Answer
$f'\left( t \right) = \frac{{{e^{1/t}}}}{{{t^2}\sqrt {{t^2} - 1} }}\left( {{t^3} - {t^2} +1} \right)$
Work Step by Step
$$\eqalign{
& f\left( t \right) = {e^{1/t}}\sqrt {{t^2} - 1} \cr
& {\text{Differentiating}} \cr
& f'\left( t \right) = \frac{d}{{dt}}\left[ {{e^{1/t}}\sqrt {{t^2} - 1} } \right] \cr
& {\text{Use the product rule for derivatives}} \cr
& f'\left( t \right) = {e^{1/t}}\frac{d}{{dt}}\left[ {\sqrt {{t^2} - 1} } \right] + \sqrt {{t^2} - 1} \frac{d}{{dt}}\left[ {{e^{1/t}}} \right] \cr
& {\text{Use the general power rule to }}\frac{d}{{dt}}\left[ {\sqrt {{t^2} - 1} } \right]{\text{ and the chain rule}} \cr
& f'\left( t \right) = {e^{1/t}}\left( {\frac{1}{2}{{\left( {{t^2} - 1} \right)}^{ - 1/2}}\frac{d}{{dt}}\left[ {{t^2} - 1} \right]} \right) + \sqrt {{t^2} - 1} \left( {{e^{1/t}}} \right)\frac{d}{{dt}}\left[ {\frac{1}{t}} \right] \cr
& {\text{Compute derivatives and simplify}} \cr
& f'\left( t \right) = {e^{1/t}}\left( {\frac{1}{{2\sqrt {{t^2} - 1} }}\left( {2t} \right)} \right) + \sqrt {{t^2} - 1} \left( {{e^{1/t}}} \right)\left( { - \frac{1}{{{t^2}}}} \right) \cr
& f'\left( t \right) = \frac{{t{e^{1/t}}}}{{\sqrt {{t^2} - 1} }} - \frac{{{e^{1/t}}}}{{{t^2}}}\sqrt {{t^2} - 1} \cr
& {\text{Factoring}} \cr
& f'\left( t \right) = \frac{{{e^{1/t}}}}{{{t^2}\sqrt {{t^2} - 1} }}\left( {{t^3} - {t^2}+ 1} \right) \cr} $$
