xsin 2y = y cos 2x

Answer

To find \(\frac{dy}{dx}\) for the equation \(x \sin 2y = y \cos 2x\), we will differentiate both sides with respect to \(x\).

### Step-by-Step Solution:

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

2. **Apply Product Rule on the Left Side:**
For the left side \(x \sin 2y\), we apply the product rule:
\[ \frac{d}{dx}(x \sin 2y) = \frac{dx}{dx} \cdot \sin 2y + x \cdot \frac{d}{dx}(\sin 2y) \]
The derivative of \(\sin 2y\) is \(2 \cos 2y \cdot \frac{dy}{dx}\) (using the chain rule). Thus, we have:
\[ \sin 2y + x \cdot 2 \cos 2y \cdot \frac{dy}{dx} \]

3. **Apply Product Rule on the Right Side:**
For the right side \(y \cos 2x\), we again apply the product rule:
\[ \frac{d}{dx}(y \cos 2x) = \frac{dy}{dx} \cdot \cos 2x + y \cdot \frac{d}{dx}(\cos 2x) \]
The derivative of \(\cos 2x\) is \(-2 \sin 2x\). Thus, we have:
\[ \frac{dy}{dx} \cdot \cos 2x - 2y \sin 2x \]

4. **Set the Derivatives Equal:**
Now we can set the derivatives from both sides equal to each other:
\[ \sin 2y + 2x \cos 2y \cdot \frac{dy}{dx} = \frac{dy}{dx} \cdot \cos 2x - 2y \sin 2x \]

5. **Rearranging the Equation:**
We want to isolate \(\frac{dy}{dx}\). Rearranging gives us:
\[ 2x \cos 2y \cdot \frac{dy}{dx} - \frac{dy}{dx} \cdot \cos 2x = -\sin 2y - 2y \sin 2x \]
Factoring out \(\frac{dy}{dx}\):
\[ \frac{dy}{dx}(2x \cos 2y - \cos 2x) = -\sin 2y - 2y \sin 2x \]

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

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