Answer
$\displaystyle \frac{1-\ln t}{t^{2}}$
Work Step by Step
We us the quotient rule, along with $\displaystyle \frac{d}{dt}[\ln t]=\frac{1}{t}$:
$h^{\prime}(t)=\displaystyle \frac{(\frac{1}{t})\cdot t-\ln t\cdot(1)}{t^{2}}=\frac{1-\ln t}{t^{2}}$
