If `y=a^(x)x^(a),"then "dy/dx` is equal to :

Options

A. `a^(x)x^(a-1)(a-xloga)`
B. `a^(x)x^(a-1)(a+xloga)`
C. `a^(x)x^(a)(a+xloga)`
D. `a^(x)x^(a-1)(x+aloga)`

Answer

To find the derivative of the function \( y = a^x x^a \), we will use the product rule and the chain rule of differentiation.

### Step-by-Step Solution:

1. **Identify the function**:
\[ y = a^x \cdot x^a \]

2. **Apply the product rule**:
The product rule states that if you have two functions \( u \) and \( v \), then the derivative \( \frac{d}{dx}(uv) = u \frac{dv}{dx} + v \frac{du}{dx} \). Here, let:
- \( u = a^x \)
- \( v = x^a \)

3. **Differentiate \( u \)**:
\[ \frac{du}{dx} = \frac{d}{dx}(a^x) = a^x \ln(a) \]

4. **Differentiate \( v \)**:
\[ \frac{dv}{dx} = \frac{d}{dx}(x^a) = a x^{a-1} \]

5. **Substitute into the product rule**:
\[ \frac{dy}{dx} = u \frac{dv}{dx} + v \frac{du}{dx} \]
Substituting \( u \) and \( v \):
\[ \frac{dy}{dx} = a^x \cdot (a x^{a-1}) + x^a \cdot (a^x \ln(a)) \]

6. **Simplify the expression**:
\[ \frac{dy}{dx} = a^{x+1} x^{a-1} + a^x x^a \ln(a) \]

7. **Final result**:
\[ \frac{dy}{dx} = a^x \left( a x^{a-1} + x^a \ln(a) \right) \]