Proof of the derivative of $\ln(x)$ 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?
 A: 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
A: If you can use the chain rule and the fact that the derivative of $e^x$ is $e^x$ and the fact that $\ln(x)$ is differentiable,  then we have:
$$\frac{\mathrm{d} }{\mathrm{d} x} x = 1$$
$$\frac{\mathrm{d} }{\mathrm{d} x} e^{\ln(x)} =  e^{\ln(x)} \frac{\mathrm{d} }{\mathrm{d} x} \ln(x) = 1$$
$$e^{\ln(x)} \frac{\mathrm{d} }{\mathrm{d} x} \ln(x) = 1$$
$$x \frac{\mathrm{d} }{\mathrm{d} x} \ln(x) = 1$$
$$\frac{\mathrm{d} }{\mathrm{d} x} \ln(x) = \frac{1}{x}$$
A: The simplest way is to use the inverse function theorem for derivatives:
If $f$ is a bijection from an interval $I$ onto an interval $J=f(I)$, which has a derivative at $x\in I$, and if $f'(x)\neq 0$, then $f^{-1}\colon J\to I$ has a derivative at $y=f(x)$, and
$$\bigl(f^{-1}\bigr)'(y)=\frac1{f'(x)}=\frac1{f'\bigl(f^{-1}(y)\bigr)}.$$
As $(\mathrm e^x)'=\mathrm e^x\neq 0\,$ for all $x$, we know that $\,\ln\,$  has a derivative at each point of its domain, and
$$(\ln)'(y)=\frac1{\mathrm e^{\,\ln y}}=\frac1y.$$
A: If you can use the definition of $e$ as: $$e:=\lim_{n\rightarrow∞}\left(1+\frac{1}{n}\right)^n$$
and the slightly modified form: $\displaystyle e^x=\lim_{n\rightarrow∞}\left(1+\frac{x}n\right)^n$ 
then, by setting $h=\frac1{x}$ you can calculate the desired limit.
A: Define $$e=\lim_{h\to 0} \left(1+h\right)^{1/h}.$$ Then change variables $h\mapsto h/x$ giving $$e=\lim_{h/x\to 0} \left(1+\frac{h}{x}\right)^{\frac{x}{h}}=\lim_{h\to 0} \left(1+\frac{h}{x}\right)^{\frac{x}{h}},$$ where the limit in the second equality follows since $h$ approaches $0$ as $h/x$ does. Since $x$ is constant w.r.t. $h$, we can simplify by raising both sides to the power $1/x$, giving you the desired identity.
