Answer
$$f\left( x \right) = \frac{1}{{{{\left( {x + 1} \right)}^2}}}$$
Work Step by Step
$$\eqalign{
& f\left( x \right) = \frac{x}{{x + 1}} \cr
& {\text{Write the function as}} \cr
& f\left( x \right) = x{\left( {x + 1} \right)^{ - 1}} \cr
& {\text{Differentiate by using the product rule}} \cr
& f\left( x \right) = x\frac{d}{{dx}}\left[ {{{\left( {x + 1} \right)}^{ - 1}}} \right] + {\left( {x + 1} \right)^{ - 1}}\frac{d}{{dx}}\left[ x \right] \cr
& \cr
& {\text{By the formula in the exercice 75}}{\text{. }} \cr
& f\left( x \right) = {\left( {mx + b} \right)^n}\,\, \Rightarrow \,\,\,f'\left( x \right) = nm{\left( {mx + b} \right)^{n - 1}} \cr
& {\text{We obtain}} \cr
& f\left( x \right) = x\left[ { - 1{{\left( {x + 1} \right)}^{ - 2}}} \right] + {\left( {x + 1} \right)^{ - 1}}\left( 1 \right) \cr
& f\left( x \right) = - x{\left( {x + 1} \right)^{ - 2}} + {\left( {x + 1} \right)^{ - 1}} \cr
& f\left( x \right) = - \frac{x}{{{{\left( {x + 1} \right)}^2}}} + \frac{1}{{x + 1}} \cr
& {\text{Simplifying}} \cr
& f\left( x \right) = \frac{{ - x + x + 1}}{{{{\left( {x + 1} \right)}^2}}} \cr
& f\left( x \right) = \frac{1}{{{{\left( {x + 1} \right)}^2}}} \cr} $$
