Answer
$h’(x) = 1 $ $for$ $t\ne1$
Work Step by Step
$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 $ $for$ $t\ne1$
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