Question

If \( y(x) = x^x, \, x > 0 \), then \( y''(2) - 2y'(2) \) is equal to:

• A. \( 4 (\log_e 2)^2 - 2 \)
• B. \( 8 \log_e 2 - 2 \)
• C. \( 4 (\log_e 2)^2 + 2 \)
• D. \( 4 \log_e 2 + 2 \)

Hint:

When working with functions like x^x, take the natural logarithm for simplification. This often helps to differentiate effectively using logarithmic properties.

Answer 1

The Correct Option is A

Given, \( y = x^x \).
\[y' = x^x (1 + \ln x)\]
\[y'' = x^x (1 + \ln x)^2 + x^x \cdot \frac{1}{x}\]
Substituting \( x = 2 \):
\[y'(2) = 4(1 + \ln 2)\]
\[y''(2) = 4(1 + \ln 2)^2 + 2\]
Now, calculate \( y''(2) - 2y'(2) \):
\[y''(2) - 2y'(2) = 4(1 + \ln 2)^2 + 2 - 8(1 + \ln 2)\]
Simplify:
\[= 4(1 + \ln 2)[1 + \ln 2 - 2] + 2\]
\[= 4(1 + \ln 2)(\ln 2 - 1) + 2\]
\[= 4(\ln 2)^2 - 4 \ln 2 + 4 \ln 2 - 4 + 2\]
\[= 4(\ln 2)^2 - 2\]
Final Result: \( y''(2) - 2y'(2) = 4(\ln 2)^2 - 2 \).