If $$y(x) = (x^x)^x, x > 0$$ then $$\frac{d^2x}{dy^2} + 20$$ at $$x = 1$$ is equal to ______

Given $$y(x) = (x^x)^x = x^{x^2}$$ for $$x > 0$$, we need to evaluate $$\frac{d^2x}{dy^2} + 20$$ at $$x = 1$$.

Taking the natural logarithm gives $$\ln y = x^2 \ln x$$, and differentiating yields $$\frac{1}{y}\frac{dy}{dx} = 2x \ln x + x$$. Therefore $$\frac{dy}{dx} = y \cdot x(2\ln x + 1) = x^{x^2} \cdot x(2\ln x + 1)$$, and at $$x = 1$$ we have $$y = 1$$ and $$\frac{dy}{dx} = 1\cdot1\cdot1 = 1$$.

To find the second derivative, let $$u = x^{x^2}$$ and $$v = x(2\ln x + 1) = 2x\ln x + x$$ so that $$\frac{dy}{dx} = uv$$. Then $$\frac{du}{dx} = u\cdot(2x\ln x + x) = uv$$ and $$\frac{dv}{dx} = 2\ln x + 3$$, which implies $$\frac{d^2y}{dx^2} = \frac{du}{dx}\,v + u\,\frac{dv}{dx} = uv^2 + u(2\ln x + 3)$$. Evaluating at $$x = 1$$ where $$u = 1$$ and $$v = 1$$ gives $$\frac{d^2y}{dx^2} = 1 + 3 = 4$$.

Using the relation $$\frac{d^2x}{dy^2} = -\frac{\frac{d^2y}{dx^2}}{\bigl(\frac{dy}{dx}\bigr)^3}$$ leads to $$\frac{d^2x}{dy^2} = -\frac{4}{1^3} = -4$$, so $$\frac{d^2x}{dy^2} + 20 = -4 + 20 = 16$$.

The answer is $$16$$.