`y=(sinx)^(logx)`

Answer

To find the derivative of the function \( y = (\sin x)^{\log x} \), we can follow these steps:

### Step 1: Take the logarithm of both sides
We start by taking the natural logarithm of both sides to simplify the expression:
\[ \log y = \log((\sin x)^{\log x}) \]

### Step 2: Use the power rule of logarithms
Using the property of logarithms that states \( \log(a^b) = b \cdot \log a \), we can rewrite the equation:
\[ \log y = \log x \cdot \log(\sin x) \]

### Step 3: Differentiate both sides with respect to \( x \)
Now, we differentiate both sides with respect to \( x \). We will use implicit differentiation on the left side and the product rule on the right side:
\[ \frac{1}{y} \frac{dy}{dx} = \frac{d}{dx}(\log x \cdot \log(\sin x)) \]

Using the product rule, we have:
\[ \frac{d}{dx}(\log x \cdot \log(\sin x)) = \log(\sin x) \cdot \frac{d}{dx}(\log x) + \log x \cdot \frac{d}{dx}(\log(\sin x)) \]

### Step 4: Compute the derivatives
Now we compute the derivatives:
- The derivative of \( \log x \) is \( \frac{1}{x} \).
- The derivative of \( \log(\sin x) \) is \( \frac{\cos x}{\sin x} = \cot x \).

Substituting these derivatives back into our equation gives:
\[ \frac{1}{y} \frac{dy}{dx} = \log(\sin x) \cdot \frac{1}{x} + \log x \cdot \cot x \]

### Step 5: Solve for \( \frac{dy}{dx} \)
Now, we can solve for \( \frac{dy}{dx} \):
\[ \frac{dy}{dx} = y \left( \frac{\log(\sin x)}{x} + \log x \cdot \cot x \right) \]

### Step 6: Substitute back for \( y \)
Recall that \( y = (\sin x)^{\log x} \), so we substitute this back into our expression:
\[ \frac{dy}{dx} = (\sin x)^{\log x} \left( \frac{\log(\sin x)}{x} + \log x \cdot \cot x \right) \]

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

---