Euler's number #e# makes our lives a little easier since the slope of #f(x)=e^x# at #x=0# is exactly one. In other words, #e# is a base such that
#f'(0)=lim_{h to 0}{e^(0+h}-e^0}/h=lim_{h to 0}{e^h-1}/h=1#,
which is quite useful in finding #f'(x)#. Let us use the definition to find the derivative of #f(x)#.
#f'(x)=lim_{h to 0}{e^{x+h}-e^x}/{h}=lim_{h to 0}{e^x cdot e^h-e^x}/h#
by pulling #e^x# out of the limit,
#=e^x lim_{h to 0}{e^h-1}/h=e^x cdot 1=e^x#.
Hence, the derivative of #e^x# is itself, which is very convenient.
I hope that this was helpful.
