# 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: $\displaystyle{{\rm f}'\left(x\right) = \lim_{h \to 0}{2^{x}\left(2^{h} - 1\right) \over h}}$.

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.

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 $e^x \ln{e} = e^x$.

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 $$\lim_{x \to 0} \dfrac{e^x-1}{x} = 1$$

$a>0$

$a^x := e^{x\ln a}$

$f:x\mapsto e^{x\ln a}$

$g:x\mapsto x\ln a$

$f=\exp \circ g$

$f'=g' \times(\exp '\circ g)=(x\mapsto \ln a)\times(\exp\circ g)=(\ln a) \times f$

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 $(e^x)'_x=e^x$