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

1. anonymous

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

2. anonymous

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

3. DanJS

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

4. anonymous

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

5. anonymous

okay let me start over

6. DanJS

either is good, it isnt too much work

7. anonymous

okay so this is what iv got so far $(24(6x-3)^3)((3x+4)^{-1/3}) + ((6x-3)^4)(-(3x+4)^{-4/3})$

8. anonymous

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

9. DanJS

no

10. anonymous

so i can distribute the 24?

11. DanJS

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

12. DanJS

use any number for x

13. anonymous

okay sorry lost internet connection,

14. DanJS

maybe do the quotient rule and see if that is easier

15. anonymous

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

16. anonymous

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

17. anonymous

??

18. DanJS

that is the derivative, done..

19. DanJS

i am sure they want it all simplified though i assume

20. anonymous

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

21. 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)

22. DanJS

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

23. 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

24. DanJS

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

25. DanJS

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

26. DanJS

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

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

28. DanJS

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

29. anonymous

you mean 4/3 for your exponent right?

30. 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

31. 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)

32. DanJS

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