Question

If \(y(x)=x^x,\ x>0\), then \[ y''(2)-2y'(2)= \]

• A. \(4\log_e2-2\)
• B. \(4\log_e2+2\)
• C. \(4(\log_e2)^2+2\)
• D. \(4(\log_e2)^2-2\)

Hint:

For \(x^x\), always use logarithmic differentiation: write \(\log y=x\log x\).

Answer 1

The Correct Option is D

Concept: For \(y=x^x\), use logarithmic differentiation.

Step 1: Given: \[ y=x^x \] Taking logarithm: \[ \log y=x\log x \]

Step 2: Differentiate both sides. \[ \frac{1}{y}y'=\log x+1 \] \[ y'=x^x(\log x+1) \]

Step 3: Differentiate again. \[ y''=\frac{d}{dx}\left[x^x(\log x+1)\right] \] Using product rule: \[ y''=x^x(\log x+1)^2+x^x\cdot \frac{1}{x} \]

Step 4: Find \(y'(2)\). \[ y'(2)=2^2(\log_e2+1) \] \[ y'(2)=4(\log_e2+1) \]

Step 5: Find \(y''(2)\). \[ y''(2)=4(\log_e2+1)^2+4\cdot\frac{1}{2} \] \[ y''(2)=4(\log_e2+1)^2+2 \]

Step 6: Now calculate: \[ y''(2)-2y'(2) \] \[ =4(\log_e2+1)^2+2-2[4(\log_e2+1)] \] \[ =4(\log_e2+1)^2+2-8(\log_e2+1) \] Let \(L=\log_e2\). Then: \[ =4(L+1)^2+2-8(L+1) \] \[ =4(L^2+2L+1)+2-8L-8 \] \[ =4L^2+8L+4+2-8L-8 \] \[ =4L^2-2 \] Thus: \[ y''(2)-2y'(2)=4(\log_e2)^2-2 \] Therefore, \[ \boxed{4(\log_e2)^2-2} \]