Step 1: Understanding the Concept
Slope is given by $f'(x)$. To maximize the slope, find the critical points of $f'(x)$ by setting $f''(x) = 0$.
Step 2: Evaluation
$f'(x) = e^x(sin~x + cos~x)$. $f''(x) = e^x(sin~x + cos~x + cos~x - sin~x) = 2e^x cos~x$.
Step 3: Final Calculation
Set $f''(x) = 0 \Rightarrow 2e^x cos~x = 0 \Rightarrow cos~x = 0 \Rightarrow x = \pi/2$ in the given interval.
Step 4: Conclusion
Check $f'''(x)$ at $x = \pi/2$: $f'''(x) = 2e^x(cos~x - sin~x)$. At $x = \pi/2$, $f'''(\pi/2) = -2e^{\pi/2}<0$. Thus, slope is maximum at $x = \pi/2$.
Hence, the Answer is: (b)