Step 1: Understanding the Concept:
The function is a sum of two terms where both the base and exponent are variables ($f(x)^{g(x)}$ form).
We cannot directly use standard power or exponential rules. We must use logarithmic differentiation for each term separately.
Step 2: Key Formula or Approach:
Let $y = u + v$, then $\frac{dy}{dx} = \frac{du}{dx} + \frac{dv}{dx}$.
For $u = f(x)^{g(x)}$, take natural logarithm: $\ln u = g(x) \ln f(x)$.
Differentiate implicitly: $\frac{1}{u} \frac{du}{dx} = g'(x) \ln f(x) + g(x) \frac{f'(x)}{f(x)}$.
Step 3: Detailed Explanation:
Let $y = u + v$ where $u = x^x$ and $v = x^{\frac{1}{x}}$.
Part 1: Differentiate $u = x^x$
Take natural logarithm on both sides:
\[ \log u = \log(x^x) = x \log x \]
Differentiate with respect to $x$ using the product rule:
\[ \frac{1}{u} \frac{du}{dx} = (1) \cdot \log x + x \cdot \left(\frac{1}{x}\right) \]
\[ \frac{1}{u} \frac{du}{dx} = \log x + 1 \]
\[ \frac{du}{dx} = u (1 + \log x) = x^x(1 + \log x) \]
Part 2: Differentiate $v = x^{\frac{1}{x}}$
Take natural logarithm on both sides:
\[ \log v = \log\left(x^{\frac{1}{x}}\right) = \frac{1}{x} \log x \]
Differentiate with respect to $x$ using the quotient rule (or product rule with $x^{-1}$):
Let's use quotient rule on $\frac{\log x}{x}$:
\[ \frac{1}{v} \frac{dv}{dx} = \frac{x \cdot \frac{d}{dx}(\log x) - \log x \cdot \frac{d}{dx}(x)}{x^2} \]
\[ \frac{1}{v} \frac{dv}{dx} = \frac{x \cdot \left(\frac{1}{x}\right) - \log x \cdot (1)}{x^2} \]
\[ \frac{1}{v} \frac{dv}{dx} = \frac{1 - \log x}{x^2} \]
\[ \frac{dv}{dx} = v \left( \frac{1 - \log x}{x^2} \right) = x^{\frac{1}{x}} \frac{1}{x^2} (1 - \log x) \]
Now, add the derivatives together:
\[ \frac{dy}{dx} = \frac{du}{dx} + \frac{dv}{dx} \]
\[ \frac{dy}{dx} = x^x(1 + \log x) + x^{\frac{1}{x}} \frac{1}{x^2} (1 - \log x) \]
Step 4: Final Answer:
The derivative is $x^x(1 + \log x) + x^{\frac{1}{x}} \frac{1}{x^2} (1 - \log x)$.