Complex Derivatives !!

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

Question

Complex Derivatives !!

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

anonymous · Answer

$$y = \frac{ (6x-3)^4 }{ \sqrt[3]{3x+4} }$$

anonymous · Answer

okay so let me show you what i got and then tell me what iv done wrong

danjs · Answer

did you move the root to the top as -1/3 power, and apply the product rule

anonymous · Answer

!!! no i was doing the quotient rule like the fool i am!!! very clever!!

anonymous · Answer

okay let me start over

danjs · Answer

either is good, it isnt too much work

anonymous · Answer

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 · Answer

quick question : i cant distribute the 24 into the (6x-3) because its cubed right?

danjs · Answer

no

anonymous · Answer

so i can distribute the 24?

danjs · Answer

test and see if you forget algebra things... 

24(6x - 3)^3  = the distributed value?

danjs · Answer

use any number for x

anonymous · Answer

okay sorry lost internet connection,

danjs · Answer

maybe do the quotient rule and see if that is easier

anonymous · Answer

not sure where to go now, do i just bring down the negative exponents?

anonymous · Answer

$$(24(6x-3)^3 + (6x-3)^4)/(-(3x+4)^{4/3}+(3x+4)^{1/3})$$

anonymous · Answer

??

danjs · Answer

that is the derivative, done..

danjs · Answer

i am sure they want it all simplified though i assume

anonymous · Answer

yep it seems so , it dosent appear to be any of the answer choices

danjs · Answer

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 · Answer

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

anonymous · Answer

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 · Answer

notice the parenthesis value appears in both terms, for both the different parenthesis, 
maybe try some factoring

danjs · Answer

$$(24(6x-3)^3)((3x+4)^{-1/3}) + ((6x-3)^4)(-(3x+4)^{-4/3})$$

danjs · Answer

or, put the negative exponents back to the bottom, then you would have to do the common denominator to add the two fractions

danjs · Answer

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

danjs · Answer

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

anonymous · Answer

you mean 4/3 for your exponent right?

anonymous · Answer

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 · Answer

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 · Answer

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