Question

How do you find the second derivative of `y=(3t-1)^10`?

Answer 1

I got: `y"=810(3t-1)^8`

Explanation:

The first derivative can be found using the Chain Rule to get:
`y'=10(3t-1)^(10-1)*3=30(3t-1)^9`
using it again:
`y"=30*9(3t-1)^(9-1)*3=810(3t-1)^8`