\[\begin{align*} \boxed{\dfrac{d}{dx} \ln x \cdot \ln x = \dfrac{2\ln x}{x}} \end{align*}\]

Solving for the Derivative

To find the derivative, we will use the Product Rule:

\[\begin{align*} \frac{d}{dx} f(x) \cdot g(x) = f(x) \cdot \frac{d}{dx} g(x) + g(x) \cdot \frac{d}{dx} f(x) \end{align*}\]

In our case both functions are the same, $f(x) = g(x) = \ln x$, we get that $f'(x) = g'(x) = \dfrac1x$. Using the product rule, we obtain:

\[\begin{align*} \dfrac{d}{dx} \ln x \cdot \ln x &= \ln x \cdot \dfrac1x + \ln x \cdot \dfrac1x\\ &=\dfrac{\ln x}{x} + \dfrac{\ln x}{x}\\ &=\boxed{\dfrac{2\ln x}{x}} \end{align*}\]

Alternatively, we can use the Chain Rule:

\[\begin{align*} \dfrac{d}{dx}(f(g(x))) = f'(g(x)) \cdot g'(x)) \end{align*}\]

We can rewrite our derivative $\dfrac{d}{dx} \ln x \cdot \ln x = \ln^2 x$. Thus, we get that $f(x) = x^2$ and $g(x) = \ln x$. Now using the chain rule we obtain:

\[\begin{align*} \dfrac{d}{dx} \ln x \cdot \ln x &= \ln^2 x\\ &=2\ln x \cdot \dfrac{1}{x}\\ &=\boxed{\dfrac{2\ln x}{x}} \end{align*}\]