Find `(dy)/(dx)`, when:
`xsin2y=ycos2x`

Answer

To find \(\frac{dy}{dx}\) for the equation \(x \sin(2y) = y \cos(2x)\), we will use implicit differentiation and the product rule. Here’s a step-by-step solution:

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

### Step 2: Apply the product rule on the left side
Using the product rule on the left side \(x \sin(2y)\):
\[ \frac{d}{dx}(x \sin(2y)) = \sin(2y) \cdot \frac{d}{dx}(x) + x \cdot \frac{d}{dx}(\sin(2y)) \]
This simplifies to:
\[ \sin(2y) + x \cdot \cos(2y) \cdot \frac{d(2y)}{dx} \]
Using the chain rule, \(\frac{d(2y)}{dx} = 2 \frac{dy}{dx}\), we get:
\[ \sin(2y) + 2x \cos(2y) \frac{dy}{dx} \]

### Step 3: Apply the product rule on the right side
Now, differentiate the right side \(y \cos(2x)\):
\[ \frac{d}{dx}(y \cos(2x)) = \cos(2x) \cdot \frac{dy}{dx} + y \cdot \frac{d}{dx}(\cos(2x)) \]
Using the chain rule, \(\frac{d}{dx}(\cos(2x)) = -2 \sin(2x)\), we have:
\[ \cos(2x) \frac{dy}{dx} - 2y \sin(2x) \]

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

### Step 5: Collect all 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 \(\frac{dy}{dx}\)
Thus, we can express \(\frac{dy}{dx}\) as:
\[ \frac{dy}{dx} = \frac{-2y \sin(2x) - \sin(2y)}{2x \cos(2y) - \cos(2x)} \]

### Final Answer
\[ \frac{dy}{dx} = \frac{-2y \sin(2x) - \sin(2y)}{2x \cos(2y) - \cos(2x)} \]