Answer
\[\bar c{\,^,}\left( x \right) = \frac{{x - \,\left( {x + 5} \right)\ln \,\left( {x + 5} \right)}}{{{x^2}\,\left( {x + 5} \right)}}\]
Work Step by Step
\[\begin{gathered}
c\,\left( x \right) = \ln \,\left( {x + 5} \right) \hfill \\
The\,\,average\,\,cost\,\,is\,\,defined\,\,by \hfill \\
\bar c\,\left( x \right) = \frac{{c\,\left( x \right)}}{x} \hfill \\
\bar c\,\left( x \right) = \frac{{\ln \,\left( {x + 5} \right)}}{x} \hfill \\
The\,\,marginal\,\,cost\,\,is\,\,given\,\,by \hfill \\
\frac{d}{{dx}}\,\,\left[ {\bar c\,\left( x \right)} \right] = \frac{d}{{dx}}{\left[ {\frac{{\ln \,\left( {x + 5} \right)}}{x}} \right]^,} \hfill \\
Use\,\,the\,\,quotient\,\,rule \hfill \\
\bar c{\,^,}\left( x \right) = \frac{{x\,\,{{\left[ {\ln \,\left( {x + 5} \right)} \right]}^,} - \ln \,\left( {x + 5} \right)\,{{\left( x \right)}^,}}}{{{x^2}}} \hfill \\
\bar c{\,^,}\left( x \right) = \frac{{x\,\left( {\frac{1}{{x + 5}}} \right) - \ln \,\left( {x + 5} \right)}}{{{x^2}}} \hfill \\
Simplifying \hfill \\
\bar c{\,^,}\left( x \right) = \frac{{\frac{x}{{x + 5}} - \ln \,\left( {x + 5} \right)}}{{{x^2}}} \hfill \\
\bar c{\,^,}\left( x \right) = \frac{{x - \,\left( {x + 5} \right)\ln \,\left( {x + 5} \right)}}{{{x^2}\,\left( {x + 5} \right)}} \hfill \\
\hfill \\
\end{gathered} \]
