## Calculus 10th Edition

$h''(1)=24.$
$h'(x)=(3)(3)\frac{1}{9}(3x+1)^{3-1}=(3x+1)^2$ $h''(x)=(3)(2)(3x+1)^{2-1}=6(3x+1)=18x+6.$ $h''(1)=18(1)+6=24.$ A computer algebra system has been used to verify this result.