Which of the following is the result of differentiating both sides of the equation \[ x \sin (y)=\ln (x)+y^{2} \] with respect to $x$ ? ( $y^{\prime}$ denotes the derivative with respect to $x$.) $\cos (y)+x \sin (y) \cdot y^{\prime}=\frac{1}{x}+2 y \cdot y^{\prime}$ $\cos (y) \cdot y^{\prime}=\frac{1}{x}+2 y \cdot y^{\prime}$ $x \cos (y) \cdot y^{\prime}+\sin (y)=\frac{1}{x}+2 y \cdot y^{\prime}$ $\sin (y) \cdot y^{\prime}=\frac{1}{x}+2 y \cdot y^{\prime}$ $x \cos \left(y^{\prime}\right)=\frac{1}{x}+2 y \cdot y^{\prime}$

Step 1 ：The given equation is \(x \sin(y) = \ln(x) + y^{2}\).

Step 2 ：We differentiate both sides of the equation with respect to \(x\).

Step 3 ：The derivative of \(x \sin(y)\) with respect to \(x\) is \(\sin(y) + x \cos(y) y'\) by applying the product rule.

Step 4 ：The derivative of \(\ln(x)\) with respect to \(x\) is \(\frac{1}{x}\).

Step 5 ：The derivative of \(y^{2}\) with respect to \(x\) is \(2y y'\) by applying the chain rule.

Step 6 ：Combining these results, the derivative of the entire equation with respect to \(x\) is \(\sin(y) + x \cos(y) y' = \frac{1}{x} + 2y y'\).

Step 7 ：Rearranging terms, we get \(\sin(y) - \frac{1}{x} + x \cos(y) y' - 2y y' = 0\).

Step 8 ：Final Answer: \(\boxed{\sin(y) - \frac{1}{x} + x \cos(y) y' - 2y y'}\)