To find #f'(x)# (and later #f''(x)#), you can use the Product Rule for derivatives, which says the following:
Given #f(x) = g(x) * h(x)#, then #f'(x) = g(x) * h'(x) + h(x) * g'(x)#.
In this case, we can break up #f(x) = x^(5/2)e^x# into two separate functions:
#{(g(x) = x^(5/2)),(h(x) = e^x):}#
Following the Product Rule pattern, we then have:
#f'(x) = x^(5/2)e^x = x^(5/2) * e^x + e^x * (5/2)x^(3/2)#
# = x^(5/2)e^x + 5/2 * x^(3/2)e^x #
To find #f''(x)#, you will technically need to use the Product Rule two more times, once for each term in the preceding #f'(x)#. However, the savvy reader should note that the first term of #f'(x)# is an exact match for the original #f(x)#, and thus there's no need to manually find the derivative of that term again since we already know it (it's #f'(x)# itself!
#f''(x) = underbrace(d / dx (x^(5/2)e^x))_(f'(x)) + d/dx (5/2 * x^(3/2)e^x)#
# = x^(5/2)e^x + 5/2 * x^(3/2)e^x + ( 5/2x^(3/2) * e^x + e^x * (5/2)(3/2)x^(1/2) )#
# = x^(5/2)e^x + 5/2 * x^(3/2)e^x + ( 5/2x^(3/2)e^x + 15/4 x^(1/2)e^x )#
# = x^(5/2)e^x + 5x^(3/2)e^x + 15/4 x^(1/2)e^x #
# = x^(1/2)e^x (x^2 + 5x + 15/4 ) #
