`y=x^(sinx)+a^(sinx)`

Answer

To find the derivative of the function \( y = x^{\sin x} + a^{\sin x} \), we will differentiate each term separately and then combine the results. Here’s the step-by-step solution:

### Step 1: Differentiate the first term \( x^{\sin x} \)

We will use the logarithmic differentiation method for the term \( x^{\sin x} \).

1. Let \( t = x^{\sin x} \).
2. Taking the natural logarithm of both sides, we have:
\[ \ln t = \sin x \cdot \ln x \]
3. Differentiate both sides with respect to \( x \):
\[ \frac{1}{t} \frac{dt}{dx} = \cos x \cdot \ln x + \sin x \cdot \frac{1}{x} \]
4. Rearranging gives:
\[ \frac{dt}{dx} = t \left( \cos x \cdot \ln x + \frac{\sin x}{x} \right) \]
5. Substituting back \( t = x^{\sin x} \):
\[ \frac{d}{dx}(x^{\sin x}) = x^{\sin x} \left( \cos x \cdot \ln x + \frac{\sin x}{x} \right) \]

### Step 2: Differentiate the second term \( a^{\sin x} \)

For the term \( a^{\sin x} \), we will use the chain rule:

1. The derivative of \( a^{\sin x} \) is given by:
\[ \frac{d}{dx}(a^{\sin x}) = a^{\sin x} \cdot \ln a \cdot \cos x \]

### Step 3: Combine the derivatives

Now, we combine the derivatives from both terms:

\[ \frac{dy}{dx} = \frac{d}{dx}(x^{\sin x}) + \frac{d}{dx}(a^{\sin x}) \]

Substituting the derivatives we found:

\[ \frac{dy}{dx} = x^{\sin x} \left( \cos x \cdot \ln x + \frac{\sin x}{x} \right) + a^{\sin x} \cdot \ln a \cdot \cos x \]

### Final Answer

Thus, the derivative of the function \( y = x^{\sin x} + a^{\sin x} \) is:

\[ \frac{dy}{dx} = x^{\sin x} \left( \cos x \cdot \ln x + \frac{\sin x}{x} \right) + a^{\sin x} \cdot \ln a \cdot \cos x \]

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