Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 3 - Section 3.2 - The Product and Quotient Rules - 3.2 Exercises - Page 188: 18

Answer

$h'(r)=\dfrac{abe^{r}}{(b+e^{r})^{2}}$

Work Step by Step

$h(r)=\dfrac{ae^{r}}{b+e^{r}}$ Differentiate using the quotient rule $h'(r)=\dfrac{(b+e^{r})(ae^{r})'-(ae^{r})(b+e^{r})'}{(b+e^{r})^{2}}=\dfrac{(b+e^{r})(ae^{r})-(ae^{r})(e^{r})}{(b+e^{r})^{2}}=...$ Evaluate the products and simplify: $...=\dfrac{abe^{r}+ae^{2r}-ae^{2r}}{(b+e^{r})^{2}}=\dfrac{abe^{r}}{(b+e^{r})^{2}}$
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