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To find \( f'(x) \) for the given function \( f(x) = (\ln x)^8 \), we need to differentiate it with respect to \( x \).

### Step 1: Differentiate \( f(x) = (\ln x)^8 \)

We can apply the chain rule here. Let \( u = \ln x \), so \( f(x) = u^8 \).

1. Differentiate \( u^8 \) with respect to \( u \):
   \[
   \frac{d}{du}(u^8) = 8u^7
   \]
   
2. Differentiate \( u = \ln x \) with respect to \( x \):
   \[
   \frac{du}{dx} = \frac{1}{x}
   \]

Now, using the chain rule:
\[
f'(x) = 8(\ln x)^7 \cdot \frac{1}{x} = \frac{8(\ln x)^7}{x}
\]

So, we have:
\[
f'(x) = \frac{8(\ln x)^7}{x}
\]

### Step 2: Evaluate \( f'(e^4) \)

To find \( f'(e^4) \), substitute \( x = e^4 \) into the derivative:

1. Calculate \( \ln(e^4) \):
   \[
   \ln(e^4) = 4
   \]

2. Substitute into \( f'(x) \):
   \[
   f'(e^4) = \frac{8(4)^7}{e^4}
   \]

3. Simplify \( 4^7 \):
   \[
   4^7 = 16384
   \]

4. Substitute the values:
   \[
   f'(e^4) = \frac{8 \times 16384}{e^4} = \frac{131072}{e^4}
   \]

### Final Answer

- \( f'(x) = \frac{8(\ln x)^7}{x} \)
- \( f'(e^4) = \frac{131072}{e^4} \)

Would you like a more detailed explanation of any part? 

Here are some related questions for further exploration:
1. How would the derivative change if the exponent were different, say \( (\ln x)^n \)?
2. How can we use logarithmic differentiation to simplify functions with exponents?
3. What is the process for differentiating composite functions like \( (\ln x)^8 \)?
4. How would you interpret the derivative \( f'(x) \) in terms of the rate of change of \( f(x) \)?
5. Can we use approximation methods to evaluate \( f'(e^4) \) without an exact answer?

**Tip:** The chain rule is essential for differentiating nested functions; practice identifying inner and outer functions to apply it effectively.

Find the derivative of f(x) = (ln x)^8 and evaluate it at x = e^4.

Learn how to find the derivative of f(x) = (ln x)^8 using the chain rule and evaluate it at x = e^4. This solution covers the differentiation of composite functions, applying the chain rule, and handling logarithmic functions. Suitable for high school students in Grades 11-12.

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