The simplest way to find the derivative of the natural logarithm is to use the Inverse Function Theorem (or the Chain Rule), but since you say you only recently started, you may not know it yet.
So instead, we begin with two ingredients. One is that
$\ln(u)$ is continuous. That means that if $\lim\limits_{x\to a}f(x)$ exists, then
$$\lim_{x\to a}\ln(f(x)) = \ln\left(\lim_{x\to a}f(x)\right).$$
The second ingredient (which you may or may not know yet) is that
$$\lim_{h\to\infty}\left(1 + \frac{a}{h}\right)^h = e^a.$$
To see this, note that this is immediate if $a=0$; if $a\gt 0$, then just do a quick rewrite:
$$\begin{align*}
\lim_{h\to\infty}\left(1 + \frac{a}{h}\right)^h &= \lim_{h\to\infty}\left( 1 + \frac{1}{(h/a)}\right)^h\\
&=\lim_{h\to\infty}\left(\left(1 + \frac{1}{(h/a)}\right)^{h/a}\right)^a\\
&= \left(\lim_{h\to\infty}\left(1 + \frac{1}{(h/a)}\right)^{h/a}\right)^a.
\end{align*}$$
If $a\gt 0$, then $h/a\to\infty$ as $h\to\infty$, so by the definition of $e$ you get that
$$\lim_{h\to\infty}\left(1+\frac{a}{h}\right)^h = \left(\lim_{(h/a)\to\infty}\left(1 + \frac{1}{(h/a)}\right)^{h/a}\right)^a = (e)^a = e^a.$$
If $a\lt 0$, then replacing $a$ with $-a$ we can do the same trick as above after proving that
$$\lim_{h\to\infty}\left(1 - \frac{1}{h}\right)^h = e^{-1}.$$
Indeed, though it takes a bit more algebraic trickery:
$$\begin{align*}
\lim_{h\to\infty}\left(1 - \frac{1}{h}\right)^h &= \lim_{h\to\infty}\left(\frac{h-1}{h}\right)^h = \lim_{h\to\infty}\left(\frac{h}{h-1}\right)^{-h}\\
&= \left(\lim_{h\to\infty}\left(\frac{(h-1)+1}{h-1}\right)^h\right)^{-1}\\
&= \left(\lim_{h\to\infty}\left(1 + \frac{1}{h-1}\right)^h\right)^{-1}\\
&=\left(\lim_{h\to\infty}\left(1 + \frac{1}{h-1}\right)^{h-1}\left(1 + \frac{1}{h-1}\right)^1\right)^{-1}\\
&= \left(\lim_{h\to\infty}\left(1 + \frac{1}{h-1}\right)^{h-1}\lim_{h\to\infty}\left(1 + \frac{1}{h-1}\right)\right)^{-1}\\
&= \Bigl((e)(1)\Bigr)^{-1} = e^{-1}.\end{align*}$$
Then, in the previous limit, if $a\lt 0$ then replace it with $-a$ and change the $+$ to a $-$, to get that the limit equals $(e^{-a})^{-1} = e^a$ as well.
And finally, with these ingredients in hand, we are ready. We have:
$$\begin{align*}
\frac{d}{dx}\ln x &= \lim_{\Delta\to 0}\frac{\ln(x+\Delta)-\ln(x)}{\Delta}\\
&= \lim_{\Delta\to 0}\frac{1}{\Delta}\left(\ln(x+\Delta)-\ln(x)\right)\\
&=\lim_{\Delta\to 0}\frac{1}{\Delta}\ln\left(\frac{x+\Delta}{x}\right)\\
&=\lim_{\Delta\to 0}\frac{1}{\Delta}\ln\left(1 +\frac{\Delta}{x}\right)\\
&=\lim_{\Delta\to 0}\ln\left(\left(1 + \frac{\Delta}{x}\right)^{1/\Delta}\right)\\
&= \lim_{\Delta\to 0}\ln\left(\left(1 + \frac{1/x}{1/\Delta}\right)^{1/\Delta}\right).
\end{align*}$$
If $\Delta\to 0^+$, then $\frac{1}{\Delta}\to\infty$, so letting $h=\frac{1}{\Delta}$ we have:
$$\lim_{\Delta\to 0^+}\left(1 + \frac{1/x}{1/\Delta}\right)^{1/\Delta} =
\lim_{h\to\infty}\left( 1 + \frac{1/x}{h}\right)^h = e^{1/x}.$$
If $\Delta\to 0^-$, then $\frac{1}{\Delta}\to-\infty$, so letting $h=-\frac{1}{\Delta}$, we have:
$$\begin{align*}
\lim_{\Delta\to 0^-}\left(1 + \frac{1/x}{1/\Delta}\right)^{1/\Delta} &= \lim_{h\to\infty}\left(1 - \frac{1/x}{h}\right)^{-h}\\
&= \lim_{h\to\infty}\left(\left(1 - \frac{1/x}{h}\right)^{h}\right)^{-1}\\
&= \left(e^{-1/x}\right)^{-1} = e^{1/x}.
\end{align*}$$
Therefore, we have:
$$\begin{align*}
(\ln x)' &= \lim_{\Delta\to 0}\frac{\ln(x+\Delta)-\ln(x)}{\Delta}\\
&= \ln\left(\lim_{\Delta\to 0}\left(1 + \frac{1/x}{1/\Delta}\right)^{1/\Delta}\right)\\
&= \ln\left(e^{1/x}\right) = \frac{1}{x}.
\end{align*}$$
And this is why the Chain Rule or the Inverse Function Theorem are such a better way of proving this...
