TITLE: Proof of the derivative of $\ln(x)$
QUESTION [34 upvotes]: I'm trying to prove that $\frac{\mathrm{d} }{\mathrm{d} x}\ln x = \frac{1}{x}$.
Here's what I've got so far:
$$
\begin{align}
\frac{\mathrm{d}}{\mathrm{d} x}\ln x &= \lim_{h\to0} \frac{\ln(x + h) - \ln(x)}{h} \\
&= \lim_{h\to0} \frac{\ln(\frac{x + h}{x})}{h} \\
&= \lim_{h\to0} \frac{\ln(1 + \frac{h}{x})}{h} \\
\end{align}
$$
To simplify the logarithm:
$$
\lim_{h\to0}\left (1 + \frac{h}{x}\right )^{\frac{1}{h}} = e^{\frac{1}{x}}
$$
This is the line I have trouble with. I can see that it is true by putting numbers in, but I can't prove it. I know that $e^{\frac{1}{x}} = \lim_{h\to0}\left (1 + h \right )^{\frac{h}{x}}$, but I can't work out how to get from the above line to that.
$$
\lim_{h\to0}\left ( \left (1 + \frac{h}{x}\right )^{\frac{1}{h}}\right )^{h} = e^{\frac{h}{x}}
$$
Going back to the derivative:
$$
\begin{align}
\frac{\mathrm{d}}{\mathrm{d} x}\ln x &= \lim_{h\to0} \frac{\ln(e^{\frac{h}{x}})}{h} \\
&= \lim_{h\to0} \frac{\frac{h}{x}\ln(e)}{h} \\
&= \lim_{h\to0} \frac{h}{x} \div h\\
&= \frac{1}{x} \\
\end{align}
$$
This proof seems fine, apart from the middle step to get $e^{\frac{1}{x}}$. How could I prove that part?

REPLY [6 votes]: Just throwing it out there for you to see, I also like this proof:
$$y=\ln x$$
$$e^y=x$$
after differentiating, 
$$e^y \frac{dy}{dx}=1$$
$$ \begin{align} 
\frac{dy}{dx}&=\frac{1}{e^y}\\
&= \frac{1}{e^{\ln x}}\\
&= \frac{1}{x} \end{align}
$$
of course, that assumes you already know the derivative of $e^x$ and the chain rule