1. Derivatives (checking work)

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

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

Step 1: Chain rule?

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

Step 2: Simplify

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

so ...

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

2. Nearly. Look closely at the power of the function.

It should be $y' = (3x+4)^{-\frac{2}{3}}$

Remember you're taking 1 off the power, and $\frac{1}{3} -1 = -\frac{2}{3}$