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Let's solve the problem step-by-step!

Given:
\[
f(x) = \frac{\ln(x)}{x^5}.
\]

### Step 1: Find \( f'(x) \) to determine the critical number(s).
We need to apply the **quotient rule**:
\[
\left( \frac{u}{v} \right)' = \frac{u'v - uv'}{v^2},
\]
where \( u = \ln(x) \) and \( v = x^5 \).

- \( u' = \frac{1}{x} \) (the derivative of \( \ln(x) \)).
- \( v' = 5x^4 \) (the derivative of \( x^5 \)).

Now apply the quotient rule:
\[
f'(x) = \frac{\frac{1}{x} \cdot x^5 - \ln(x) \cdot 5x^4}{(x^5)^2}.
\]
Simplify the numerator:
\[
\frac{x^5}{x} = x^4, \quad \text{so} \quad x^4 - 5x^4 \ln(x) = x^4(1 - 5 \ln(x)).
\]
Thus,
\[
f'(x) = \frac{x^4(1 - 5 \ln(x))}{x^{10}} = \frac{1 - 5 \ln(x)}{x^6}.
\]

### Step 2: Set \( f'(x) = 0 \) to find critical points.
\[
\frac{1 - 5 \ln(x)}{x^6} = 0 \quad \Rightarrow \quad 1 - 5 \ln(x) = 0.
\]
Solve for \( x \):
\[
5 \ln(x) = 1 \quad \Rightarrow \quad \ln(x) = \frac{1}{5}.
\]
\[
x = e^{1/5}.
\]
So the **critical number** is:
\[
A = e^{1/5}.
\]

### Step 3: Find \( f''(x) \) and evaluate it at \( A = e^{1/5} \).
We need to take the derivative of \( f'(x) = \frac{1 - 5 \ln(x)}{x^6} \).

Using the **product rule and chain rule**, the second derivative involves some computation, but we are only interested in the sign of \( f''(e^{1/5}) \). After some detailed differentiation, we get:

\[
f''(x) = \frac{-30 + 25 \ln(x)}{x^7}.
\]
Now evaluate at \( A = e^{1/5} \):
\[
\ln(e^{1/5}) = \frac{1}{5}, \quad \text{so} \quad f''(e^{1/5}) = \frac{-30 + 25 \cdot \frac{1}{5}}{(e^{1/5})^7}.
\]
Simplify:
\[
f''(e^{1/5}) = \frac{-30 + 5}{e^{7/5}} = \frac{-25}{e^{7/5}}.
\]

Since \( f''(e^{1/5}) < 0 \), the function has a **local maximum** at \( A = e^{1/5} \).

### Final Answer:
- Critical number: \( A = e^{1/5} \).
- \( f''(A) = \frac{-25}{e^{7/5}} \).
- Conclusion: \( f(x) \) has a **local maximum** at \( A = e^{1/5} \).

---

Would you like more details about any part of this process?  

Here are 5 related questions for further exploration:
1. How do we simplify expressions involving natural logarithms?
2. What other methods can we use to confirm local maxima or minima?
3. How does the quotient rule work for more complex derivatives?
4. What is the behavior of \( f(x) \) as \( x \to \infty \) or \( x \to 0^+ \)?
5. Why is the second derivative test useful for classifying critical points?

**Tip:** The second derivative test can fail if \( f''(A) = 0 \), so checking the first derivative's sign change around critical points is a good backup plan!

Consider the function f(x) = ln(x) / x^5. f(x) has a critical number A = __. f''(A) = __. Thus we conclude that f(x) has a local __ at A (type in MAX or MIN).

Learn how to find the critical points and classify the local extrema of the function f(x) = ln(x) / x^5 using derivatives. This solution involves applying the quotient rule and the second derivative test, suitable for college-level calculus students.

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