Intuition why the derivative of $e^x$ is itself Is there an intuitive reason why the constant $e$ to the power of $x$ has a derivative that equals the value of the function? I know that this is the result of differentiating, and I've seen several proofs of how you work out the derivative, I was just wondering why this is and why it is the case for a number estimated from constant growth? 
 A: Maybe a nice way to see it is the power series expansion:
$$\forall x\in \mathbb{R} \colon e^x =1 + x + \frac {x^2} {2!} + \frac {x^3} {3!} + \cdots
$$
A: By definition,$$\frac{d}{dx}a^x=\lim_{h\to0}\frac{a^{x+h}-a^x}{h}.$$Each exponential function has the nice property $a^{x+h}=a^xa^h$, so we can take out an $h$-independent $e^x$ factor, making the above limit the original function $a^x$ times just a number, namely $\lim_{h\to0}\frac{a^h-1}{h}$. This limit is in turn, by definition, the derivative of $a^x$ at $x=0$. Now if we gradually increase $a$ from just above $0$ to not quite $\infty$, $a^x$ will get steeper and steeper at $x=0$. And $e$ is just the choice of $a$ for which the slope is $1$, so that $e^x$ is its own derivative.
A: Something that might lend some intuition, is that
$$
e^x=\lim_{n\to\infty}\left(1+\frac xn\right)^n=\lim_{n\to\infty}\left(1+\frac xn\right)^{n-1}
$$
and
$$
\frac{\mathrm{d}}{\mathrm{d}x}\left(1+\frac xn\right)^n=\left(1+\frac xn\right)^{n-1}
$$
It takes a little bit to make this rigorous, but intuition and rigor often follow different paths.
A: The slope of the tangent line at point $x$ is always equal to the value of $f(x) = e^x$. This may be the intuition you're looking for. And this is true for every point $x$. And it is true for every point $x$ only for functions of the form $e^x + const$. That makes $f(x) = e^x$ quite "special" if you think about it from geometrical perspective.  
How to see this? Draw in one plane: 1) the graph of $e^x$, 2) the unit circle, 3) the line S with the equation $x=1$ (on which values of $\tan$ are measured). 
Take any point $x_0$ and draw the tangent line $L_0$ to the graph (of $e^x$) at the point $(x_0, e^{x_0})$. Now draw a line $L_1$ passing through $(0,0)$ and parallel to the tangent line $L_0$. Note the height at which $L_1$ intersects the line $S$. The special ting here is that this height is always equal to $e^{x_0}$ i.e. to the value (at point $x_0$) of the function itself. In a way that property is quite remarkable especially since it holds for every value of $x$ ($x_0$ was taken arbitrary here). 
Why remarkable? Because as we know no other function (apart from the functions $g(x) = e^x + const$) has this property (no other function has it for every $x$, I mean).  
A: If $x$ changes by $1$, then the function changes from $e^x$ to $e^{x+1} = e\cdot e^x.$ 
If $x$ changes by $2$, then the function changes from $e^x$ to $e^{x+2} = e^2 e^x.$ 
If $x$ changes by $h$, then the function changes from $e^x$ to $e^{x+2} = e^h e^x.$  In each case the change in the function is $e^x(e^h-1)$, which is $e^x$ times some number.  This is true for any base, but for $e$, the "some number" over $h$ happens to tend to $1$.   
