Answer
${\bf{r}}'\left( t \right) = \left( { - 1, - 2{t^{ - 3}},\frac{1}{t}} \right)$
Work Step by Step
We have ${\bf{r}}\left( t \right) = \left( {1 - t,{t^{ - 2}},\ln t} \right)$.
So,
${\bf{r}}'\left( t \right) = \left( { - 1, - 2{t^{ - 3}},\frac{1}{t}} \right)$.
