If $y = x\sin y$, then $\frac{dy}{dx}$ is:

Given the equation $y = x\sin y$, where $y$ appears on both sides. Since $y$ cannot be isolated, implicit differentiation is needed.

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Differentiating both sides with respect to $x$:

$\frac{d}{dx}(y) = \frac{d}{dx}(x \sin y)$

For the right side, applying the product rule to $x \sin y$:

$\frac{dy}{dx} = x \cdot \frac{d}{dx}(\sin y) + \frac{d}{dx}(x) \cdot \sin y$

$\frac{dy}{dx} = x \cdot \cos y \cdot \frac{dy}{dx} + 1 \cdot \sin y$

$\frac{dy}{dx} = \sin y + x\cos y \cdot \frac{dy}{dx}$

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Collecting all terms with $\frac{dy}{dx}$ on the left side:

$\frac{dy}{dx} - x\cos y \cdot \frac{dy}{dx} = \sin y$

$\frac{dy}{dx}(1 - x\cos y) = \sin y$

$\frac{dy}{dx} = \frac{\sin y}{1 - x\cos y}$

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From the original equation $y = x\sin y$:

$x = \frac{y}{\sin y}$

Substituting into the expression for $\frac{dy}{dx}$:

$\frac{dy}{dx} = \frac{\sin y}{1 - \frac{y}{\sin y} \cdot \cos y}$

$\frac{dy}{dx} = \frac{\sin y}{1 - \frac{y\cos y}{\sin y}}$

$\frac{dy}{dx} = \frac{\sin y}{\frac{\sin y - y\cos y}{\sin y}}$

$\frac{dy}{dx} = \frac{\sin y \cdot \sin y}{\sin y - y\cos y}$

$\frac{dy}{dx} = \frac{\sin^2 y}{\sin y - y\cos y}$