We will be trying to find the derivative of $\ln(x^3)$.
Using the Power Rule for logarithms, which tells us $\ln(x^n) = n \ln (x)$, we get $\ln(x^3) = 3\ln(x)$. Using the fact that $\dfrac{d}{dx} (\ln(x)) = \dfrac{1}{x}$, we have
\[\begin{align*} \dfrac{d}{dx}(\ln(x^3)) = \dfrac{d}{dx}(3\ln(x)) = 3 \cdot \dfrac{d}{dx} (\ln (x)) = 3 \cdot \dfrac{1}{x} = \dfrac{3}{x}. \end{align*}\]
Another way we could approach this problem is using the formula of the Chain Rule. This states that
\[\begin{align*} \dfrac{d}{dx}(f(g(x))) = f'(g(x)) \cdot g'(x). \end{align*}\]
Using this formula, we let $f(x) = \ln (x)$ and $g(x) = x^3$ such that $f(g(x)) = \ln(x^3)$.
To plug these values into the chain rule, we need to calculate their derivatives. We can use the formula for the derivative of $\ln(x)$ and the Power Rule:
\[\begin{align*} \dfrac{d}{dx}(\ln(x)) = \dfrac1x \;\;\;\;\;\;\;\;\;\;\;\;\;\;\; \dfrac{d}{dx}(x^n)= nx^{(n-1)} \end{align*}\]
Using this, we find:
\[\begin{align*} f(x) = \ln(x) &\longrightarrow f'(x) = \dfrac1x g(x) = x^3 &\longrightarrow g'(x) = 3x^2. \end{align*}\]
Plugging these values into the Chain Rule formula, we find:
\[\begin{align*} \dfrac{d}{dx}(ln(x^3)) = \dfrac{1}{g(x)} \cdot 3x^2. \dfrac{d}{dx}(ln(x^3)) = \dfrac{1}{x^3} \cdot 3x^2 \dfrac{d}{dx}(ln(x^3)) = \boxed{\dfrac{3}{x^2}}. \end{align*}\]
