Answer
\[ = \frac{1}{{\,\left( {x + h + 1} \right)\,\left( {x + 1} \right)}}\]
Work Step by Step
\[\begin{gathered}
f\,\left( x \right) = \frac{x}{{x + 1}} \hfill \\
\hfill \\
Use{\text{ }}the{\text{ }}definition{\text{ }}of{\text{ }}derivative \hfill \\
\hfill \\
= \frac{{f\,\left( {x + y} \right) - f\,\left( x \right)}}{h} = \frac{{\frac{{x + h}}{{x + h + 1}} - \frac{x}{{x + 1}}}}{h} \hfill \\
\hfill \\
combine\,\,fractions \hfill \\
\hfill \\
= \frac{{\frac{{\,\left( {x + h} \right)\,\left( {x + 1} \right) - x\,\left( {x + h + 1} \right)}}{{\,\left( {x + h + 1} \right)\,\left( {x + 1} \right)}}}}{h} \hfill \\
\hfill \\
multiply \hfill \\
\hfill \\
= \frac{{\frac{{{x^2} + x + xh + h - {x^2} - xh - x}}{{\,\left( {x + h + 1} \right)\,\left( {x + 1} \right)}}}}{h} \hfill \\
\hfill \\
simplify \hfill \\
\hfill \\
= \frac{{\frac{h}{{\,\left( {x + h + 1} \right)\,\left( {x + 1} \right)}}}}{h} \hfill \\
\hfill \\
= \frac{h}{{\,\left( {x + h + 1} \right)\,\left( {x + 1} \right)h}} \hfill \\
\hfill \\
= \frac{1}{{\,\left( {x + h + 1} \right)\,\left( {x + 1} \right)}} \hfill \\
\hfill \\
\end{gathered} \]
