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The problem asks for the derivative of the function:

\[
f(x) = \log_5\left( \sqrt[3]{\frac{-7x - 5}{-2x - 2}} \right)
\]

### Step-by-Step Solution:

1. **Apply Logarithmic Properties**:
   Use the property of logarithms: \(\log_b(a^c) = c \cdot \log_b(a)\). We can rewrite the function as:
   \[
   f(x) = \frac{1}{3} \log_5\left( \frac{-7x - 5}{-2x - 2} \right)
   \]

2. **Differentiate**:
   To find \( f'(x) \), we use the chain rule. The derivative of \( \log_b(u) \) is:
   \[
   \frac{d}{dx} \log_b(u) = \frac{1}{\ln(b)} \cdot \frac{1}{u} \cdot \frac{du}{dx}
   \]
   So, applying this formula with \( b = 5 \) and \( u = \frac{-7x - 5}{-2x - 2} \), we have:
   \[
   f'(x) = \frac{1}{3} \cdot \frac{1}{\ln(5)} \cdot \frac{1}{\frac{-7x - 5}{-2x - 2}} \cdot \frac{d}{dx}\left( \frac{-7x - 5}{-2x - 2} \right)
   \]

3. **Differentiate the Inner Function**:
   Use the quotient rule for the derivative of \( \frac{-7x - 5}{-2x - 2} \). The quotient rule states:
   \[
   \frac{d}{dx} \left( \frac{g(x)}{h(x)} \right) = \frac{g'(x) h(x) - g(x) h'(x)}{h(x)^2}
   \]
   Here:
   - \( g(x) = -7x - 5 \), so \( g'(x) = -7 \)
   - \( h(x) = -2x - 2 \), so \( h'(x) = -2 \)

   Applying the quotient rule:
   \[
   \frac{d}{dx} \left( \frac{-7x - 5}{-2x - 2} \right) = \frac{(-7)(-2x - 2) - (-7x - 5)(-2)}{(-2x - 2)^2}
   \]

   Simplifying the numerator:
   \[
   (-7)(-2x - 2) = 14x + 14
   \]
   \[
   (-7x - 5)(-2) = 14x + 10
   \]
   So the numerator becomes:
   \[
   14x + 14 - (14x + 10) = 4
   \]

   The denominator is:
   \[
   (-2x - 2)^2 = 4(x + 1)^2
   \]

   Therefore, the derivative of the inner function is:
   \[
   \frac{d}{dx} \left( \frac{-7x - 5}{-2x - 2} \right) = \frac{4}{4(x + 1)^2} = \frac{1}{(x + 1)^2}
   \]

4. **Final Derivative**:
   Now substitute everything back into the original expression:
   \[
   f'(x) = \frac{1}{3} \cdot \frac{1}{\ln(5)} \cdot \frac{1}{\frac{-7x - 5}{-2x - 2}} \cdot \frac{1}{(x + 1)^2}
   \]

   Simplify:
   \[
   f'(x) = \frac{1}{3 \ln(5)} \cdot \frac{-2x - 2}{(x + 1)^2} = \frac{-2(x + 1)}{3 \ln(5) (x + 1)^2}
   \]

   Finally:
   \[
   f'(x) = \frac{-2}{3 \ln(5) (x + 1)}
   \]

Thus, the derivative is:

\[
f'(x) = \frac{-2}{3 \ln(5) (x + 1)}
\]

---

Would you like further details or clarifications on any step?

Here are 5 related questions:
1. How do logarithmic properties simplify differentiation of complex functions?
2. What is the quotient rule and how is it applied in derivatives?
3. How can you generalize the derivative of a logarithmic function with different bases?
4. Why is the natural logarithm \(\ln\) used when differentiating logarithmic functions?
5. How can chain rule help when differentiating nested functions like in this problem?

**Tip**: When dealing with logarithmic functions, simplifying the argument using logarithmic properties can often make differentiation easier!

Find the derivative of f(x) = log5(cube root[(-7x - 5) / (-2x - 2)]) using the properties of logarithms.

Find the derivative of f(x) = log5(cube root[(-7x - 5) / (-2x - 2)]) using logarithmic properties and differentiation rules. This solution covers the application of the chain rule and quotient rule in calculus. Suitable for advanced high school or college-level calculus students.

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