How do you find the derivative of #y=e^x#?

1 Answer
Wataru
Aug 31, 2014

This is one of the favorite function to take the derivatives of.
#y'=e^x#

If you wish to find this derivative by the limit definition, then here is how we find it. First, we have to know the following property of #e#:
#lim_{h to 0}{e^h-1}/{h}=1#.
(Note: This means that the slope of #y=e^x# at #x=0# is #1#.)

By the limit definition of the derivative, we have
#y'=lim_{h to 0}{e^{x+h}-e^x}/h =lim_{h to 0}{e^x cdot e^h-e^x}/h#
by factoring out #e^x#,
#=lim_{h to 0}{e^x(e^h-1)}/h=e^x lim_{h to 0}{e^h-1}/h#
by the property of #e# mentioned above,
#=e^x cdot 1=e^x#

Hence, the derivative of #e^x# is itself.

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