Calculus
posted by Mike .
Differentiate and simplify as much as possible.
Cube root(5z+6/9z+3).
The answer should be y'=
(23(9z+23)^2/3)/9(3z1)^2(5z+6)^2/3)
So far, I'm stuck at y'=[23/3(5z+6)(3z+1)][((5z+6)^1/3)/((9z+3)^1/3)]

if you let w be the fraction, then you have
w^(1/3)
which has derivative
1/3 w^(2/3) dw/dz
Now just use the quotient rule on dw/dz, and you get
69/(9z+3)^2
That gives you
1/3 * ((5z+6)/(9z+3))^(2/3) * 69/(9z+3)^2
1 * (9z+3)^(2/3) * 69

3 * (5z+6)^(2/3) * (9z+3)^2
which you can see is the desired answer. I see a (3z+1) in your answer, which is already a typo. Try breaking it up into factors as I did, and you should see things cancel. 
using the quotient rule, by first line derivative is
y' = (1/3)[(5z+6)/(39z)]^(2/3) [ (5(39z)  (9)(5z+6) ]/(39z)^2
= (1/3)[(5z+6)/(39z)]^(2/3) [ 69/(39z)^2 ]
= 23 (39z)^(2/3) / (5z+6)^2/3 [ 1/(39z)^2 ]
= not getting your answer
let's try the product rule
y = (5z+6)^(1/3) ( 39z)^(1/3)
y' = (5z+6)^(1/3) (1/3)(39z)^(4/3) (9) + ( 39z)^(1/3) (1/3)(5z+6)^(2/3) (5)
= (9/3)(5z+6)^(1/3) (39z)^(4/3) + (5/3)(39z)^(1/3) (5z+6)^(2/3)
= (1/3)(5z+6)^(2/3) (39z)^(4/3) [ 9(5z+6) + 5(39z)]
= 23(5z+6)^(2/3) (39z)^(4/3)
check through my steps 
looks like Steve and I have the same answer
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