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anonymous

  • 5 years ago

trying to find the second derivative in this question......3x-3 over 2x+4. the Vertical Asy is -2 and the Horizontal Asy is 3/2...the xint is (1,0) and the yint is (0,-3/4) please give me step buy step

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  1. anonymous
    • 5 years ago
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    can you find the first derivative? because that is the only hard part

  2. anonymous
    • 5 years ago
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    quotient rule give \[\frac{d}{dx}\frac{3x-3}{2x+4}=\frac{(2x+4)\times 3-(3x+3)\times 2}{(2x+4)^2}\]

  3. anonymous
    • 5 years ago
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    yes i have the first one all ready

  4. anonymous
    • 5 years ago
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    so you end up with \[\frac{9}{(2x+4)^2}=9(2x+4)^{-2}\] now second should be easy

  5. anonymous
    • 5 years ago
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    \[-18(2x+4)^{-3}\times 2=-36(2x+4)^{-3}=-\frac{36}{(2x+4)^3}\]

  6. anonymous
    • 5 years ago
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    you can cancel a little because \[(2x+4)^3=(2(x+2))^3=8(x+2)^3\]

  7. anonymous
    • 5 years ago
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    im so confused

  8. anonymous
    • 5 years ago
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    ok which step? did you get the first derivative?

  9. anonymous
    • 5 years ago
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    i got (2x+4)-2(3x-3) over (2x+4)^2

  10. anonymous
    • 5 years ago
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    then you have to do the algebra in the numerator

  11. anonymous
    • 5 years ago
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    should be \[(2x+4)3-2(3x-3) \]

  12. anonymous
    • 5 years ago
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    in front of the 2x+4 there is suppose to be a 3

  13. anonymous
    • 5 years ago
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    yes it is the denominator times the derivative of the numerator as the first term

  14. anonymous
    • 5 years ago
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    minus the numerator times the derivative of the denominator

  15. anonymous
    • 5 years ago
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    if you multiply out the x terms add up to zero and you get 9

  16. anonymous
    • 5 years ago
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    so 18 over (2x+4)^2 ?

  17. anonymous
    • 5 years ago
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    yes

  18. anonymous
    • 5 years ago
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    \[(2x+4)3-2(3x-3)\] \[6x+12-6x+6=18\] yes

  19. anonymous
    • 5 years ago
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    my equation is 3(2x+4)-2(3x-3)

  20. anonymous
    • 5 years ago
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    yes multiply out and you will get 18

  21. anonymous
    • 5 years ago
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    ok then what do i do to get the second answer

  22. anonymous
    • 5 years ago
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    you have \[\frac{18}{(2x+4)^2}\] so rather than using the quotient rule again, rewrite in exponential form as \[18\times (2x+4)^{-2}\] and use the power rule (and the chain rule)

  23. anonymous
    • 5 years ago
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    ok so multiple it out?

  24. anonymous
    • 5 years ago
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    power rule right? \[\frac{d}{dx}x^n=nx^{n-1}\]

  25. anonymous
    • 5 years ago
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    i made a mistake, sorry

  26. anonymous
    • 5 years ago
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    \[\frac{d}{dx}18(2x+4)^{-2}=-2\times 18(2x+4)^{-2-1}\times 2\] \[=-72(2x+4)^{-3}=-\frac{72}{(2x+4)^3}\] but you can simplify this

  27. anonymous
    • 5 years ago
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    because \[(2x+4)^3=(2(x+2))^3=8(x+2)^3\] and \[\frac{72}{8}=9\]

  28. anonymous
    • 5 years ago
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    so your "final answer" will be \[-\frac{9}{(x+2)^3}\]

  29. anonymous
    • 5 years ago
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    thank you so much

  30. anonymous
    • 5 years ago
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    yw

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