`(d)/(dx) [x^(x) + x^(a) + a^(x) + a^(a)] = …, a ` is constant

Options

A. `x^(x) (1 + log x) + a - x^(a-1)`
B. ` x^(x) (1 + log x) + a. x^(a-1) + a^(x) log a `
C. `x^(x) (1 + log x) + a^(a) (1 + loga)`
D. `x^(x) (1 + log x) + a^(a) (1 + log a) + ax^(a-1) + a^(a) (1 + log a)`

Explanation

To solve the problem \(\frac{d}{dx} \left[ x^x + x^a + a^x + a^a \right]\), where \(a\) is a constant, we will differentiate each term separately.

### Step-by-Step Solution:

1. **Differentiate \(x^x\)**:
- Let \(u = x^x\).
- Taking the natural logarithm: \(\ln u = x \ln x\).
- Differentiate both sides:
\[ \frac{1}{u} \frac{du}{dx} = \ln x + 1 \]
- Therefore,
\[ \frac{du}{dx} = u(\ln x + 1) = x^x(\ln x + 1). \]

2. **Differentiate \(x^a\)**:
- Let \(v = x^a\).
- The derivative is straightforward:
\[ \frac{dv}{dx} = ax^{a-1}. \]

3. **Differentiate \(a^x\)**:
- Let \(w = a^x\).
- Taking the natural logarithm: \(\ln w = x \ln a\).
- Differentiate both sides:
\[ \frac{1}{w} \frac{dw}{dx} = \ln a. \]
- Therefore,
\[ \frac{dw}{dx} = w \ln a = a^x \ln a. \]

4. **Differentiate \(a^a\)**:
- Let \(z = a^a\).
- Since \(a^a\) is a constant, its derivative is:
\[ \frac{dz}{dx} = 0. \]

5. **Combine all derivatives**:
- Now, we can combine all the derivatives:
\[ \frac{d}{dx} \left[ x^x + x^a + a^x + a^a \right] = \frac{du}{dx} + \frac{dv}{dx} + \frac{dw}{dx} + \frac{dz}{dx}. \]
- Substituting the values we found:
\[ \frac{d}{dx} \left[ x^x + x^a + a^x + a^a \right] = x^x(\ln x + 1) + ax^{a-1} + a^x \ln a + 0. \]

6. **Final Result**:
- Thus, the final answer is:
\[ \frac{d}{dx} \left[ x^x + x^a + a^x + a^a \right] = x^x(\ln x + 1) + ax^{a-1} + a^x \ln a. \]