Step 1: Understanding the Concept:
This problem involves the limit of a sum as a Riemann integral. We use the property that for large \( n \), \( [A] \approx A \). Specifically, the limit of the sum of floor functions divided by \( n^k \) is equal to the integral of the underlying continuous function.
Step 2: Key Formula or Approach:
1. Riemann Sum: \( \lim_{n \to \infty} \frac{1}{n} \sum f(\frac{r}{n}) = \int_0^1 f(s) ds \).
2. Logarithmic Differentiation for \( y = f(x)^{g(x)} \).
Step 3: Detailed Explanation:
The limit is:
\[ x = \lim_{n \to \infty} \frac{\sum_{k=1}^{n} k^2 (f(x))^x}{n^3} \]
(The floor function symbols \( [ \dots ] \) vanish in the limit as \( n \to \infty \) when divided by \( n^3 \)).
Rearranging:
\[ x = (f(x))^x \lim_{n \to \infty} \frac{1}{n} \sum_{k=1}^{n} \left( \frac{k}{n} \right)^2 \]
The limit corresponds to the integral \( \int_0^1 s^2 ds = \frac{1}{3} \).
So, \( x = (f(x))^x \cdot \frac{1}{3} \implies 3x = (f(x))^x \).
Taking natural log on both sides:
\[ \ln(3x) = x \ln(f(x)) \implies \ln(f(x)) = \frac{\ln(3x)}{x} \]
\[ f(x) = e^{\frac{\ln(3x)}{x}} = (3x)^{1/x} \]
Now, differentiate \( f(x) \) using logarithmic differentiation. Let \( y = f(x) \):
\[ \ln y = \frac{1}{x} \ln(3x) \]
\[ \frac{1}{y} y' = -\frac{1}{x^2} \ln(3x) + \frac{1}{x} \left( \frac{1}{3x} \cdot 3 \right) \]
\[ \frac{y'}{y} = \frac{1 - \ln(3x)}{x^2} \]
\[ f'(x) = y \left( \frac{1 - \ln(3x)}{x^2} \right) = (3x)^{1/x} \left[ \frac{1 - \ln 3x}{x^2} \right] \]
Step 4: Final Answer:
The derivative is \( (3x)^{1/x} \left[ \frac{1 - \ln 3x}{x^2} \right] \).