If $$y(x) = x^x, x > 0$$, then $$y''(2) - 2y'(2)$$ is equal to :

Given $$y(x) = x^x$$ for $$x > 0$$, we need to find $$y''(2) - 2y'(2)$$. Taking logarithms gives $$\ln y = x \ln x$$, and differentiating yields $$\frac{y'}{y} = \ln x + 1$$ so $$y' = x^x(\ln x + 1)$$.

Differentiating again using the product rule, we have $$y'' = \frac{d}{dx}[x^x(\ln x + 1)] = y'(\ln x + 1) + x^x \cdot \frac{1}{x} = x^x(\ln x + 1)^2 + x^{x-1}$$.

Evaluating at $$x = 2$$ gives $$y(2) = 4$$, $$y'(2) = 4(\ln 2 + 1)$$, and $$y''(2) = 4(\ln 2 + 1)^2 + 2$$. Therefore,

$$y''(2) - 2y'(2) = 4(\ln 2 + 1)^2 + 2 - 8(\ln 2 + 1) = 4(\ln^2 2 + 2\ln 2 + 1) + 2 - 8\ln 2 - 8 = 4\ln^2 2 + 8\ln 2 + 4 + 2 - 8\ln 2 - 8 = 4(\log_e 2)^2 - 2$$.

4(\log_e 2)^2 - 2