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

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

Mathematics
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\[y = \frac{ (6x-3)^4 }{ \sqrt[3]{3x+4} }\]
okay so let me show you what i got and then tell me what iv done wrong
did you move the root to the top as -1/3 power, and apply the product rule

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

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