TITLE: Finding the derivative of $2^{x}$ from first terms?
QUESTION [5 upvotes]: 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:
$\displaystyle{{\rm f}'\left(x\right)
= \lim_{h \to 0}{2^{x}\left(2^{h} - 1\right) \over h}}$.

REPLY [2 votes]: 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 $e^x \ln{e} = e^x$.