Proof of $\frac{d}{dx}e^x = e^x$ I'm working through the proof of $\frac{d}{dx}e^x = e^x$, and trying to understand it, but my mind has gotten stuck at the last step.
Starting with the definition of a derivative, we can formulate it like so:
$$\frac{d}{dx} e^x = \lim_{h \to 0} \frac{e^{x+h}-e^x}{h}$$
After some algebra, we arrive at:
$$\frac{d}{dx} e^x = e^x \lim_{h \to 0} \frac{e^h-1}{h}$$
As $h\to0$, the expression approaches $\frac{0}{0}$, which makes it indeterminate. And, this is where my understanding ends. I've tried looking at wikipedia and other descriptions of the proof, but couldn't understand those explanations. It has usually been something along the lines of, "plot $\frac{e^x - 1}{x}$ and see the function's behavior at $0$," which ends up approaching $1$, which can substitute the limit to give the result of the derivative:
$$\frac{d}{dx} e^x = e^x \cdot 1 = e^x$$
I vaguely understand the concept of indeterminate forms, and why it is difficult to know what is happening with the function. But is there a better explanation of how the result of $1$ is obtained?
 A: If you take the series definition of $e^x$ then you have
$$\begin{eqnarray*} e^x &=& 1 + x + \frac{x^2}{2!} + \ldots \\
\implies \frac{e^x - 1}{x} &=&1 + \frac{x}{2!} + \frac{x^2 }{3!} + \ldots  \\
\implies \lim_{x \rightarrow 0} \frac{e^x - 1}{x} &=& \lim_{x\rightarrow 0} 1 + \frac{x}{2!} + \frac{x^2 }{3!} + \ldots \\
&=& 1.\end{eqnarray*}$$
A: $$e^x := \lim_{n \to \infty} \left(1+\frac{x}{n} \right)^n \implies \\
\frac{d}{dx} e^x = \lim_{n \to \infty} n \cdot \frac{1}{n} \cdot\left(1+\frac{x}{n} \right)^{n-1}=\lim_{n \to \infty} \left(1+\frac{x}{n} \right)^{n-1}=e^x$$
A: Note that even though the expression $\lim_{h\to 0}{e^h-1 \over h}$ is indeterminate, you can rewrite it as
$$\lim_{h\to 0}{e^h-1\over h} = \lim_{h\to 0}{f(h)-f(0)\over h} = f'(0)$$
where $f(x)=e^x$, and where we have used the definition of $f'(0)$. The same can be done with any other base; if $g(x)=a^x$, then the same calculation as you have given shows that $$g'(x)=a^x\lim_{h\to 0}{a^h-1\over h}=a^x\lim_{h\to 0}{g(h)-g(0)\over h}=a^x\cdot g'(0)$$
So it is the slope at the origin which appears here. The formula for $g'(x)$ is simplest when the slope at the origin is equal to $1$, and this is one of the many possible ways to define $e$.
A: Everything in the "proof" will depend on your definition of the function $e^x$. I will choose the definition 
$$
e^h \overset{\text{def}}{=} \sum_{k=0}^\infty \frac{h^k}{k!}.
$$
Using this, one sees that 
$$
\frac{e^h - 1}{h} = \frac{\sum_{k=0}^\infty \frac{h^k}{k!} - 1}{h} = \sum_{k=1}^\infty \frac{h^{k-1}}{k!} = 1 + h \sum_{k=0}^\infty \frac{h^k}{(k+2)!}.
$$
If you have studied convergence tests, you know that the last series on the right converges for all $h \in \mathbb R$, hence taking the limit when $h \to 0$, the RHS goes to $1$ because the series will converge to something and the $h$ factored out will make the product go to zero.
Another way to do this would be to show that this series is an analytic function and is its own Taylor expansion around zero (this is not hard to do using convergence tests), so to differentiate it you can go term by term and readily see that its derivative is itself.
A third approach, which will sound a little stupid and meaningless but is nonetheless funny, is choosing another definition for $e^x$ : consider the differential equation 
$$
f'(x) = f(x), \quad f(0) = 1.
$$
Using differential equation theory it is really not hard at all to show that the solutions to the equation (without the initial condition) is a one-dimensional vector space and there exists an unique element of this vector space which satisfies the initial condition $f(0) = 1$, because the solutions are of the form $Cg(x)$ for some solution $g(x)$. Let $exp(x)$ be defined as a solution to this differential equation satisfying the initial condition. Then clearly $exp'(x) = exp(x)$. Then you can easily see that $exp'(x)$ is a differentiable function, so by induction $exp(x)$ is an infinitely differentiable function with Taylor expansion
$$
\exp(x) = \sum_{k=0}^\infty \frac{x^k}{k!}.
$$
I mentioned it to show the importance of which definition one decides to choose ; it can change the whole structure of an argument.
Hope that helps,
A: Suppose you have an exponential function, like $f(x)=2^x$.
The derivative is
$$
f'(x) = \lim_{h\to0} \frac{f(x+h)-f(x)}{h} = \lim_{h\to0}\frac{2^{x+h}-2^x}{h} = \lim_{h\to0}\left(2^x\cdot\frac{2^h-1}{h}  \right).
$$
So far, just algebra.  Now watch this:
$$
= 2^x\lim_{h\to0}\frac{2^h-1}{h}.
$$
This can be done because $2^x$ is "constant" and "constant" means "not depending on $h$".
But this is equal to $(2^x\cdot\text{constant})$.  But in this case "constant" means "not depending on $x$".  "Constant" always means "not depending on something", but "something" varies with the context.
What's the "constant"?  In the case above, it's not hard to show that the constant is somewhere between $1/2$ and $1$.  It we'd started with $4^x$ instead, then it would be fairly easy to show that the "constant" would be more than $1$.  For a base somewhere between $2$ and $4$, the "constant" is $1$.  That base is $e$.
If you want to talk about how it is knonw that $2$ is too small and $4$ is too big, to be the base of the "natural" exponential function (i.e., the one for which the "constant" is $1$), I can post further on this.
Later edit: OK, how do we know that $2$ is too small and $4$ is too big, to serve in the role of the base of the "natural" exponential function?  Look at the graph of $y=2^x$.  It gets steeper as you go from left to right.  As $x$ goes from $0$ to $1$, $y$ goes from $1$ to $2$.  So the average slope between $x=1$ and $x=2$ is
$$\dfrac{\text{rise}}{\text{run}} = \frac{2-1}{1-0} = 1.$$
Since it gets steeper going from left to right, the slope at the left end of this interval, i.e. at $x=0$, must be less than that.  Thus we have $\dfrac{d}{dx}2^x = (2^x\cdot\text{constant})$ and the "constant" is less than $1$.  (Thinking about the interval from $x=-1$ to $x=0$ in the same way shows that the "constant" is more than $1/2$.)
Now look at $y=4^x$, and use the interval from $x=-1/2$ to $x=0$, and do the same thing, and you see that when you write $\dfrac{d}{dx}4^x = (4^x\cdot\text{constant})$, then that "constant" is more than $1$.
This should suggest that $4$ is too big, and $2$ is too small.
You can show that $3$ is too big by using the interval from $x=-1/6$ to $x=0$ and doing the same thing.  But the arithmetic is messy.
A: Apostol's approach in his Calculus text is to define the natural logarithm by $$\log x=\int_1^xt^{-1}\,dt$$ from which you immediately get that the derivative of $\log x$ is $1/x$. Then Apostol defines the exponential function as the functional inverse of the logarithm. So if $y=e^x$, then $x=\log y$; now differentiate with respect to $x$, using the chain rule, to get $$1={1\over y}{dy\over dx}$$ So, ${dy\over dx}=y=e^x$, as desired. 
A: just use the series definition of $e^x$, put in $h$ and subtract 1 and divide by $h$ to get the desired result.
using the series definition of $e^x$ the only question that needs to be answered is that the derivative of each term summed up is same as summing up and then differentiating.
A: Well, $\ln e^x = x$
So $\frac{d}{dx} \ln e^x = \frac{d}{dx} x \implies \frac{1}{e^x} \frac{d}{dx} e^x = 1$ by chain rule.
Then multiply both sides by $e^x$ and you get $\frac{d}{dx} e^x = e^x$
A: Let's define  $b^x$ as 
$$ b^x = a_0 + a_1x +a_2 \frac{x^2}{2!} + \ldots =\sum_{k=0}^{\infty} a_k\frac{x^k}{k!}$$
for $x=0$, 
$ b^0 =1 $  Thus  $a_0=1$
$$ b^{x} =  1 + a_1x +a_2 \frac{x^2}{2!} + \ldots = 1 + \sum_{k=1}^{\infty} a_k\frac{x^k}{k!} \tag 1$$
$$\frac{d}{dx} b^x = \lim_{h \to 0} \frac{b^{x+h}-b^x}{h}=b^x \lim_{h \to 0} \frac{b^{h}-1}{h}=b^x \lim_{h \to 0} \frac{(1 + a_1h +a_2 \frac{h^2}{2!} + \ldots)-1}{h}$$
$$\frac{d}{dx} b^x=b^x \lim_{h \to 0}  (a_1 +a_2 \frac{h}{2!} + \ldots)=a_1b^x $$
Let's select $a_1=1$  then we define $b=e$
and then we need to find all $a_k$ values and $e$.
After selecting $a_1=1$ and $b=e$, we have :
$$\frac{d}{dx} e^x=e^x \tag 2$$
According to this defination, $a_k=1$ for $k \geq 2$
Proof:
for $a_1=1$ and $b=e$ , 
And using relation (1)
$$ e^x = 1 + x +a_2 \frac{x^2}{2!} + \ldots =1 + x+ \sum_{k=2}^{\infty} a_k\frac{x^k}{k!}$$
$$ \frac{d}{dx} e^x = 1 + a_2 x +a_3 \frac{x^2}{2!} + \ldots =$$
According to the result (2),
$$ \frac{d}{dx} e^x =e^x= 1 + x +a_2 \frac{x^2}{2!} + \ldots= 1 + a_2 x +a_3 \frac{x^2}{2!} + \ldots $$
Now We need to equal all cooeffients of $x^k$, 
we find $a_{2}=1$ and $a_{k+1}=a_{k}$  for $k \geq 2$  
If we solve that relation, we get: $a_{k}=1 $ for $k \geq 2$ 
Thus $ e^x$ can be written as power series as shown below
$$ e^x= 1 + x + \frac{x^2}{2!} + \ldots= \sum_{k=0}^{\infty} \frac{x^k}{k!} $$
and to find $e$: put $x=1$, $e^1=1 + 1 + \frac{1^2}{2!} + \ldots= \sum_{k=0}^{\infty} \frac{1}{k!}$

Note:
If we select $a_1=m$ and follow the same way as shown in the proof, we will get
$$ b^{x}= 1 + mx + \frac{(mx)^2}{2!} + \ldots= \sum_{k=0}^{\infty} \frac{(mx)^k}{k!} =e^{mx}=(e^m)^x$$
$$ b^{x}=(e^m)^x$$
Thus
$ b=e^m$
$\ln(b)=\ln(e^m)=m$
$$\frac{d}{dx} b^x=a_1 b^x =m b^x=\ln(b) b^x $$
$$\frac{d}{dx} (b^x)=\ln(b) b^x $$
A: Using the series definition just differentiate the series.  You can regard all terms of the sum in the series as functions and thus use the sum rule ("1" consists of a constant function).  In other words, where $D$ indicates the derivative, $D[f(x)+g(x)]=D(f(x))+D(g(x))$.  So $$D(1+x+\frac{x^2}{2!}+\ldots)=D(1)+D(x)+D\left(\frac{x^2}{2!}\right)+\ldots=0+1+x+\frac{x^2}{2!}+\ldots$$
A: We know that 
$\frac{d}{dx} \ln(e^{x}) = 1$ (1)
 since $e^{x}$ and $\ln(x)$ are inverse functions. We also know
 $\frac{d}{dx} e^{x} = \frac{1}{e^{x}}$ by (1).
That also means $\frac{d}{dx} e^{x}  {\frac{e^{x}}{e^{x}}} = e^{x}$ by the chain rule.
Source: Khan Academy 
A: l'Hôpital's rule applies to indeterminate form of the limit you have in
$$
\frac{d}{dx} e^x = e^x \lim_{h \to 0} \frac{e^h-1}{h}
$$
So you get:
$$
\lim_{h \to 0} \frac{e^h-1}{h} = \lim_{h \to 0} \frac{\frac{d}{dh}(e^h-1)}{\frac{dh}{dh}}=\lim_{h \to 0}\frac{e^h}{1}=1
$$
