Answer
$$\frac{d}{{dx}}\left[ {\overline C \left( x \right)} \right] = \frac{{x{e^{ - x}} - 10 + {e^{ - x}}}}{{{x^2}}}$$
Work Step by Step
$$\eqalign{
& C\left( x \right) = 10 - {e^{ - x}} \cr
& {\text{The marginal average cost is the derivative of the average cost function}}{\text{. then}} \cr
& {\text{the marginal average cost is given by}} \cr
& \frac{d}{{dx}}\left[ {\overline C \left( x \right)} \right] = \frac{d}{{dx}}\left[ {\frac{{10 - {e^{ - x}}}}{x}} \right] \cr
& {\text{by using the quotient rule}} \cr
& \frac{d}{{dx}}\left[ {\overline C \left( x \right)} \right] = \frac{{x\frac{d}{{dx}}\left[ {10 - {e^{ - x}}} \right] - \left( {10 - {e^x}} \right)\frac{d}{{dx}}\left[ x \right]}}{{{x^2}}} \cr
& \frac{d}{{dx}}\left[ {\overline C \left( x \right)} \right] = \frac{{x\left( {{e^{ - x}}} \right) - \left( {10 - {e^{ - x}}} \right)\left( 1 \right)}}{{{x^2}}} \cr
& {\text{simplify}} \cr
& \frac{d}{{dx}}\left[ {\overline C \left( x \right)} \right] = \frac{{x{e^{ - x}} - 10 + {e^{ - x}}}}{{{x^2}}} \cr
& \cr} $$
