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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

!!! 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

rather it is just (a/b) + (c/d)

\[(24(6x-3)^3)((3x+4)^{-1/3}) + ((6x-3)^4)(-(3x+4)^{-4/3})\]

\[\frac{ 24(6x-3)^3 }{ (3x+4)^{1/3} } + \frac{ (6x-3)^4 }{ (3x+4)^{4/3}}\]

need to multiply first term by (3x+4)^{3/3} / (3x+4)^{3/3}

you mean 4/3 for your exponent right?

\[\frac{ (3x+4)*24*(6x-3)^2 + (6x-3)^4 }{ (3x+4)^{4/3} }\]