1. The problem is to find the derivative $y'(x)$ of the function $y(x) = x^x$. 2. To differentiate $y = x^x$, first rewrite it using logarithms: $$y = x^x \implies \ln y = \ln(x^x) = x \ln x$$ 3. Differentiate both sides with respect to $x$ using implicit differentiation: $$\frac{d}{dx}(\ln y) = \frac{d}{dx}(x \ln x)$$ 4. Calculate derivatives: $$\frac{1}{y} y' = 1 \cdot \ln x + x \cdot \frac{1}{x} = \ln x + 1$$ 5. Solve for $y'$: $$y' = y (\ln x + 1) = x^x (\ln x + 1)$$ 6. Therefore, the derivative of $y = x^x$ is: $$y'(x) = x^x (\ln x + 1)$$ Note: The symbol in the bottom-right quadrant you mentioned is not a standard mathematical symbol and is not relevant to the differentiation problem given. Final answer: $$y'(x) = x^x (\ln x + 1)$$