Answer
\[{h^,}\,\left( z \right) = \frac{{{e^z}}}{{\,\left( {1 + {e^z}} \right)\ln 10}}\]
Work Step by Step
\[\begin{gathered}
h\,\left( z \right) = \log \,\left( {1 + {e^z}} \right) \hfill \\
Find\,\,the\,\,derivative\,\,of\,\,the\,\,\,function \hfill \\
{h^,}\,\left( z \right) = \,\,\left[ {\log \,\left( {1 + {e^z}} \right)} \right] \hfill \\
Use\,\,the\,\,formula \hfill \\
\frac{d}{{dx}}\,\,\left[ {{{\log }_a}g\,\left( x \right)} \right] = \frac{1}{{\ln a}}\,\left( {\frac{{{g^,}\,\left( x \right)}}{{g\,\left( x \right)}}} \right)\, \hfill \\
Then \hfill \\
{h^,}\,\left( z \right) = \frac{1}{{\ln 10}}\,\left( {\frac{{\,\left( {1 + {e^z}} \right)}}{{1 + {e^z}}}} \right) \hfill \\
{h^,}\,\left( z \right) = \frac{1}{{\ln 10}}\,\left( {\frac{{{e^z}}}{{1 + {e^z}}}} \right) \hfill \\
Multiplying \hfill \\
{h^,}\,\left( z \right) = \frac{{{e^z}}}{{\,\left( {1 + {e^z}} \right)\ln 10}} \hfill \\
\end{gathered} \]
