Answer
\[{f^,}\,\left( x \right) = - \frac{{0.05{e^{ - 0.05x}}}}{{\,{{\left( {1 - {e^{ - 0.05x}}} \right)}^2}}}\]
Work Step by Step
\[\begin{gathered}
f\,\left( x \right) = \,{\left( {1 - {e^{ - 0.05x}}} \right)^{ - 1}} \hfill \\
\hfill \\
Chain\,\,rule \hfill \\
\hfill \\
\frac{{dy}}{{dx}} = {f^,}\,\left( {g\,\left( t \right)} \right) \cdot {g^,}\,\left( t \right) \hfill \\
\hfill \\
{f^,}\,\left( x \right) = - 1\,{\left( {1 - {e^{ - 0.05x}}} \right)^{ - 2}}\,{\left( {1 - {e^{ - 0.05x}}} \right)^,} \hfill \\
\hfill \\
therefore \hfill \\
\hfill \\
{f^,}\,\left( x \right) = - 1\,{\left( {1 - {e^{ - 0.05x}}} \right)^{ - 2}}\,\left( {0.05{e^{ - 0.05x}}} \right) \hfill \\
\hfill \\
simplify \hfill \\
\hfill \\
{f^,}\,\left( x \right) = - \frac{{0.05{e^{ - 0.05x}}}}{{\,{{\left( {1 - {e^{ - 0.05x}}} \right)}^2}}} \hfill \\
\hfill \\
\end{gathered} \]
