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