If $$x^2 + y^2 + \sin y = 4$$, then the value of $$\frac{d^2y}{dx^2}$$ at the point (-2, 0) is:

We are given the implicit relation $$x^2 + y^2 + \sin y = 4$$ that connects the variables $$x$$ and $$y$$. Our objective is to find the second derivative $$\dfrac{d^2y}{dx^2}$$ at the specific point $$(-2,\,0)$$ lying on the curve.

First we differentiate the whole equation with respect to $$x$$. Using the rule $$\dfrac{d}{dx}(u+v+w)=\dfrac{du}{dx}+\dfrac{dv}{dx}+\dfrac{dw}{dx}$$ together with the basic derivatives $$\dfrac{d}{dx}(x^2)=2x$$, $$\dfrac{d}{dx}(y^2)=2y\dfrac{dy}{dx}$$ (chain rule), and $$\dfrac{d}{dx}(\sin y)=\cos y\dfrac{dy}{dx}$$ (again chain rule), we obtain

$$2x \;+\; 2y\dfrac{dy}{dx} \;+\; \cos y\,\dfrac{dy}{dx} \;=\; 0.$$

Now we collect the terms that contain $$\dfrac{dy}{dx}$$:

$$2x \;+\;\left(2y+\cos y\right)\dfrac{dy}{dx}=0.$$

Rearranging to isolate the first derivative gives

$$\left(2y+\cos y\right)\dfrac{dy}{dx}=-2x,$$

so that

$$\dfrac{dy}{dx}=\dfrac{-2x}{\,2y+\cos y\,}.$$

Next, we evaluate this first derivative at the point $$(-2,\,0)$$. Substituting $$x=-2$$ and $$y=0$$, while noting that $$\cos0=1$$, we get

$$\dfrac{dy}{dx}\Big|_{(-2,0)}=\dfrac{-2(-2)}{\,2(0)+1\,}=\dfrac{4}{1}=4.$$

With the first derivative in hand, we proceed to the second derivative. Write the first derivative in a compact quotient form

$$\dfrac{dy}{dx}=\dfrac{N}{D},\quad\text{where }N=-2x\text{ and }D=2y+\cos y.$$

To differentiate a quotient we recall the quotient rule: if $$u=\dfrac{p}{q}$$, then $$\dfrac{du}{dx}=\dfrac{q\dfrac{dp}{dx}-p\dfrac{dq}{dx}}{q^2}.$$ Applying this rule to $$\dfrac{dy}{dx}=\dfrac{N}{D}$$ we have

$$\dfrac{d^2y}{dx^2}=\dfrac{D\dfrac{dN}{dx}-N\dfrac{dD}{dx}}{D^2}.$$

We now compute the required building blocks one by one.

1. The derivative of $$N=-2x$$ with respect to $$x$$ is

$$\dfrac{dN}{dx}=-2.$$

2. The derivative of $$D=2y+\cos y$$ with respect to $$x$$ requires the chain rule again:

$$\dfrac{dD}{dx}=2\dfrac{dy}{dx}-\sin y\,\dfrac{dy}{dx}=\left(2-\sin y\right)\dfrac{dy}{dx}.$$

We now evaluate every symbol at the point $$(-2,0)$$. We already have

$$x=-2,\quad y=0,\quad \dfrac{dy}{dx}=4,$$

and we also know $$\sin0=0$$ and $$\cos0=1$$. Hence,

$$N= -2(-2)=4,\qquad D=2(0)+1=1,$$

$$\dfrac{dN}{dx}=-2,\qquad \dfrac{dD}{dx}=\left(2-0\right)\cdot4 = 8.$$

Substituting everything into the quotient-rule expression gives

$$\dfrac{d^2y}{dx^2}\Big|_{(-2,0)}=\dfrac{(1)(-2)-\,(4)(8)}{1^2}=(-2)-32=-34.$$

Hence, the correct answer is Option A.