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Let's work through each of the problems you've uploaded:

### Problem 6: Differentiate the function \( F(s) = \ln(\ln(2s)) \)

1. **Outer derivative (ln):** The derivative of \( \ln(u) \) with respect to \( u \) is \( \frac{1}{u} \).
   - Here, \( u = \ln(2s) \), so the first part of the derivative is \( \frac{1}{\ln(2s)} \).

2. **Inner derivative (ln(2s)):** The derivative of \( \ln(2s) \) is \( \frac{1}{2s} \), using the chain rule.

Thus, the derivative of \( F(s) \) is:
\[
F'(s) = \frac{1}{\ln(2s)} \cdot \frac{1}{2s} = \frac{1}{2s \ln(2s)}
\]

### Problem 7: Find \( y' \) and \( y'' \) for \( y = x^2 \ln(5x) \)

1. **First derivative \( y' \):** 
   - This is a product rule problem. The product rule is \( (uv)' = u'v + uv' \).
   - Let \( u = x^2 \) and \( v = \ln(5x) \).
   - Derivative of \( u = x^2 \) is \( u' = 2x \).
   - Derivative of \( v = \ln(5x) \) is \( v' = \frac{1}{x} \) (since \( \ln(5x) = \ln(5) + \ln(x) \)).

   Applying the product rule:
   \[
   y' = 2x \ln(5x) + x^2 \cdot \frac{1}{x} = 2x \ln(5x) + x
   \]

2. **Second derivative \( y'' \):**
   - Use the product rule again for \( 2x \ln(5x) \) and combine with the derivative of \( x \).

   The second derivative is:
   \[
   y'' = 2 \ln(5x) + 2x \cdot \frac{1}{x} + 1 = 2 \ln(5x) + 2 + 1 = 2 \ln(5x) + 3
   \]

### Problem 8: Differentiate \( f(x) = \frac{x}{1 - \ln(x - 7)} \) and find the domain.

1. **Quotient rule for derivative:** 
   The quotient rule is \( \left( \frac{u}{v} \right)' = \frac{u'v - uv'}{v^2} \).

   - \( u = x \), so \( u' = 1 \).
   - \( v = 1 - \ln(x - 7) \), so \( v' = -\frac{1}{x - 7} \) (derivative of \( \ln(x - 7) \)).

   Applying the quotient rule:
   \[
   f'(x) = \frac{(1)(1 - \ln(x - 7)) - x\left(-\frac{1}{x - 7}\right)}{(1 - \ln(x - 7))^2}
   = \frac{1 - \ln(x - 7) + \frac{x}{x - 7}}{(1 - \ln(x - 7))^2}
   \]

2. **Domain:** 
   - The domain is restricted by the natural logarithm, so \( x - 7 > 0 \), meaning \( x > 7 \).
   - Additionally, \( 1 - \ln(x - 7) \neq 0 \), so \( \ln(x - 7) \neq 1 \), or \( x - 7 \neq e \), which gives \( x \neq 7 + e \).

   Therefore, the domain is \( (7, 7 + e) \cup (7 + e, \infty) \).

### Problem 9: Use logarithmic differentiation to find the derivative of \( y = x^8 \cos(x) \)

1. **Take the natural logarithm of both sides:**
   \[
   \ln(y) = \ln(x^8 \cos(x)) = \ln(x^8) + \ln(\cos(x)) = 8 \ln(x) + \ln(\cos(x))
   \]

2. **Differentiate implicitly:**
   \[
   \frac{d}{dx} [\ln(y)] = \frac{1}{y} \cdot y' = 8 \cdot \frac{1}{x} + \frac{-\sin(x)}{\cos(x)} = 8 \cdot \frac{1}{x} - \tan(x)
   \]

   Therefore, multiplying both sides by \( y = x^8 \cos(x) \), we get:
   \[
   y' = x^8 \cos(x) \left( \frac{8}{x} - \tan(x) \right)
   \]

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Let me know if you'd like more details or have any questions! Here are five related questions:

1. How is the chain rule applied to logarithmic functions?
2. Can you find the second derivative of \( F(s) = \ln(\ln(2s)) \)?
3. What are the conditions for applying the quotient rule?
4. Can you derive the product rule for a more complex function, like \( y = x^n e^x \)?
5. How does logarithmic differentiation simplify the process for differentiating products of functions?

**Tip:** Always double-check the domain of logarithmic functions to ensure the argument is positive!

Differentiate the functions, find derivatives, and determine the domain for various logarithmic functions as shown in the math problems.

Detailed step-by-step solutions to differentiate functions involving logarithms, including using the chain rule, product rule, and quotient rule. Topics include finding the domain of logarithmic functions and implicit differentiation. Suitable for Grade 12 and college-level calculus students.

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