Step 1: Understanding the Concept:
To differentiate a function of the form $f(x)^{g(x)}$, where both the base and the exponent are functions of x, we must use logarithmic differentiation. The power rule or exponential rule cannot be applied directly.
Step 2: Key Formula or Approach:
The process of logarithmic differentiation involves these steps:
1. Let $y = f(x)^{g(x)}$.
2. Take the natural logarithm of both sides: $\ln y = g(x) \ln(f(x))$.
3. Differentiate both sides implicitly with respect to x, using the product rule on the right side.
4. Solve for $\frac{dy}{dx}$.
Step 3: Detailed Explanation:
Let $y = x^x$.
Take the natural logarithm of both sides:
\[ \ln y = \ln(x^x) \]
Using the property of logarithms, bring the exponent down:
\[ \ln y = x \ln x \]
Now differentiate both sides with respect to x. Use the chain rule on the left and the product rule on the right.
\[ \frac{d}{dx}(\ln y) = \frac{d}{dx}(x \ln x) \]
\[ \frac{1}{y} \frac{dy}{dx} = \left(\frac{d}{dx}(x)\right) \ln x + x \left(\frac{d}{dx}(\ln x)\right) \]
\[ \frac{1}{y} \frac{dy}{dx} = (1) \ln x + x \left(\frac{1}{x}\right) \]
\[ \frac{1}{y} \frac{dy}{dx} = \ln x + 1 \]
To find $\frac{dy}{dx}$, multiply both sides by y:
\[ \frac{dy}{dx} = y (1 + \ln x) \]
Substitute back $y = x^x$:
\[ \frac{dy}{dx} = x^x(1 + \ln x) \]
(Note: In this context, $\log x$ usually means $\ln x$).
Step 4: Final Answer:
The derivative of $x^x$ is $x^x(1 + \log x)$. Therefore, option (D) is correct.