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

The correct answer is Option (3) → \( 4(\log_e 2)^2-2 \)

By logarithmic differentiation,

    \( \displaystyle \log y=x\log x \)

    \( \displaystyle \frac{1}{y}\,y'=\log x+1 \)

    \( \displaystyle y'=x^x(1+\log x) \)

Differentiating again by the product rule,

    \( \displaystyle y''=x^x(1+\log x)^2+x^x\cdot\frac{1}{x} \)

At x = 2, \( x^x=4 \). Writing L for \( \log_e 2 \):

    \( \displaystyle y'(2)=4(1+L) \)

    \( \displaystyle y''(2)=4(1+L)^2+\frac{4}{2} \)

    \( \displaystyle =4(1+L)^2+2 \)

    \( \displaystyle y''(2)-2y'(2)=4(1+L)^2+2-8(1+L) \)

    \( \displaystyle =4+8L+4L^2+2-8-8L \)

    \( \displaystyle =4L^2-2 \)

    \( \displaystyle =4(\log_e 2)^2-2 \)