JEE MAIN - Mathematics (2023 - 1st February Evening Shift - No. 6)
If $$y(x)=x^{x},x > 0$$, then $$y''(2)-2y'(2)$$ is equal to
$$4(\log_{e}2)^{2}+2$$
$$8\log_{e}2-2$$
$$4\log_{e}2+2$$
$$4(\log_{e}2)^{2}-2$$
Explanation
$\begin{aligned} & y=x^x \\\\ & y^{\prime}=x^x(1+\ln x) \\\\ & y^{\prime \prime}=x^x(1+\ln x)^2+\frac{x^x}{x} \\\\ & f^{\prime \prime}(2)-2 f^{\prime}(2)=\left(4(1+\ln 2)^2+2\right)-(2)(4(1+\ln 2)) \\\\ & =4\left(1+(\ln 2)^2\right)+2-8 \\\\ & =4(\ln 2)^2-2 \\\\ & \end{aligned}$
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