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The problem you're working with is:

\[
y = \log_6\left( (x^4 - 4x^3 + 1)^5 + 4x^2 - 1200x \right)
\]
and you're asked to find \( y'(2) \) and round your answer to three decimal places.

### Steps to solve:

1. **Differentiate using the chain rule and logarithmic differentiation:**

   - Recall the derivative of \( \log_b(u) \) is \( \frac{1}{u \ln(b)} \times u' \), where \( u \) is the argument of the logarithm, and \( b \) is the base.
   - In this case, \( u = (x^4 - 4x^3 + 1)^5 + 4x^2 - 1200x \) and the base \( b = 6 \).

   So, using the chain rule, we have:

   \[
   y' = \frac{1}{u \ln(6)} \cdot u'
   \]
   
   where \( u' \) is the derivative of the inside function \( u \).

2. **Find \( u' \):**

   - \( u = (x^4 - 4x^3 + 1)^5 + 4x^2 - 1200x \)
   - We need to differentiate each term:
   
     a. \( \frac{d}{dx} \left( (x^4 - 4x^3 + 1)^5 \right) \) using the chain rule:
        \[
        5 \cdot (x^4 - 4x^3 + 1)^4 \cdot (4x^3 - 12x^2)
        \]
        
     b. \( \frac{d}{dx} \left( 4x^2 \right) = 8x \)
     
     c. \( \frac{d}{dx} \left( -1200x \right) = -1200 \)

     Now, combine these derivatives:
     \[
     u' = 5 \cdot (x^4 - 4x^3 + 1)^4 \cdot (4x^3 - 12x^2) + 8x - 1200
     \]

3. **Substitute \( x = 2 \) into \( u \) and \( u' \):**

   - First, calculate \( u(2) \):
     \[
     u(2) = (2^4 - 4(2)^3 + 1)^5 + 4(2)^2 - 1200(2)
     \]
     \[
     u(2) = (16 - 32 + 1)^5 + 16 - 2400 = (-15)^5 + 16 - 2400 = -759375 + 16 - 2400 = -761759
     \]

   - Now, calculate \( u'(2) \):
     \[
     u'(2) = 5 \cdot (-15)^4 \cdot (4(2)^3 - 12(2)^2) + 8(2) - 1200
     \]
     \[
     u'(2) = 5 \cdot 50625 \cdot (32 - 48) + 16 - 1200 = 5 \cdot 50625 \cdot (-16) + 16 - 1200
     \]
     \[
     u'(2) = 5 \cdot 50625 \cdot (-16) + 16 - 1200 = -4050000 + 16 - 1200 = -4051184
     \]

4. **Finally, calculate \( y'(2) \):**

   \[
   y'(2) = \frac{-4051184}{-761759 \cdot \ln(6)}
   \]

   Approximate \( \ln(6) \approx 1.79176 \):
   \[
   y'(2) = \frac{-4051184}{-761759 \cdot 1.79176} = \frac{-4051184}{-1364562.45} \approx 2.967
   \]

Thus, \( y'(2) \approx 2.967 \), rounded to three decimal places.

Would you like more details on any step?

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Here are 5 related questions to expand this topic:
1. How does logarithmic differentiation work in general cases?
2. Can you explain the chain rule in the context of composite functions?
3. What would change if the logarithm had a different base, say base 10?
4. How do you differentiate more complex nested functions?
5. What is the significance of the derivative in real-world applications?

**Tip:** Always remember to apply the chain rule carefully when differentiating composite functions, especially in logarithmic cases!

Given y = log_6((x^4 - 4x^3 + 1)^5 + 4x^2 - 1200x), find y'(2) and round your answer to three decimal places.

Learn how to differentiate the function y = log_6((x^4 - 4x^3 + 1)^5 + 4x^2 - 1200x) and find y'(2). This problem involves logarithmic differentiation and the chain rule, and is suitable for students in advanced high school or college calculus.

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