The Derivative of $\exp(x)=e^x$ at $x=0$: $\exp'(0)=1$.
Estimation Argument - "Proof":
We'll follow the four step approach to estimate the derivative at $x =0$ by examining $h>0$ chosen close to $0$.
Recall from ELM.NEL.1 that for large values of $n =2^k, (1+\frac1n)^ n
\approx e$.
For $h$ in the difference quotient, we choose $h =
\frac1n$, for the same large value of $n$. This guarantees the value
of $h>0$ will be close to $0$.
Then at Step 2 of the four step method (subtract) we consider $e^h -1$
which is approximately $((1+ \frac 1n)^n )^h - 1= (1+ \frac 1n )
^{ n h} - 1$.
Since $n h =
1$, $(1+ \frac 1n)^{ n h} - 1= ((1+ \frac 1n)) - 1=
\frac 1n = h$.
Now we go to Step 3 (divide) and we find that when $h=\frac1n$, $ \frac {((1+ \frac 1n)^n )^h - 1}h =1$
Thus at Step 4 (Think), with $n$ getting larger, $h$ gets closer
to 0 and the estimate (which is $1$) is closer to the derivative.
So the
derivative of $f$ at $0$ must be $1$, i.e., $f '(0) = 1$
You can explore this further with the following mapping diagram.
CCD.DEest.0