We are tasked with differentiating the expression: \[ f(x) = x^x + x^a + a^x + a^a \] We will differentiate each term separately. 1. Differentiating \( x^x \): The derivative of \( x^x \) can be computed using logarithmic differentiation. Take the natural logarithm of \( x^x \): \[ \ln(y) = \ln(x^x) = x \ln(x) \] Now differentiate both sides with respect to \( x \): \[ \frac{d}{dx} \left( \ln(y) \right) = \frac{d}{dx} \left( x \ln(x) \right) \] Using the product rule: \[ \frac{1}{y} \frac{dy}{dx} = \ln(x) + 1 \] So: \[ \frac{dy}{dx} = x^x (1 + \log x) \] 2. Differentiating \( x^a \): The derivative of \( x^a \) (where \( a \) is constant) is: \[ \frac{d}{dx} \left( x^a \right) = a x^{a-1} \] 3. Differentiating \( a^x \): The derivative of \( a^x \) (where \( a \) is constant) is: \[ \frac{d}{dx} \left( a^x \right) = a^x \log a \] 4. **Differentiating \( a^a \): Since \( a^a \) is a constant, its derivative is 0. Now, summing all the derivatives: \[ \frac{d}{dx} \left( x^x + x^a + a^x + a^a \right) = x^x (1 + \log x) + a x^{a-1} + a^x \log a \]
Therefore, the correct answer is (B).