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anonymous

  • one year ago

Find y" by implicit differentiation 9x^2 + y^2 =9

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  1. anonymous
    • one year ago
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    bye girl dont go posting questions on our chat log

  2. anonymous
    • one year ago
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    Why can't I post it here?

  3. amistre64
    • one year ago
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    what is your workings for y'?

  4. zepdrix
    • one year ago
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    \(\Large\bf \color{#CC0033}{\text{Welcome to OpenStudy! :)}}\) Starting with a circle? Oo these are fun. Oh wait that's not a circle :d

  5. anonymous
    • one year ago
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    i was joking

  6. anonymous
    • one year ago
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    I got my y' as (-9x)/y And y" as (-9y-9xy')/y^2 I was trying to substitute y' back into y" But then I got (81xy+81x^2)/y^3 However the answer is ((9)(9x^2+y^2))/y^3

  7. amistre64
    • one year ago
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    9x^2 + y^2 =9 9(2)x + 2y y' =0 9x + y y' =0 ; your y' is fine now derive again; notice that yy' is a product tho

  8. anonymous
    • one year ago
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    And I don't understand why is there a y^2 on top

  9. amistre64
    • one year ago
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    9x + y y' =0 9 + y'y' + yy'' = 0 yy'' = -(9 + y'^2) y'' = -(9 + y'^2)/y

  10. DanJS
    • one year ago
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    y is a function of x, to take the derivative of y w.r.t. x, you have to use the chain rule

  11. amistre64
    • one year ago
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    y'' = -(9 + y'^2)/y y'' = -(9 + (-9x/y )^2)/y y'' = -(9 + 81x^2/y^2)/y

  12. amistre64
    • one year ago
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    multiply top and bottom by y^2 to clear the fraction in the numerator

  13. DanJS
    • one year ago
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    \[\frac{ d }{ dx }y(x) = \frac{ d }{ dy }y(x)*\frac{ dy }{ dx }\]

  14. zepdrix
    • one year ago
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    `I got my y' as (-9x)/y` `And y" as (-9y-9xy')/y^2` These steps look correct ^ Maybe you just ran into a little trouble plugging in your y'. Hmm.

  15. zepdrix
    • one year ago
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    Oh there should be another negative in the numerator, I missed that. -(-9x)y'

  16. zepdrix
    • one year ago
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    It looks like you chose the quotient rule route, so remember that it's subtraction in the numerator, and we're carrying around a (-9x).

  17. anonymous
    • one year ago
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    As my y' is (-9x/y) I used quotient rule to derive my y" That's y I came up with : Y"=(-9y-9xy')/y^2 y"= (-9y-9x)(-9x/y) / y^2 Y"= (81xy+81x^2)/y^3 However the answer shows: 9(9x^2+y^2)/y^3

  18. amistre64
    • one year ago
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    using product (it tends to be simpler to follow) y' = -9xy^(-1) y'' = -9y^(-1) + 9xy'y^(-2) -9y^(-1) + 9x(-9xy^(-1))y^(-2) -9 y^(-1) -81x^2 y^(-3)

  19. amistre64
    • one year ago
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    you sure the answer is without a negative?

  20. anonymous
    • one year ago
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    There is a negative sign I'm sorry I missed it

  21. zepdrix
    • one year ago
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    \[\large\rm y''=\frac{-9y-(-9x)\color{orangered}{y'}}{y^2}=y''=\frac{-9y-(-9x)\color{orangered}{\frac{(-9x)}{y}}}{y^2}\]Depending on how you proceeded from here.... I would probably choose to factor a 1/y from each term in the numerator,\[\large\rm y''=\frac{1}{y}\cdot\frac{-9y^2+81x^2}{y^2}\]That last step is one giving you trouble? :o

  22. zepdrix
    • one year ago
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    I missed a negative again :) lol\[\large\rm y''=\frac{1}{y}\cdot\frac{-9y^2-81x^2}{y^2}\]

  23. anonymous
    • one year ago
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    Ok I'm trying to figure it out right now. Thanks for all of your help:)!!!!!!!!!!!! Thanks!

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