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

Question

1388 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-03-01T00:21:10

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

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:
$y"=30*9(3t-1)^(9-1)*3=810(3t-1)^8$

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