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