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

(C) $4\left(\log _{e} 2\right)^{2}-2$

(D) $4\left(\log _{e} 2\right)^{2}+2$

JEE Main · 2023 · 01 Feb 2023 · Shift 2 · Differentiation

y (x) = x^{x}, x > 0

, then

y^{''} (2) - 2 y^{'} (2)

is equal to :

Answer: (C)

From

y = x^{x}

, taking logarithms gives

ln y = x ln x

, so

\frac{y ^{'}}{y} = 1 + ln x

and

y^{'} = x^{x} (1 + ln x)

. Differentiating again,

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

. At

x = 2

this gives

y^{'} (2) = 4 (1 + ln 2)

and

y^{''} (2) = 4 (1 + ln 2)^{2} + 2

. Hence

y^{''} (2) - 2 y^{'} (2) = 4 (1 + ln 2)^{2} + 2 - 8 (1 + ln 2) = 4 (1 + ln 2) [(1 + ln 2) - 2] + 2 = 4 (ln 2)^{2} - 2

JEE Main · 2023 · 01 Feb 2023 · Shift 2 · Single correct option