Step 1: Understanding the Concept:
To differentiate a function where both the base and exponent are variables, we use logarithmic differentiation[cite: 1]. We first find the first derivative \( y' \), and then differentiate again to find the second derivative \( y'' \).
Step 2: Key Formula or Approach:
1. For \( y = x^x \), \( y' = x^x(1 + \log_e x) \).
2. Use the product rule for \( y'' \): \( \frac{d}{dx}[uv] = u'v + uv' \).
Step 3: Detailed Explanation:
Given \( y = x^x \), the first derivative is \( y'(x) = x^x(1 + \log_e x) \).
At \( x = 2 \):
\[ y'(2) = 2^2(1 + \log_e 2) = 4(1 + \log_e 2) = 4 + 4 \log_e 2 \]
Now find \( y''(x) \) using the product rule on \( y'(x) \):
\[ y''(x) = \frac{d}{dx}[x^x] \cdot (1 + \log_e x) + x^x \cdot \frac{d}{dx}(1 + \log_e x) \]
\[ y''(x) = x^x(1 + \log_e x)^2 + x^x(\frac{1}{x}) \]
At \( x = 2 \):
\[ y''(2) = 2^2(1 + \log_e 2)^2 + 2^2(\frac{1}{2}) = 4(1 + 2\log_e 2 + (\log_e 2)^2) + 2 \]
\[ y''(2) = 4 + 8\log_e 2 + 4(\log_e 2)^2 + 2 = 6 + 8\log_e 2 + 4(\log_e 2)^2 \]
Now calculate \( y''(2) - 2y'(2) \):
\[ (6 + 8\log_e 2 + 4(\log_e 2)^2) - 2(4 + 4\log_e 2) \]
\[ = 6 + 8\log_e 2 + 4(\log_e 2)^2 - 8 - 8\log_e 2 \]
\[ = 4(\log_e 2)^2 - 2 \]
Step 4: Final Answer:
The value of the expression is \( 4 (\log_e 2)^2 - 2 \).