`y=x^(sin2x)`

Answer

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

### Step 1: Rewrite the function
We start with the function:
\[ y = x^{\sin(2x)} \]

### Step 2: Take the natural logarithm of both sides
Taking the natural logarithm helps us simplify the expression:
\[ \log y = \log(x^{\sin(2x)}) \]

### Step 3: Apply the logarithmic identity
Using the property of logarithms, we can rewrite the right-hand side:
\[ \log y = \sin(2x) \cdot \log x \]

### Step 4: Differentiate both sides with respect to \( x \)
Now we differentiate both sides. For the left-hand side, we use implicit differentiation:
\[ \frac{1}{y} \frac{dy}{dx} = \frac{d}{dx}(\sin(2x) \cdot \log x) \]

### Step 5: Apply the product rule on the right-hand side
Using the product rule \( (u \cdot v)' = u'v + uv' \), where \( u = \sin(2x) \) and \( v = \log x \):
- \( u' = 2 \cos(2x) \) (since the derivative of \( \sin(2x) \) is \( 2\cos(2x) \))
- \( v' = \frac{1}{x} \) (since the derivative of \( \log x \) is \( \frac{1}{x} \))

Thus, we have:
\[ \frac{d}{dx}(\sin(2x) \cdot \log x) = \sin(2x) \cdot \frac{1}{x} + \log x \cdot 2 \cos(2x) \]

### Step 6: Substitute back into the equation
Now substituting back into our equation gives:
\[ \frac{1}{y} \frac{dy}{dx} = \sin(2x) \cdot \frac{1}{x} + 2 \cos(2x) \cdot \log x \]

### Step 7: Solve for \( \frac{dy}{dx} \)
Multiplying both sides by \( y \):
\[ \frac{dy}{dx} = y \left( \sin(2x) \cdot \frac{1}{x} + 2 \cos(2x) \cdot \log x \right) \]

### Step 8: Substitute back for \( y \)
Since \( y = x^{\sin(2x)} \), we substitute back:
\[ \frac{dy}{dx} = x^{\sin(2x)} \left( \sin(2x) \cdot \frac{1}{x} + 2 \cos(2x) \cdot \log x \right) \]

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

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