## Calculus: Early Transcendentals 8th Edition

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