## Calculus with Applications (10th Edition)

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