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bmelyk

  • 3 years ago

just wondering how to proceed with this problem.

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  1. bmelyk
    • 3 years ago
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    Use the appropriate rule or combination of rules to find the derivative of the function defined below. y=\[(16x-4)\sqrt{8x-2}\]

  2. bmelyk
    • 3 years ago
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    i know i have to use my product rule, when i do that i get: (16x-4)(1/2)(8x-2)^-1/2 + (8x-2)^1/2 *16

  3. hartnn
    • 3 years ago
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    u forgot chain rule for (8x-2)

  4. bmelyk
    • 3 years ago
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    oh times 8 in there lol. i had that wrote down, lol just forgot to type it.

  5. hartnn
    • 3 years ago
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    then u are correct

  6. bmelyk
    • 3 years ago
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    okay so that i got: \[\frac{ 4(16x-4) }{ \sqrt{8x-2} }+16\sqrt{8x-2}\]

  7. hartnn
    • 3 years ago
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    absolutely correct!

  8. bmelyk
    • 3 years ago
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    would i take my 4 out of the numerator so i got 16(4x-1) ?

  9. hartnn
    • 3 years ago
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    thats not much of a simplification, but u can !

  10. bmelyk
    • 3 years ago
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    is tehre a way to further simplify this> my answer isn't matching any of the answers on my assingment (this one is a multiple choice)

  11. hartnn
    • 3 years ago
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    what are the options ? u can factor out 16 from both the terms..

  12. bmelyk
    • 3 years ago
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    a) 16- 4/squrt8x-2 B)16(squrt8x-2)+32 times 16x-4/squrt8x-2 c)16-16/squrt8x-2 d)64/squrt8x-2 e)32*2^1/2*x^1/2 -16*2^1/2+(16-4x)/squrt8x-2 f)8*csgn(4x-1)*2^1/2 g)16*squrt8x-2+4(16x-4/squrt8x-2) h)8(16x-4)(squrt8x-2)

  13. hartnn
    • 3 years ago
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    ohh, u can cross-multiply to get the common denominator as \(\sqrt{}8x-2\)

  14. bmelyk
    • 3 years ago
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    so then i have\[4(16x-4)+16\sqrt{8x-2}(\sqrt{8x-2})\]

  15. bmelyk
    • 3 years ago
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    ??

  16. hartnn
    • 3 years ago
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    which is? \(\sqrt{8x-2}\sqrt{8x-2}=8x-2\)

  17. bmelyk
    • 3 years ago
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    i had 320x-16 as my numerator but that's not there as an answer either.

  18. bmelyk
    • 3 years ago
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    is the answer c?

  19. hartnn
    • 3 years ago
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    checking, wait..your c option is weird...16-16=0

  20. bmelyk
    • 3 years ago
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    \[16-\frac{ 16 }{ \sqrt{8x-2} }\]

  21. bmelyk
    • 3 years ago
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    that's what it actually is, it's a little weird when i wrote it the other way lol..

  22. bmelyk
    • 3 years ago
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    so is that right tho?

  23. Chlorophyll
    • 3 years ago
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    My answer's the denom. cancel out! However the way your post the option too complicated to read !!!

  24. bmelyk
    • 3 years ago
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    how did your answer look? i can re-write them out.

  25. bmelyk
    • 3 years ago
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    \[a) 16-\frac{ 4 }{ \sqrt{8x-2} }\]

  26. Chlorophyll
    • 3 years ago
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    = 24 √ ( 8x -1)

  27. bmelyk
    • 3 years ago
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    \[b) 16\sqrt{8x-2}+32*\frac{ 16x-4 }{ \sqrt{8x-2} }\]

  28. bmelyk
    • 3 years ago
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    \[c)16-\frac{ 16 }{ 8x-2 }\]

  29. bmelyk
    • 3 years ago
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    \[d)\frac{ 64 }{ \sqrt{8x-2} }\]

  30. bmelyk
    • 3 years ago
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    \[e) 32*2^{\frac{ 1 }{ 2 }}*x^\frac{ 1 }{ 2 }-16*2^\frac{ 1 }{ 2 }+(16x-4)*\frac{ 2^\frac{ 1 }{ 2 } }{ x \frac{ 1 }{ 2 } }\]

  31. bmelyk
    • 3 years ago
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    \[f)8*csgn(4x-1)*2^\frac{ 1 }{ 2 }\]

  32. bmelyk
    • 3 years ago
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    are you sure?

  33. bmelyk
    • 3 years ago
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    \[g)16\sqrt{8x-2}+4\frac{ 16-4x }{ \sqrt{8x-2}}\]

  34. bmelyk
    • 3 years ago
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    no idea hah.

  35. bmelyk
    • 3 years ago
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    \[h)8(16-4x)(\sqrt{8x-2}\]

  36. hartnn
    • 3 years ago
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    isn't g) option the same thing what we got ?

  37. bmelyk
    • 3 years ago
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    omg.. look at that.

  38. bmelyk
    • 3 years ago
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    hhaha so it is!!!

  39. bmelyk
    • 3 years ago
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    if you wanna check out my lastest/last question that would be great!

  40. Chlorophyll
    • 3 years ago
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    So it turns out that there's no need to simplified =)

  41. bmelyk
    • 3 years ago
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    i had the right answer all along haha.

  42. bmelyk
    • 3 years ago
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    http://openstudy.com/study#/updates/50902c28e4b043900ad9134d please : )

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