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Find $y^{\prime}$ by implicit differentiation. Match the equations defining $y$ implicitly with the letters labeling the expressions for $y^{\prime}$.
1. $7 x \cos y+2 \sin 2 y=4 \sin y$
2. $7 x \cos y+2 \cos 2 y=4 \sin y$
3. $7 x \sin y+2 \sin 2 y=4 \cos y$
4. $7 x \sin y+2 \cos 2 y=4 \cos y$
A. $y^{\prime}=\frac{7 \cos y}{7 x \sin y-4 \cos 2 y+4 \cos y}$
B. $y^{\prime}=\frac{7 \sin y}{-7 x \cos y-4 \cos 2 y-4 \sin y}$
C. $y^{\prime}=\frac{7 \sin y}{4 \sin 2 y-7 x \cos y-4 \sin y}$
D. $y^{\prime}=\frac{7 \cos y}{7 x \sin y+4 \sin 2 y+4 \cos y}$
Note: You can earn partial credit on this problem.

Step 1 ：Differentiate both sides of the first equation $7 x \cos y+2 \sin 2 y=4 \sin y$ with respect to $x$ using the chain rule and the product rule.

Step 2 ：The derivative of $7x \cos y$ with respect to $x$ is $7 \cos y - 7x \sin y \cdot y'$.

Step 3 ：The derivative of $2 \sin 2y$ with respect to $x$ is $4 \cos 2y \cdot y'$.

Step 4 ：The derivative of $4 \sin y$ with respect to $x$ is $4 \cos y \cdot y'$.

Step 5 ：Setting these equal gives us $7 \cos y - 7x \sin y \cdot y' + 4 \cos 2y \cdot y' = 4 \cos y \cdot y'$.

Step 6 ：Solving for $y'$ gives us $y' = \frac{7 \cos y}{7x \sin y + 4 \cos 2y - 4 \cos y}$, which matches with option D.

Step 7 ：Differentiate both sides of the third equation $7 x \sin y+2 \sin 2 y=4 \cos y$ with respect to $x$ using the chain rule and the product rule.

Step 8 ：The derivative of $7x \sin y$ with respect to $x$ is $7 \sin y + 7x \cos y \cdot y'$.

Step 9 ：The derivative of $2 \sin 2y$ with respect to $x$ is $4 \cos 2y \cdot y'$.

Step 10 ：The derivative of $4 \cos y$ with respect to $x$ is $-4 \sin y \cdot y'$.

Step 11 ：Setting these equal gives us $7 \sin y + 7x \cos y \cdot y' + 4 \cos 2y \cdot y' = -4 \sin y \cdot y'$.

Step 12 ：Solving for $y'$ gives us $y' = \frac{7 \sin y}{-7x \cos y - 4 \cos 2y + 4 \sin y}$, which matches with option B.

Step 13 ：So, the first equation matches with option D and the third equation matches with option B.