How many way ,we can show $(e^x)'=e^x$ It is famous enough that $f(x)=\exp(x) \to f'(x)=f(x)$ . For example we can show it by taylor series like below :
$$\quad{e^x=1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\frac{x^4}{4!}+...\\
(e^x)'=0+1+\frac{2x}{2!}+\frac{3x^2}{3!}+\frac{4x^3}{4!}+\frac{5x^4}{5!}+...(\text{simplify})\\(e^x)'=1+\frac{x}{1!}+\frac{x^2}{2!}+\frac{x^3}{3!}+\frac{x^4}{4!}+...=e^x}$$
or by definition we have:
$$\quad{\large e=\lim_{\Delta x \to 0 }(1+\Delta x)^{\frac{1}{\Delta x}}\\
e^x=\lim_{\Delta x \to 0 }(1+\Delta x)^{\frac{x}{\Delta x}}\\
f'(x)=\lim_{\Delta x \to 0 }\frac{f(x+\Delta x)-f(x)}{\Delta x }=\\
\large \lim_{\Delta x \to 0 }\frac{(1+\Delta x)^{\frac{x+\Delta x}{\Delta x}}-(1+\Delta x)^{\frac{x}{\Delta x}}}{\Delta x }=\\
\large \lim_{\Delta x \to 0 }\frac{(1+\Delta x)^{\frac{x}{\Delta x}}((1+\Delta x)^{\frac{\Delta x}{\Delta x}}-1)}{\Delta x }=\\
\large \lim_{\Delta x \to 0 }\frac{(1+\Delta x)^{\frac{x}{\Delta x}}((1+\Delta x)^1-1)}{\Delta x }=\\
\large \lim_{\Delta x \to 0 }\frac{(1+\Delta x)^{\frac{x}{\Delta x}}((\Delta x)}{\Delta x }=\\
\large \lim_{\Delta x \to 0 }(1+\Delta x)^{\frac{x}{\Delta x}}=\\e^x}$$
And now, my question is about other way or ways to show this fact ...
It is very interesting if there was a visual proof ...
Is there any idea to show $(\exp(x))'=\exp(x)$
Thanks in advanced .
 A: Yes, you can show this visually.
I don't know if this could be called a proof, strictly speaking, but it will indeed show it visually.
Choose any point on the graph of $y=e^x$.  Mark that point.  (Call it point $B$.)
Now mark the point on the $x$ axis with the same $x$ coordinate as point $B$.  (Call it point $C$.)
Now mark the point on the $x$ axis one unit to the left of point $C$.  Call this point $A$.
Draw a line from point $A$ to point $B$.
The line will be exactly tangent to the graph of $y=e^x$ at point $B$.
A: 
ATTENTION: No cucumbers were harmed in the making of the graphics!

Here's a proof that I thought was kind of fun:
You can try this experiment to prove $(e^x)'=e^x$.
Materials:


*

*A cucumber

*A knife

*A cutting board

*An almost infinite amount of time



Place the cucumber on the cutting board and assume its radius to be $1$ unit in length. Now, using the knife, make a small incision approximately $45^{\circ}$.

Now, continue cutting and making the cut longer and longer as you rotate the cucumber counter-clockwise (on its axis) and tilting the knife of the blade backwards while keeping the guard of the knife at the edge of the board.

The carves a somewhat accurate representation of the graph of the exponential curve, in cylindrical coordinates: $y=\exp\theta$.

Define the $x$ axis to be the bottom of the cucumber and the $y$ axis to be the central line through the cucumber.

If you change the perpendicular view, and draw an imaginary tangent line (or you can draw an actual one. Just be careful though!), the curve intersects the $y$ axis at different heights. However, the tangent will always intersect the $x$ axis at the same point. Thus,
$$(e^x)'=\dfrac {\Delta y}{\Delta x}=\dfrac {e^\theta-0}{0+1}=e^\theta$$
A: Here is a completion of the idea given by Wildcard, turning the visual demonstration into a proof.
We claim that for any $x_0$, the line 
$$y(x)=e^{x_0}(x-(x_0-1))$$
 is tangent to $e^x$ at $x_0$. This tells us, as Wildcard suggested, that the derivative of $e^x$ at $x_0$ is the slope $e^{x_0}$.
To see that this line is tangent, it suffices to show because $e^x$ is concave (a whole other story), that $y(x)\leq e^x$ for every $x$.
First note that $e^x=e^{x_0}e^{x-x_0}$, and since
 $$x-(x_0-1)=1+(x-x_0)\leq e^{x-x_0},$$
 It follows that $$y(x)\leq e^x,$$
completing our proof.
