Applications of Derivatives
Logarithmic differentiation of $x^x$
nta_pyq_2023_jan
Grade 12
Question:
If $y(x) = x^{x}$, $x > 0$, then $y''(2) - 2y'(2)$ is equal to
(1) $8\log_{e}2 - 2$
(2) $4\log_{e}2 + 2$
(3) $4(\log_{e}2)^{2} - 2$
(4) $4(\log_{e}2)^{2} + 2$
$8\log_{e}2 - 2$
$4\log_{e}2 + 2$
$4(\log_{e}2)^{2} - 2$
$4(\log_{e}2)^{2} + 2$
Step-by-Step Solution
Key Concept: Use logarithmic differentiation: $\ln y = x \ln x$, giving $y' = x^x(1 + \ln x)$ and $y'' = x^x(1+\ln x)^2 + x^{x-1}$. Evaluate at $x=2$.
$y = x^x \Rightarrow y' = x^x(1+\ln x)$. $y'' = x^x(1+\ln x)^2 + x^{x-1}$. At $x=2$: $y'(2) = 4(1+\ln 2)$, $y''(2) = 4(1+\ln 2)^2 + 2$. $y''(2) - 2y'(2) = 4(1+\ln 2)^2 + 2 - 8(1+\ln 2) = 4(1+\ln 2)(1+\ln 2 - 2) + 2 = 4(\ln 2)^2 - 4\ln 2 \cdot 1 + 2$. Simplifying: $= 4(\ln 2)^2 - 4\ln 2 + 4\ln 2 - 4 + 2 = 4(\ln 2)^2 - 2$. Answer: (3).
Correct Answer: 3