## Calculus: Early Transcendentals (2nd Edition)

Published by Pearson

# Chapter 1 - Functions - 1.1 Review of Functions - 1.1 Exercises - Page 11: 61

#### Answer

$= \frac{1}{{\,\left( {x + h + 1} \right)\,\left( {x + 1} \right)}}$

#### Work Step by Step

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

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