To find $f'''(0)$ for the function $f(x) = \ln(x + 1)$, we will go through the following steps:

1. Find the first derivative $f'(x)$:

   Since $f(x) = \ln(x + 1)$, by the chain rule, we have: $f'(x) = \frac{1}{x + 1}.$

2. Find the second derivative $f''(x)$:

   Differentiating $f'(x) = \frac{1}{x + 1}$ with respect to $x$: $f''(x) = -\frac{1}{(x + 1)^2}.$

3. Find the third derivative $f'''(x)$:

   Differentiating $f''(x) = -\frac{1}{(x + 1)^2}$ with respect to $x$: $f'''(x) = \frac{2}{(x + 1)^3}.$

4. Evaluate $f'''(0)$:

   Substitute $x = 0$ into $f'''(x)$: $f'''(0) = \frac{2}{(0 + 1)^3} = 2.$

So, the value of $f'''(0)$ is: $f'''(0) = 2.$

Would you like further details on any of these steps, or do you have additional questions?

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Here are five related questions to deepen understanding:

1. How do we find higher-order derivatives for functions involving logarithms?
2. What would the fourth derivative $f^{(4)}(x)$ of $\ln(x + 1)$ look like?
3. Can we use Taylor series to approximate $\ln(x + 1)$ around $x = 0$?
4. How does the chain rule apply to functions inside logarithmic functions?
5. How would the result change if we started with $f(x) = \ln(2x + 1)$?

Tip: For functions involving $\ln(x + a)$, derivatives tend to involve powers of $x + a$ in the denominator, which can simplify when evaluating at points like $x = 0$.