## Calculus: Early Transcendentals (2nd Edition)

Published by Pearson

# Chapter 3 - Derivatives - 3.4 The Product and Quotient Rules - 3.4 Exercises: 59

#### Answer

$h'(x)=\dfrac{e^{x}(x^{2}+x+1)}{(x+1)^{2}}$

#### Work Step by Step

$h(x)=\dfrac{xe^{x}}{x+1}$ Start the differentiation process by using the quotient rule: $h'(x)=\dfrac{(x+1)(xe^{x})'-(xe^{x})(x+1)'}{(x+1)^{2}}=...$ Evaluate the derivatives indicated in the numerator and simplify. Use the product rule to evaluate $(xe^{x})'$: $...=\dfrac{(x+1)[x(e^{x})'+(x)'e^{x}]-xe^{x}(1)}{(x+1)^{2}}=...$ $...=\dfrac{(x+1)(xe^{x}+e^{x})-xe^{x}}{(x+1)^{2}}=...$ Take out common factor $e^{x}$ from $(xe^{x}+e^{x})$: $...=\dfrac{e^{x}(x+1)(x+1)-xe^{x}}{(x+1)^{2}}=\dfrac{e^{x}(x+1)^{2}-xe^{x}}{(x+1)^{2}}=...$ Take out common factor $e^{x}$ from the numerator: $...=\dfrac{e^{x}[(x+1)^{2}-x]}{(x+1)^{2}}=...$ Simplify again: $...=\dfrac{e^{x}(x^{2}+2x+1-x)}{(x+1)^{2}}=\dfrac{e^{x}(x^{2}+x+1)}{(x+1)^{2}}$

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