The problem asks for the second derivative of $y = x^x$. Here's the solution:

1. Rewrite $y$ in logarithmic terms: $y = x^x = e^{x \ln x}.$ Differentiating $y = e^{x \ln x}$ using the chain rule gives: $y' = e^{x \ln x} \cdot \frac{d}{dx}(x \ln x).$ Since $e^{x \ln x} = x^x$, this simplifies to: $y' = x^x \cdot \left( \ln x + 1 \right).$

2. Find the second derivative $y''$: Differentiate $y'$ again: $y' = x^x \cdot \left( \ln x + 1 \right).$ Using the product rule $(uv)' = u'v + uv'$, we calculate:

   • $u = x^x$, so $u' = x^x (\ln x + 1)$,
   • $v = (\ln x + 1)$, so $v' = \frac{1}{x}$.

   Now substitute into the product rule: $y'' = \left( x^x (\ln x + 1) \right)(\ln x + 1) + x^x \cdot \frac{1}{x}.$ Simplify terms: $y'' = x^x (\ln x + 1)^2 + x^{x-1}.$

Final Answer:

The correct option is: $\boxed{(c) \ y(1 + \ln x)^2 + x^{x-1}}$

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Related Questions:

1. How do you derive the derivative of exponential functions with variable exponents?
2. Can logarithmic differentiation be applied to other forms of $y = u(x)^{v(x)}$?
3. What are the differences between the product rule and chain rule?
4. Why does $e^{x \ln x}$ simplify to $x^x$?
5. How do second derivatives help in analyzing functions?

Tip:

For functions like $x^x$, rewriting them in terms of exponential functions simplifies differentiation.