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anonymous

  • one year ago

Complex Derivatives !! if y = ((6x-3)^4)/((3x+4)^(1/3)) what is dy/dx ?

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  1. anonymous
    • one year ago
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    \[y = \frac{ (6x-3)^4 }{ \sqrt[3]{3x+4} }\]

  2. anonymous
    • one year ago
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    okay so let me show you what i got and then tell me what iv done wrong

  3. DanJS
    • one year ago
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    did you move the root to the top as -1/3 power, and apply the product rule

  4. anonymous
    • one year ago
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    !!! no i was doing the quotient rule like the fool i am!!! very clever!!

  5. anonymous
    • one year ago
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    okay let me start over

  6. DanJS
    • one year ago
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    either is good, it isnt too much work

  7. anonymous
    • one year ago
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    okay so this is what iv got so far \[(24(6x-3)^3)((3x+4)^{-1/3}) + ((6x-3)^4)(-(3x+4)^{-4/3})\]

  8. anonymous
    • one year ago
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    quick question : i cant distribute the 24 into the (6x-3) because its cubed right?

  9. DanJS
    • one year ago
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    no

  10. anonymous
    • one year ago
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    so i can distribute the 24?

  11. DanJS
    • one year ago
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    test and see if you forget algebra things... 24(6x - 3)^3 = the distributed value?

  12. DanJS
    • one year ago
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    use any number for x

  13. anonymous
    • one year ago
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    okay sorry lost internet connection,

  14. DanJS
    • one year ago
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    maybe do the quotient rule and see if that is easier

  15. anonymous
    • one year ago
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    not sure where to go now, do i just bring down the negative exponents?

  16. anonymous
    • one year ago
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    \[(24(6x-3)^3 + (6x-3)^4)/(-(3x+4)^{4/3}+(3x+4)^{1/3})\]

  17. anonymous
    • one year ago
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    ??

  18. DanJS
    • one year ago
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    that is the derivative, done..

  19. DanJS
    • one year ago
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    i am sure they want it all simplified though i assume

  20. anonymous
    • one year ago
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    yep it seems so , it dosent appear to be any of the answer choices

  21. DanJS
    • one year ago
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    you can move the - exponents , but you can not do that, a * b^-1 + c*d^-1 is not same as (a +c) / (b + d)

  22. DanJS
    • one year ago
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    rather it is just (a/b) + (c/d)

  23. anonymous
    • one year ago
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    okay don't worry about the simplification it is just tedious work i think iv got it , thank you very much for your help

  24. DanJS
    • one year ago
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    notice the parenthesis value appears in both terms, for both the different parenthesis, maybe try some factoring

  25. DanJS
    • one year ago
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    \[(24(6x-3)^3)((3x+4)^{-1/3}) + ((6x-3)^4)(-(3x+4)^{-4/3})\]

  26. DanJS
    • one year ago
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    or, put the negative exponents back to the bottom, then you would have to do the common denominator to add the two fractions

  27. DanJS
    • one year ago
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    \[\frac{ 24(6x-3)^3 }{ (3x+4)^{1/3} } + \frac{ (6x-3)^4 }{ (3x+4)^{4/3}}\]

  28. DanJS
    • one year ago
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    need to multiply first term by (3x+4)^{3/3} / (3x+4)^{3/3}

  29. anonymous
    • one year ago
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    you mean 4/3 for your exponent right?

  30. anonymous
    • one year ago
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    not that it matters much now but we did it completely wrong, we were suppose to multiply both sides by Ln at the beginning and work from there

  31. DanJS
    • one year ago
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    yes, i just skipped a couple things, the net effect is what i said the common denominator to multiply everything by is the product of the two, so you get (3x+4)^{5/3} that puts the denominator to 5/3 power, and adds a factor of (3x+4)^{1/3} that can factor from the top two terms, so overall you are just multiplying the first fraction by (3x+4)^(3/3)

  32. DanJS
    • one year ago
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    \[\frac{ (3x+4)*24*(6x-3)^2 + (6x-3)^4 }{ (3x+4)^{4/3} }\]

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