`e^x.log (sin 2 x)`

Answer

To differentiate the function \( y = e^x \log(\sin(2x)) \) with respect to \( x \), we will use the product rule and the chain rule.

### Step-by-Step Solution:

1. **Identify the function**:
We have \( y = e^x \log(\sin(2x)) \).

2. **Apply the Product Rule**:
The product rule states that if you have two functions \( u \) and \( v \), then the derivative \( \frac{d}{dx}(uv) = u'v + uv' \). Here, let:
- \( u = e^x \)
- \( v = \log(\sin(2x)) \)

3. **Differentiate \( u \)**:
The derivative of \( u = e^x \) is:
\[ u' = e^x \]

4. **Differentiate \( v \)**:
To differentiate \( v = \log(\sin(2x)) \), we will use the chain rule:
\[ v' = \frac{1}{\sin(2x)} \cdot \frac{d}{dx}(\sin(2x)) \]
Now, differentiate \( \sin(2x) \):
\[ \frac{d}{dx}(\sin(2x)) = 2\cos(2x) \]
Therefore, substituting this back, we get:
\[ v' = \frac{1}{\sin(2x)} \cdot 2\cos(2x) = \frac{2\cos(2x)}{\sin(2x)} = 2\cot(2x) \]

5. **Combine using the Product Rule**:
Now we can combine the derivatives using the product rule:
\[ \frac{dy}{dx} = u'v + uv' = e^x \log(\sin(2x)) + e^x \cdot 2\cot(2x) \]

6. **Final expression**:
Thus, the derivative of \( y \) is:
\[ \frac{dy}{dx} = e^x \log(\sin(2x)) + 2e^x \cot(2x) \]

### Final Answer:
\[ \frac{dy}{dx} = e^x \log(\sin(2x)) + 2e^x \cot(2x) \]