By def. of the derivative, we have for y = ln(x),
[tex]\displaystyle \frac{dy}{dx} = \lim_{h\to0} \frac{\ln(x+h)-\ln(x)}{h}[/tex]
[tex]\displaystyle \frac{dy}{dx} = \lim_{h\to0} \frac1h \ln\left(\frac{x+h}{x}\right)[/tex]
[tex]\displaystyle \frac{dy}{dx} = \lim_{h\to0} \ln\left(1+\frac hx\right)^{\frac1h}[/tex]
Substitute y = h/x, so that as h approaches 0, so does y. We then rewrite the limit as
[tex]\displaystyle \frac{dy}{dx} = \lim_{y\to0} \ln\left(1+y\right)^{\frac1{xy}}[/tex]
[tex]\displaystyle \frac{dy}{dx} = \frac1x \lim_{y\to0} \ln\left(1+y\right)^{\frac1y}[/tex]
Recall that the constant e is defined by the limit,
[tex]\displaystyle e = \lim_{y\to0} \left(1+y\right)^{\frac1y}[/tex]
Then in our limit, we end up with
[tex]\displaystyle \frac{dy}{dx} = \frac1x \ln(e) = \boxed{\frac1x}[/tex]
In Mathematica, use
D[Log[x], x]
