We are tasked with finding $\frac{dy}{dx}$ from the equation:

$\sin(xy) - \sin(2x) = 1$

Let’s differentiate both sides of the equation implicitly with respect to $x$.

Step 1: Differentiate each term

1. Differentiate $\sin(xy)$:

   • We need to apply the product rule because $xy$ involves both $x$ and $y$. $\frac{d}{dx} \left( \sin(xy) \right) = \cos(xy) \cdot \frac{d}{dx}(xy)$ Using the product rule on $xy$, we get: $\frac{d}{dx}(xy) = y + x \frac{dy}{dx}$ So, the derivative of $\sin(xy)$ becomes: $\cos(xy) \cdot (y + x \frac{dy}{dx})$

2. Differentiate $\sin(2x)$: $\frac{d}{dx} \left( \sin(2x) \right) = 2 \cos(2x)$

3. Differentiate the constant $1$: $\frac{d}{dx}(1) = 0$

Step 2: Write the full derivative equation

After differentiating both sides, we have: $\cos(xy) \cdot (y + x \frac{dy}{dx}) - 2\cos(2x) = 0$

Step 3: Solve for $\frac{dy}{dx}$

Rearrange the equation to isolate $\frac{dy}{dx}$: $\cos(xy) \cdot (y + x \frac{dy}{dx}) = 2 \cos(2x)$

Expand the left-hand side: $\cos(xy) \cdot y + \cos(xy) \cdot x \frac{dy}{dx} = 2 \cos(2x)$

Now, isolate $\frac{dy}{dx}$: $\cos(xy) \cdot x \frac{dy}{dx} = 2 \cos(2x) - \cos(xy) \cdot y$ $\frac{dy}{dx} = \frac{2 \cos(2x) - \cos(xy) \cdot y}{x \cos(xy)}$

Step 4: Match with the options

Now compare this expression with the given choices:

• Option C matches this result: $\frac{2 \cos(2x) - y \cos(xy)}{x \cos(xy)}$

Thus, the correct answer is C.

Would you like any further details or have any questions? Here are some related questions for deeper understanding:

1. How do you apply implicit differentiation in other contexts?
2. What is the product rule, and how is it applied here?
3. Why do we differentiate $\sin(2x)$ using the chain rule?
4. Can you explain how to isolate $\frac{dy}{dx}$ in a different equation?
5. What would happen if the equation had more complex trigonometric functions?

Tip: Always apply the product rule carefully when differentiating expressions like $\sin(xy)$, which involve products of variables.