Here is the equation and how I worked it out. Am I correct?

$\displaystyle y=(3x+4)^\frac{1}{3}$ what is $\displaystyle y'$

Step 1: Chain rule?

$\displaystyle y=(3) \frac{1}{3}(3x+4)^\frac{-1}{3}$

Step 2: Simplify

$\displaystyle (x+1)^\frac{-1}{3}$. Since I believe the 3 will take care of the 1/3 multiple.

so ...

$\displaystyle y'=(x+1)^\frac{-1}{3}$?