anonymous
  • anonymous
Complex Derivatives !! if y = ((6x-3)^4)/((3x+4)^(1/3)) what is dy/dx ?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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anonymous
  • anonymous
\[y = \frac{ (6x-3)^4 }{ \sqrt[3]{3x+4} }\]
anonymous
  • anonymous
okay so let me show you what i got and then tell me what iv done wrong
DanJS
  • DanJS
did you move the root to the top as -1/3 power, and apply the product rule

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anonymous
  • anonymous
!!! no i was doing the quotient rule like the fool i am!!! very clever!!
anonymous
  • anonymous
okay let me start over
DanJS
  • DanJS
either is good, it isnt too much work
anonymous
  • anonymous
okay so this is what iv got so far \[(24(6x-3)^3)((3x+4)^{-1/3}) + ((6x-3)^4)(-(3x+4)^{-4/3})\]
anonymous
  • anonymous
quick question : i cant distribute the 24 into the (6x-3) because its cubed right?
DanJS
  • DanJS
no
anonymous
  • anonymous
so i can distribute the 24?
DanJS
  • DanJS
test and see if you forget algebra things... 24(6x - 3)^3 = the distributed value?
DanJS
  • DanJS
use any number for x
anonymous
  • anonymous
okay sorry lost internet connection,
DanJS
  • DanJS
maybe do the quotient rule and see if that is easier
anonymous
  • anonymous
not sure where to go now, do i just bring down the negative exponents?
anonymous
  • anonymous
\[(24(6x-3)^3 + (6x-3)^4)/(-(3x+4)^{4/3}+(3x+4)^{1/3})\]
anonymous
  • anonymous
??
DanJS
  • DanJS
that is the derivative, done..
DanJS
  • DanJS
i am sure they want it all simplified though i assume
anonymous
  • anonymous
yep it seems so , it dosent appear to be any of the answer choices
DanJS
  • DanJS
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)
DanJS
  • DanJS
rather it is just (a/b) + (c/d)
anonymous
  • anonymous
okay don't worry about the simplification it is just tedious work i think iv got it , thank you very much for your help
DanJS
  • DanJS
notice the parenthesis value appears in both terms, for both the different parenthesis, maybe try some factoring
DanJS
  • DanJS
\[(24(6x-3)^3)((3x+4)^{-1/3}) + ((6x-3)^4)(-(3x+4)^{-4/3})\]
DanJS
  • DanJS
or, put the negative exponents back to the bottom, then you would have to do the common denominator to add the two fractions
DanJS
  • DanJS
\[\frac{ 24(6x-3)^3 }{ (3x+4)^{1/3} } + \frac{ (6x-3)^4 }{ (3x+4)^{4/3}}\]
DanJS
  • DanJS
need to multiply first term by (3x+4)^{3/3} / (3x+4)^{3/3}
anonymous
  • anonymous
you mean 4/3 for your exponent right?
anonymous
  • anonymous
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
DanJS
  • DanJS
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)
DanJS
  • DanJS
\[\frac{ (3x+4)*24*(6x-3)^2 + (6x-3)^4 }{ (3x+4)^{4/3} }\]

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