\[\begin{align*} \boxed{\dfrac{d}{dx} \ln x \cdot \ln x = \dfrac{2\ln x}{x}} \end{align*}\]
Solving for the Derivative
To find the derivative, we should review the product rule:
\[\begin{align*} \frac{d}{dx} f(x) \cdot g(x) = f(x) \cdot \frac{d}{dx} g(x) + g(x) \cdot \frac{d}{dx} f(x) \end{align*}\]
Proceeding with our two functions, $f(x) = g(x) = \ln x$, 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, if you wanted to use the chain rule instead:
\[\begin{align*} \dfrac{d}{dx}(f(g(x))) = f'(g(x)) \cdot g'(x)) \end{align*}\]
We could set our derivative $\dfrac{d}{dx} \ln x \cdot \ln x = \ln^2 x$. Doing this, we get that $f(x) = x^2$ and $g(x) = \ln x$. With this in mind, we may use chain rule to obtain:
\[\begin{align*} \dfrac{d}{dx} \ln x \cdot \ln x &= \ln^2 x\\ &=2\ln x \cdot \dfrac{1}{x}\\ &=\boxed{\dfrac{2\ln x}{x}} \end{align*}\]
