Question

If \(y=x^x,\ x>0\) then \(y^{\prime\prime}(2)-2y^{\prime}(2)\) is equal to

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

Hint:

Use \(y'=x^x(1+\ln x)\) and \(y''=x^x(1+\ln x)^2+x^{x-1}\).

Answer 1

The Correct Option is C

Step 1: Take logarithms.
Since \(y=x^x\), we get \(\ln y = x\ln x\). Differentiate both sides.
\[ \frac{y'}{y} = \ln x + 1 \Rightarrow y' = x^x(1+\ln x) \]

Step 2: Find the second derivative.
Use the product rule on \(y' = y(1+\ln x)\).
\[ y'' = y'(1+\ln x) + y\cdot\frac1x = x^x(1+\ln x)^2 + x^{x-1} \]

Step 3: Put x = 2.
Let \(L=\ln 2\). Then \(y=2^2=4\), so
\[ y'(2) = 4(1+L), \qquad y''(2) = 4(1+L)^2 + 2 \]

Step 4: Form y'' - 2y'.
\[ y''(2)-2y'(2) = 4(1+2L+L^2)+2-8(1+L) \] \[ = 4+8L+4L^2+2-8-8L = 4L^2-2 \]

Step 5: Check the options.
The result is \(4(\log_e 2)^2-2\). This is option 3. Option 1 has no square term. Option 2 has \(+2\) instead of \(-2\). Option 4 has a plain \(\log_e 2\), which cancelled out.

Final Answer:
The value is \(4(\log_e 2)^2-2\), option 3. \[ \boxed{4(\log_e 2)^2-2} \]