Answer
$p'\left( t \right) = \frac{t}{{{t^2} + 1}}$
Work Step by Step
$$\eqalign{
& p\left( t \right) = \ln \sqrt {{t^2} + 1} \cr
& {\text{Rewrite the function}}{\text{, recall that }}\sqrt m = {m^{1/2}} \cr
& p\left( t \right) = \ln {\left( {{t^2} + 1} \right)^{1/2}} \cr
& {\text{Using logarithmic properties}} \cr
& p\left( t \right) = \frac{1}{2}\ln \left( {{t^2} + 1} \right) \cr
& {\text{Differentiate}} \cr
& p'\left( t \right) = \frac{d}{{dx}}\left[ {\frac{1}{2}\ln \left( {{t^2} + 1} \right)} \right] \cr
& p'\left( t \right) = \frac{1}{2}\frac{d}{{dx}}\left[ {\ln \left( {{t^2} + 1} \right)} \right] \cr
& {\text{Use the formula }}\frac{d}{{dt}}\left( {\ln u} \right) = \frac{1}{u}\frac{{du}}{{dt}},{\text{ }}{\text{ }} \cr
& {\text{Let }}u = {t^2} + 1,{\text{ then}} \cr
& p'\left( t \right) = \frac{1}{2}\left( {\frac{1}{{{t^2} + 1}}} \right)\frac{d}{{dt}}\left[ {{t^2} + 1} \right] \cr
& p'\left( t \right) = \frac{1}{2}\left( {\frac{1}{{{t^2} + 1}}} \right)\left( {2t} \right) \cr
& {\text{simplify}} \cr
& p'\left( t \right) = \frac{t}{{{t^2} + 1}} \cr} $$
