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When we studied the number e, we found that the function

f(x)=ex

satisfies

f′(x)=ex

as is suggested by the graph below:

You are given that

limh→0eh−1h=1

(This is hard to prove - you may use it without proof.)

Prove, from first principles, that

f(x)=ex⇒f′(x)=ex
ex+h=exeh