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

Answer

To differentiate the function \( y = x^{\sin x} + (\log x)^{x} \) with respect to \( x \), we can follow these steps:

### Step 1: Rewrite the Function
Let \( y = x^{\sin x} + (\log x)^{x} \).

### Step 2: Differentiate Each Term
We will differentiate each term separately.

1. **Differentiate \( x^{\sin x} \)**:
We can express \( x^{\sin x} \) in exponential form:
\[ x^{\sin x} = e^{\sin x \cdot \log x} \]
Now, using the chain rule:
\[ \frac{dy_1}{dx} = e^{\sin x \cdot \log x} \cdot \frac{d}{dx}(\sin x \cdot \log x) \]

Now we apply the product rule to differentiate \( \sin x \cdot \log x \):
\[ \frac{d}{dx}(\sin x \cdot \log x) = \sin x \cdot \frac{d}{dx}(\log x) + \log x \cdot \frac{d}{dx}(\sin x) \]
\[ = \sin x \cdot \frac{1}{x} + \log x \cdot \cos x \]

Therefore,
\[ \frac{dy_1}{dx} = x^{\sin x} \left( \sin x \cdot \frac{1}{x} + \log x \cdot \cos x \right) \]

2. **Differentiate \( (\log x)^{x} \)**:
Similarly, we can express \( (\log x)^{x} \) in exponential form:
\[ (\log x)^{x} = e^{x \cdot \log(\log x)} \]
Using the chain rule again:
\[ \frac{dy_2}{dx} = e^{x \cdot \log(\log x)} \cdot \frac{d}{dx}(x \cdot \log(\log x)) \]

Again, applying the product rule:
\[ \frac{d}{dx}(x \cdot \log(\log x)) = \log(\log x) + x \cdot \frac{1}{\log x} \cdot \frac{1}{x} = \log(\log x) + \frac{1}{\log x} \]

Therefore,
\[ \frac{dy_2}{dx} = (\log x)^{x} \left( \log(\log x) + \frac{1}{\log x} \right) \]

### Step 3: Combine the Results
Now, we can combine the derivatives of both terms:
\[ \frac{dy}{dx} = \frac{dy_1}{dx} + \frac{dy_2}{dx} \]
Substituting the expressions we found:
\[ \frac{dy}{dx} = x^{\sin x} \left( \sin x \cdot \frac{1}{x} + \log x \cdot \cos x \right) + (\log x)^{x} \left( \log(\log x) + \frac{1}{\log x} \right) \]

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