(1+y')}{x+y}\) \(\frac{m(x+y)}{x}+\frac{n(x+y)y'}{y}=\frac{(m+n)x}{x}+\frac{y(m+n)y'}{y}\) \(\frac{n(x+y)-(m+n)y}{y}\cdot y'=\frac{(m+n)x-m(x+y)}{x}\) \(\frac{nx-my}{y}\cdot y'=\frac{nx-my}{x}\) Hence \(y'=y/x\) It follows that \(xy'=y\) Then \(y'+xy''=y''\) \(xy''=0\) \(y''=0\) :D :D

  48. anonymous
    • 5 years ago
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    how did you proceed from this step y'+xy''=y''

  49. anonymous
    • 5 years ago
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    and this relation \[x ^{m}.y ^{n}=(x+y)^{m+n}\] is not true. u do agree with me right?

  50. watchmath
    • 5 years ago
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    The relation doesn't need to be true for all \(x,y\). Like \(x^2+y^2=1\) doesn't need to hold true for all \(x,y\). From \(xy'=y\) take the derivative implicityly (of course we need to use the product rule on the left)

  51. watchmath
    • 5 years ago
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    You have an interesting book. What is the title of your book?

  52. anonymous
    • 5 years ago
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    actually its a solution sheet of a practice question papers i attempted. y'+xy''=y'' xy''=0 plz explain these two steps

  53. watchmath
    • 5 years ago
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    There was a typo on my solution above \(xy'=y\) \(y'+xy''=y'\) \(xy''=0\) If \(x\neq 0\) then \(y''=0\) If \(x=0\), then plug in to the original equation we have \(y=0\) which implies that \(y''=0\) as well.

  54. anonymous
    • 5 years ago
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    oh.. i c now

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