We are given the equation:

$\sin(xy) = x$

We are asked to find $\frac{dy}{dx}$. Let's differentiate both sides of this equation implicitly with respect to $x$.

1. Differentiate $\sin(xy)$ using the chain rule: [ \frac{d}{dx}[\sin(xy)] = \cos(xy) \cdot \frac{d}{dx}[xy] ] Since $xy$ is a product of two functions $x$ and $y$, we apply the product rule: $\frac{d}{dx}[xy] = y + x\frac{dy}{dx}$ So: $\frac{d}{dx}[\sin(xy)] = \cos(xy) \cdot (y + x\frac{dy}{dx})$

2. Differentiate the right-hand side $x$: $\frac{d}{dx}[x] = 1$

Now we have the equation: $\cos(xy) \cdot (y + x\frac{dy}{dx}) = 1$

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

First, expand the left-hand side: $\cos(xy) \cdot y + \cos(xy) \cdot x \frac{dy}{dx} = 1$

Move the term $\cos(xy) \cdot y$ to the right side: $\cos(xy) \cdot x \frac{dy}{dx} = 1 - \cos(xy) \cdot y$

Now, solve for $\frac{dy}{dx}$: $\frac{dy}{dx} = \frac{1 - \cos(xy) \cdot y}{x \cos(xy)}$

This matches option (D) in the given choices.

Thus, the correct answer is (D).

Would you like further details on any step?

Here are 5 questions related to this:

1. How do you differentiate trigonometric functions implicitly?
2. What is the chain rule and when is it applied in differentiation?
3. How does the product rule work in calculus?
4. Can you explain implicit differentiation in greater depth?
5. How does solving for $\frac{dy}{dx}$ differ when dealing with implicit versus explicit functions?

Tip: When differentiating products of functions, always remember to apply the product rule carefully!