To find the derivative of $\ln x$, we will use implicit differentiation, which is a standard technique to find derivatives of inverse functions. Note that $\ln x$ is the inverse function of $e^x$, so it is natural for us to set $y = \ln x$. We know that $\dfrac{d}{dx} e^x = e^x$ and we will use it in below calculations. First, rewrite $e^y$ by substitution:

\[ \begin{align*} e^y &= e^{\ln x}\end{align*}\]

Now use the fact that exponentials are inverses of logarithm functions:

\[ \begin{align*} e^y &= x\end{align*}\]

Differentiate both sides in order to find $\dfrac{dy}{dx}$. This is the step where we use implicit differentiation:

\[ \begin{align*} \dfrac{d}{dx} e^y &= \dfrac{d}{dx} x\end{align*}\]

Use the fact that the derivative of $x$ is just 1:

\[ \begin{align*} \dfrac{dy}{dx} e^y &= 1\end{align*}\]

Now we can finally solve for $\dfrac{dy}{dx}$:

\[ \begin{align*} \dfrac{dy}{dx} &= \dfrac1{e^y}\end{align*}\]

However, we want it as a function of $x$, so we need to perform the substitution from the initial set up where we wrote $y = \ln x$:

\[ \begin{align*} \dfrac{dy}{dx} &= \dfrac1{e^{\ln x}}\end{align*}\]

Finally, we use the fact that $e^x$ and $\ln x$ are inverse functions to write:

\[ \begin{align*} \dfrac{dy}{dx} &= \dfrac1x\\ \end{align*} \]

We get the end result that $\boxed{\frac{d}{dx} \ln x = \dfrac1x}$.

Stated in layman’s terms, the natural logarithm function grows at the rate $\dfrac1{x}$, which means that as $x$ increases, the natural logarithm function grows slower and slower. This is expected, since the inverse function of the natural logarithm $e^x$ grows at the rate $e^x$, which is extremely fast. For example, if a population of deer grows logarithmically, that means that the population is barely increasing. On the other hand, if, for example, a population of bacteria increases exponentially, that means that the population increases extremely fast.