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

Answer

To differentiate the function \( y = x^x + (\sin x)^x \) with respect to \( x \), we will break it down into two parts and differentiate each part separately.

### Step 1: Differentiate \( y_1 = x^x \)

1. **Take the natural logarithm of both sides**:
\[ \log y_1 = \log(x^x) = x \log x \]

2. **Differentiate both sides with respect to \( x \)** using implicit differentiation:
\[ \frac{1}{y_1} \frac{dy_1}{dx} = \frac{d}{dx}(x \log x) \]

3. **Differentiate the right-hand side** using the product rule:
\[ \frac{d}{dx}(x \log x) = \log x + x \cdot \frac{1}{x} = \log x + 1 \]

4. **Multiply both sides by \( y_1 \)** to solve for \( \frac{dy_1}{dx} \):
\[ \frac{dy_1}{dx} = y_1 (\log x + 1) = x^x (\log x + 1) \]

### Step 2: Differentiate \( y_2 = (\sin x)^x \)

1. **Take the natural logarithm of both sides**:
\[ \log y_2 = \log((\sin x)^x) = x \log(\sin x) \]

2. **Differentiate both sides with respect to \( x \)**:
\[ \frac{1}{y_2} \frac{dy_2}{dx} = \frac{d}{dx}(x \log(\sin x)) \]

3. **Differentiate the right-hand side** using the product rule:
\[ \frac{d}{dx}(x \log(\sin x)) = \log(\sin x) + x \cdot \frac{1}{\sin x} \cdot \cos x = \log(\sin x) + x \cot x \]

4. **Multiply both sides by \( y_2 \)** to solve for \( \frac{dy_2}{dx} \):
\[ \frac{dy_2}{dx} = y_2 \left(\log(\sin x) + x \cot x\right) = (\sin x)^x \left(\log(\sin x) + x \cot x\right) \]

### Step 3: Combine the results

Now, we can find the total derivative \( \frac{dy}{dx} \):
\[ \frac{dy}{dx} = \frac{dy_1}{dx} + \frac{dy_2}{dx} \]
Substituting the expressions we found:
\[ \frac{dy}{dx} = x^x (\log x + 1) + (\sin x)^x \left(\log(\sin x) + x \cot x\right) \]

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