#### Answer

$15.15$

#### Work Step by Step

We are given the function:
$f(x)=e^x$
The average rate of change on $[x_0,x_1]$ is:
$\dfrac{\Delta f}{\Delta x}=\dfrac{f(x_1)-f(x_0)}{x_1-x_0}$
The instantaneous rate of change is the limit of the average rate of change.
In order to estimate the instantaneous rate of change at $x=e$, consider intervals $[x_1,x_0],[x_0,x_1]$ for $x_1$ close to $x_0$, where $x_0=e$:
$[e-0.1,e]$: $\dfrac{\Delta f}{\Delta x}=\dfrac{f(e-0.1)-f(e)}{e-0.1-e}=\dfrac{e^{e-0.1}-e^e}{-0.1}\approx 14.421187$
See table
From the table we find that the instantaneous rate of of change at $x=e$ is approximately $15.15$.