Calculus (3rd Edition)

$h’(x) = 1$ $for$ $t\ne1$
$h(x) = \frac{t^2-1}{t-1}$ $h(x) = \frac{(t+1)(t-1)}{(t-1)}$; $t\ne1$ $h(x) = t+1$; $t\ne1$ $h’(x) = 1$ $for$ $t\ne1$