`y=x^(sin2x)`

Answer

To find the derivative of the function \( y = x^{\sin(2x)} \), we can follow these steps:

### Step 1: Take the natural logarithm of both sides
We start with the equation:
\[ y = x^{\sin(2x)} \]
Taking the natural logarithm of both sides gives:
\[ \log y = \log(x^{\sin(2x)}) \]

### Step 2: Simplify using logarithmic identities
Using the property of logarithms that states \( \log(a^b) = b \log a \), we can rewrite the equation as:
\[ \log y = \sin(2x) \cdot \log x \]

### Step 3: Differentiate both sides with respect to \( x \)
Now we differentiate both sides with respect to \( x \). Using implicit differentiation on the left side and the product rule on the right side:
\[ \frac{d}{dx}(\log y) = \frac{1}{y} \frac{dy}{dx} \]
For the right side, we apply the product rule:
\[ \frac{d}{dx}(\sin(2x) \log x) = \cos(2x) \cdot 2 \cdot \log x + \sin(2x) \cdot \frac{1}{x} \]

### Step 4: Combine the derivatives
Putting it all together, we have:
\[ \frac{1}{y} \frac{dy}{dx} = \cos(2x) \cdot 2 \cdot \log x + \sin(2x) \cdot \frac{1}{x} \]

### Step 5: Solve for \( \frac{dy}{dx} \)
Now, we multiply both sides by \( y \) to isolate \( \frac{dy}{dx} \):
\[ \frac{dy}{dx} = y \left( 2 \cos(2x) \log x + \frac{\sin(2x)}{x} \right) \]

### Step 6: Substitute back for \( y \)
Since \( y = x^{\sin(2x)} \), we substitute back:
\[ \frac{dy}{dx} = x^{\sin(2x)} \left( 2 \cos(2x) \log x + \frac{\sin(2x)}{x} \right) \]

### Final Answer
Thus, the derivative \( \frac{dy}{dx} \) is:
\[ \frac{dy}{dx} = x^{\sin(2x)} \left( 2 \cos(2x) \log x + \frac{\sin(2x)}{x} \right) \]

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