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

Answer

To differentiate the function \( y = x^{\sin x} + \sin^x \) with respect to \( x \), we will break it down into two parts: \( u = x^{\sin x} \) and \( v = \sin^x \). We will find the derivatives \( \frac{du}{dx} \) and \( \frac{dv}{dx} \) separately and then combine them.

### Step-by-Step Solution:

1. **Define the Function**:
\[ y = x^{\sin x} + \sin^x \]

2. **Differentiate \( u = x^{\sin x} \)**:
- Take the natural logarithm of both sides:
\[ \log u = \sin x \cdot \log x \]
- Differentiate both sides with respect to \( x \):
\[ \frac{1}{u} \frac{du}{dx} = \cos x \cdot \log x + \sin x \cdot \frac{1}{x} \]
- Multiply through by \( u \):
\[ \frac{du}{dx} = u \left( \cos x \cdot \log x + \frac{\sin x}{x} \right) \]
- Substitute back \( u = x^{\sin x} \):
\[ \frac{du}{dx} = x^{\sin x} \left( \cos x \cdot \log x + \frac{\sin x}{x} \right) \]

3. **Differentiate \( v = \sin^x \)**:
- Take the natural logarithm of both sides:
\[ \log v = x \cdot \log(\sin x) \]
- Differentiate both sides with respect to \( x \):
\[ \frac{1}{v} \frac{dv}{dx} = \log(\sin x) + x \cdot \frac{\cos x}{\sin x} \]
- Multiply through by \( v \):
\[ \frac{dv}{dx} = v \left( \log(\sin x) + x \cdot \cot x \right) \]
- Substitute back \( v = \sin^x \):
\[ \frac{dv}{dx} = \sin^x \left( \log(\sin x) + x \cdot \cot x \right) \]

4. **Combine the Results**:
- Now, we can find \( \frac{dy}{dx} \):
\[ \frac{dy}{dx} = \frac{du}{dx} + \frac{dv}{dx} \]
- Substitute the expressions we found:
\[ \frac{dy}{dx} = x^{\sin x} \left( \cos x \cdot \log x + \frac{\sin x}{x} \right) + \sin^x \left( \log(\sin x) + x \cdot \cot x \right) \]

### Final Answer:
\[ \frac{dy}{dx} = x^{\sin x} \left( \cos x \cdot \log x + \frac{\sin x}{x} \right) + \sin^x \left( \log(\sin x) + x \cdot \cot x \right) \]