If y(x) = x^x, x 0,then y^{prime prime}(2) | vedclass

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(C) Given $y(x) = x^x$. Taking the natural logarithm on both sides,$\ln y = x \ln x$.Differentiating with respect to $x$,$\frac{1}{y} y^{\prime} = 1 \

Vedclass · Accepted Answer

$4(\log_e 2)^2 - 2$

Vedclass · Answer

(C) Given $y(x) = x^x$. Taking the natural logarithm on both sides,$\ln y = x \ln x$.Differentiating with respect to $x$,$\frac{1}{y} y^{\prime} = 1 \cdot \ln x + x \cdot \frac{1}{x} = \ln x + 1$.Thus,$y^{\prime} = y(1 + \ln x) = x^x(1 + \ln x)$.Now,differentiate $y^{\prime}$ with respect to $x$ using the product rule:$y^{\prime \prime} = \frac{d}{dx}[x^x] \cdot (1 + \ln x) + x^x \cdot \frac{d}{dx}[1 + \ln x]$$y^{\prime \prime} = x^x(1 + \ln x)(1 + \ln x) + x^x \cdot \frac{1}{x} = x^x(1 + \ln x)^2 + x^{x-1}$.At $x = 2$:$y^{\prime}(2) = 2^2(1 + \ln 2) = 4(1 + \ln 2)$.$y^{\prime \prime}(2) = 2^2(1 + \ln 2)^2 + 2^{2-1} = 4(1 + \ln 2)^2 + 2$.Now calculate $y^{\prime \prime}(2) - 2y^{\prime}(2)$:$= 4(1 + \ln 2)^2 + 2 - 2[4(1 + \ln 2)]$$= 4(1 + 2\ln 2 + (\ln 2)^2) + 2 - 8 - 8\ln 2$$= 4 + 8\ln 2 + 4(\ln 2)^2 + 2 - 8 - 8\ln 2$$= 4(\ln 2)^2 - 2$.

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

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