xsin 2y = y cos 2x

Answer

To solve the equation \( x \sin(2y) = y \cos(2x) \) and find \(\frac{dy}{dx}\), we will differentiate both sides with respect to \(x\). Here’s the step-by-step solution:

### Step 1: Differentiate both sides
We start with the equation:
\[ x \sin(2y) = y \cos(2x) \]
Now, we differentiate both sides with respect to \(x\).

### Step 2: Apply the product rule
For the left-hand side \(x \sin(2y)\), we apply the product rule:
\[ \frac{d}{dx}(x \sin(2y)) = \sin(2y) \cdot \frac{d}{dx}(x) + x \cdot \frac{d}{dx}(\sin(2y)) \]
The derivative of \(x\) is \(1\), and for \(\sin(2y)\), we use the chain rule:
\[ \frac{d}{dx}(\sin(2y)) = \cos(2y) \cdot \frac{d}{dx}(2y) = 2 \cos(2y) \cdot \frac{dy}{dx} \]
Thus, the left-hand side becomes:
\[ \sin(2y) + 2x \cos(2y) \frac{dy}{dx} \]

### Step 3: Differentiate the right-hand side
Now, for the right-hand side \(y \cos(2x)\), we again apply the product rule:
\[ \frac{d}{dx}(y \cos(2x)) = \cos(2x) \cdot \frac{d}{dx}(y) + y \cdot \frac{d}{dx}(\cos(2x)) \]
The derivative of \(\cos(2x)\) is:
\[ \frac{d}{dx}(\cos(2x)) = -\sin(2x) \cdot \frac{d}{dx}(2x) = -2 \sin(2x) \]
Thus, the right-hand side becomes:
\[ \cos(2x) \frac{dy}{dx} - 2y \sin(2x) \]

### Step 4: Set both sides equal
Now we have:
\[ \sin(2y) + 2x \cos(2y) \frac{dy}{dx} = \cos(2x) \frac{dy}{dx} - 2y \sin(2x) \]

### Step 5: Collect terms involving \(\frac{dy}{dx}\)
Rearranging gives:
\[ 2x \cos(2y) \frac{dy}{dx} - \cos(2x) \frac{dy}{dx} = -2y \sin(2x) - \sin(2y) \]
Factoring out \(\frac{dy}{dx}\):
\[ \left(2x \cos(2y) - \cos(2x)\right) \frac{dy}{dx} = -2y \sin(2x) - \sin(2y) \]

### Step 6: Solve for \(\frac{dy}{dx}\)
Now, we can solve for \(\frac{dy}{dx}\):
\[ \frac{dy}{dx} = \frac{-2y \sin(2x) - \sin(2y)}{2x \cos(2y) - \cos(2x)} \]

### Final Result
Thus, the derivative \(\frac{dy}{dx}\) is:
\[ \frac{dy}{dx} = \frac{-2y \sin(2x) - \sin(2y)}{2x \cos(2y) - \cos(2x)} \]