Differentiate the following w.r.t. x :
`(x)^(sinx)+sin(x^(x))`

Answer

To differentiate the function \( y = x^{\sin x} + \sin(x^x) \) with respect to \( x \), we will use the rules of differentiation including the product rule and chain rule. Let's go through the steps one by one.

### Step 1: Differentiate \( y = x^{\sin x} + \sin(x^x) \)

We will differentiate each term separately.

1. **Differentiate \( x^{\sin x} \)**:
- We can rewrite \( x^{\sin x} \) using the exponential function:
\[ x^{\sin x} = e^{\sin x \cdot \ln x} \]
- Now, we differentiate \( e^{u} \) where \( u = \sin x \cdot \ln x \):
\[ \frac{dy_1}{dx} = e^{u} \cdot \frac{du}{dx} \]
- To find \( \frac{du}{dx} \), we apply the product rule:
\[ u = \sin x \cdot \ln x \]
- Let \( u_1 = \sin x \) and \( u_2 = \ln x \). Then:
\[ \frac{du}{dx} = u_1 \cdot \frac{du_2}{dx} + u_2 \cdot \frac{du_1}{dx} = \sin x \cdot \frac{1}{x} + \ln x \cdot \cos x \]
- Therefore:
\[ \frac{dy_1}{dx} = x^{\sin x} \left( \sin x \cdot \frac{1}{x} + \ln x \cdot \cos x \right) \]

2. **Differentiate \( \sin(x^x) \)**:
- Again, we use the chain rule:
\[ \frac{dy_2}{dx} = \cos(x^x) \cdot \frac{d}{dx}(x^x) \]
- To differentiate \( x^x \), we can rewrite it as:
\[ x^x = e^{x \ln x} \]
- Differentiating \( e^{v} \) where \( v = x \ln x \):
\[ \frac{dv}{dx} = \ln x + 1 \]
- Thus:
\[ \frac{d}{dx}(x^x) = x^x \cdot (\ln x + 1) \]
- Therefore:
\[ \frac{dy_2}{dx} = \cos(x^x) \cdot x^x \cdot (\ln x + 1) \]

### Step 2: Combine the results

Now we combine the derivatives of both terms to get the final derivative \( \frac{dy}{dx} \):
\[ \frac{dy}{dx} = x^{\sin x} \left( \sin x \cdot \frac{1}{x} + \ln x \cdot \cos x \right) + \cos(x^x) \cdot x^x \cdot (\ln x + 1) \]

### Final Answer:
Thus, the derivative of the function \( y = x^{\sin x} + \sin(x^x) \) with respect to \( x \) is:
\[ \frac{dy}{dx} = x^{\sin x} \left( \frac{\sin x}{x} + \ln x \cdot \cos x \right) + \cos(x^x) \cdot x^x \cdot (\ln x + 1) \]