Answer
\[s\,\left( t \right) = \frac{{ - {e^{ - t}} + \frac{1}{t}}}{{2\,{{\left( {{e^{ - t}} + \ln 2t} \right)}^{1/2}}}}\]
Work Step by Step
\[\begin{gathered}
s\,\left( t \right) = \sqrt {{e^{ - t}} + \ln 2t} \hfill \\
\operatorname{Re} write\,\,the\,\,function \hfill \\
s\,\left( t \right) = \,{\left( {{e^{ - t}} + \ln 2t} \right)^{1/2}} \hfill \\
Use\,\,the\,\,chain\,\,rule \hfill \\
s\,\left( t \right) = \frac{1}{2}\,{\left( {{e^{ - t}} + \ln 2t} \right)^{1/2 - 1}}\,{\left( {{e^{ - t}} + \ln 2t} \right)^,} \hfill \\
Use\,\,the\,\,formulas \hfill \\
\frac{d}{{dx}}\,\,\left[ {{e^{g\,\left( x \right)}}} \right] = {e^{g\,\left( x \right)}}{g^,}\,\left( x \right) \hfill \\
\frac{d}{{dx}}\,\,\left[ {\ln g\,\left( x \right)} \right] = \frac{{{g^,}\,\left( x \right)}}{{g\,\left( x \right)}} \hfill \\
Then \hfill \\
s\,\left( t \right) = \frac{1}{2}\,{\left( {{e^{ - t}} + \ln 2t} \right)^{ - 1/2}}\,\left( { - {e^{ - t}} + \frac{1}{t}} \right) \hfill \\
s\,\left( t \right) = \frac{{ - {e^{ - t}} + \frac{1}{t}}}{{2\,{{\left( {{e^{ - t}} + \ln 2t} \right)}^{1/2}}}} \hfill \\
\end{gathered} \]
