Derivative Of $\ln(x)$ It is required to find the derivative of the natural logarithm of $x$: $\frac {d}{dx}\ln(x)$
My solution:
Let $f(x)=\ln(x) $ then $f'(x)=\frac {d}{dx}\ln(x)    $
By definition:$$f'(x)= \lim_{h\to 0}\frac{f(x+h)- f(x)}{h} $$
By substitution:$$f'(x) = \lim_{h\to 0}\frac{\ln(x+h)- \ln(x)}{h} $$
Since $\ln(a)-\ln(b)=\ln(\frac ab)$, then:
$$f'(x) = \lim_{h\to 0}\frac{\ln(\frac{x+h}x)}{h}$$
Since $a\times \ln(b) = \ln(a^b)$, then:
$$f'(x) = \lim_{h\to 0}\ln((\frac{x+h}x)^\frac 1h)$$
Then:$$f'(x) = \lim_{h\to 0}\ln((1+\frac hx)^\frac 1h)$$
How to continue?
Are there any other ways to find the derivative?
Thanks in advance!
Note: It is not allowed to use the fact that: $(e^x)'=e^x$
     $$ e=\lim\limits_{n\to\infty} \left(1+\frac1n\right)^n$$
 A: You're almost there. Recall the definition of $e$:
$e=\lim\limits_{n\to\infty} \left(1+\frac1n\right)^n$
Or equivalently:
$e=\lim\limits_{h\to 0^+} \left(1+h\right)^{1/h}$
You can use that 
$e^u=\lim\limits_{h\to 0^+} \left(1+hu\right)^{1/h}$
Can you take it from here?
A: Hint: You're nearly there with your approach. One method is to use substitution and let $\frac{h}{x} = u$, then consider the limit as $u \rightarrow 0$. 
Note that $\lim \limits_{u \rightarrow 0}(1+u)^{\frac{1}{u}} = e$ (Why?)
A: I usually do this via the chain rule:
\begin{align}
y & = \log x \qquad \text{(or ``$\ln x$'' if you like)} \\[10pt]
x & = e^y \\[10pt]
\frac{dx}{dy} & = e^y \\[10pt]
\frac{dx}{dy} & = x \\[10pt]
\frac{dy}{dx} & = \frac 1 x.
\end{align}
Where do you see the chain rule above?  I wasn't explicit about that.  Here's a more explicit version:
\begin{align}
x & = \exp(\log x). \\[10pt]
1 = \frac{dx}{dx} & = \underbrace{\frac d {dx} \exp\log x = (\exp'\log x) \cdot\log' x}_{\text{chain rule}} = (\exp\log x) \cdot \log' x = x\log'x. \\[10pt]
1 & = x \log' x. \\[10pt]
\frac 1 x & = \log' x.
\end{align}
