In this article, we will be showing that:

\[\begin{align*} \dfrac{d}{dx}(\ln(x^2)) = \boxed{\dfrac2x}. \end{align*}\]

To prove this, we will use the Chain Rule. Recalling the formula for the Chain Rule, we find that the Chain Rule states that:

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

Where $f(x)$ and $g(x)$ are differentiable functions. Using this formula, we let $f(x) = \ln(x)$ and $g(x) = x^2$ so that $f(g(x)) = \ln(g(x)) = \ln(x^2)$. We are able to use the Chain Rule with these functions, as they are differentiable.

To plug these values into the Chain Rule, we need to calculate their derivatives. We can use the formula for the derivative of $\ln(x)$ for $f(x)$, and we can use the Power Rule for $g(x)$:

\[\begin{align*} \dfrac{d}{dx}(\ln(x)) = \dfrac1x \;\;\;\;\;\;\;\;\;\;\;\;\;\;\; \dfrac{d}{dx}(x^n)= nx^{(n-1)} \end{align*}\]

Using this, we find:

\[\begin{align*} f(x) = \ln(x) &\longrightarrow f'(x) = \dfrac1x,\\ g(x) = x^2 &\longrightarrow g'(x) = 2x. \end{align*}\]

Plugging these values into the Chain Rule, we find:

\[\begin{align*} \dfrac{d}{dx}(\ln(x^2)) = f'(g(x)) \cdot g'(x)\\ \dfrac{d}{dx}(\ln(x^2)) = \dfrac{1}{g(x)} \cdot 2x\\ \dfrac{d}{dx}(\ln(x^2)) = \dfrac{1}{x^2} \cdot 2x\\ \dfrac{d}{dx}(\ln(x^2)) = \boxed{\dfrac{2}{x}}. \end{align*}\]