Answer is: option3

\( \frac{-\sin y}{(1 - \cos y)^3} \)

Solution:

Problem: If \( x + \sin y = y + 3 \), find \( \frac{d^2 y}{dx^2} \).

Steps:

1. Differentiate the equation implicitly with respect to \( x \):

\[ 1 + \cos y \cdot \frac{dy}{dx} = \frac{dy}{dx} \]

2. Rearrange to solve for \( \frac{dy}{dx} \):

\[ 1 = \frac{dy}{dx} - \cos y \cdot \frac{dy}{dx} \] Factor out \( \frac{dy}{dx} \): \[ 1 = \frac{dy}{dx} (1 - \cos y) \] Solve for \( \frac{dy}{dx} \): \[ \frac{dy}{dx} = \frac{1}{1 - \cos y} \]

3. Differentiate again to find \( \frac{d^2 y}{dx^2} \):

Using the chain rule: \[ \frac{d^2 y}{dx^2} = \frac{d}{dx} \left( (1 - \cos y)^{-1} \right) \] Apply the derivative of \( (1 - \cos y)^{-1} \): \[ \frac{d^2 y}{dx^2} = (-1)(1 - \cos y)^{-2} \cdot (-\sin y) \cdot \frac{dy}{dx} \]

4. Substitute \( \frac{dy}{dx} \) from step 2:

\[ \frac{d^2 y}{dx^2} = \frac{(-1) \cdot \sin y}{(1 - \cos y)^2} \cdot \frac{1}{1 - \cos y} \] Simplifying: \[ \frac{d^2 y}{dx^2} = \frac{-\sin y}{(1 - \cos y)^3} \]

Final Answer:

\[ \frac{d^2 y}{dx^2} = \frac{-\sin y}{(1 - \cos y)^3} \]