Finding the derivative of $2^{x}$ from first terms? I was trying to understand why $e^{x}$ is special by finding the derivatives of other exponential functions and comparing the results. So I tried ${\rm f}\left(x\right) = 2^{x}$, but now I'm stuck.
Here's my final step:
$\displaystyle{{\rm f}'\left(x\right)
= \lim_{h \to 0}{2^{x}\left(2^{h} - 1\right) \over h}}$. 
 A: You discovered that
\begin{equation*}
\frac{d}{dx} 2^x = c 2^x
\end{equation*}
where $c = \lim_{h \to 0} (2^h - 1)/h$.
But note that $c \neq 1$, which is kind of annoying.
If you had used $e$ instead of $2$, you would have had $c = \lim_{h \to 0} (e^h - 1)/h$, which actually is equal to $1$.  In fact, this is one definition of $e$.
So the derivative of $e^x$ is just $e^x$, the same thing you started with -- a beautiful result.
A: $a>0$
$a^x := e^{x\ln a}$
$f:x\mapsto e^{x\ln a}$
$g:x\mapsto x\ln a$
$f=\exp \circ g$
$f'=g' \times(\exp '\circ g)=(x\mapsto \ln a)\times(\exp\circ g)=(\ln a) \times f$
A: It helps here to use implicit differentiation.
$y = a^x$
Take the natural logarithm of both sides.
$\ln{y} = x \ln{a}$
Differentiate both sides.
$\frac{1}{y} dy = dx \ln{a}$
Multiply and divide.
$\frac{dy}{dx} = y \ln{a}$
Substitute the original definition of $y = a^x$.
$\frac{dy}{dx} = a^x \ln{a}$
So, the derivative of $2^x$ is $2^x \ln{2}$, and the derivative of $e^x$ is $e^x \ln{e} = e^x$.
A: Write $2^h$ as $(e^{\log(2)})^h$. Hence, $$\dfrac{2^h-1}h = \log(2) \cdot \dfrac{e^{\log(2)h}-1}{h\log 2}$$ Now finish off using the fact that
$$\lim_{x \to 0} \dfrac{e^x-1}{x} = 1$$
A: One of definitions of logarithm is (see here)
$$
\log x = \lim_{n \to \infty}\frac{x^{\frac{1}{n}}-1}{\frac{1}{n}}
$$
Hence denote $h=\frac{1}{n}$
$$
\lim_{h \to 0}\frac{2^{x+h}-2^x}{h}=2^x \lim_{h \to 0}\frac{2^h-1}{h}=2^x \log 2
$$
Hence if you replace $2$ with the base of the natural logarithm, you get $(e^x)'_x=e^x$ 
