If y ( x ) = x x , x 0 , then y ( 2 ) - 2 y ( 2 ) is equal to [2023]

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Solution

Q.
$If y (x) = x^{x}, x > 0, then y'' (2) - 2 y' (2) is equal to$ [2023]

1 $4 {(\log_{e} 2)}^{2} - 2$

2 $8 \log_{e} 2 - 2$

3 $4 {(\log_{e} 2)}^{2} + 2$

4 $4 \log_{e} 2 + 2$

Ans.
(1)

Given, $y = x^{x}$

Taking log on both sides

$\log y = x \log_{e} x,$ $y' = x^{x} (1 + \log_{e} x)$

$y'' = x^{x} {(1 + \log_{e} x)}^{2} + x^{x} \cdot \frac{1}{x}$

Now, find $y'' (2)$ and $y' (2)$

$y'' (2) = 4 {(1 + \log_{e} 2)}^{2} + 2$

$y' (2) = 4 (1 + \log_{e} 2)$

Now, $y'' (2) - 2 y' (2) = 4 {(1 + \log_{e} 2)}^{2} + 2 - 2 [4 (1 + \log_{e} 2)]$

$= 4 {(1 + \log_{e} 2)}^{2} + 2 - 8 (1 + \log_{e} 2)$

$= 4 (1 + \log_{e} 2) [1 + \log_{e} 2 - 2] + 2$

$= 4 {(\log_{e} 2)}^{2} - 1 + 2 = 4 {(\log_{e} 2)}^{2} - 2$

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