If ` x ^ 2 + y^ 2 + sin y = 4 , ` then the value of ` | (d^ 2 y ) /( dx ^ 2 ) | ` at point ` ( -2, 0 ) ` is …..

Explanation

To solve the problem, we need to find the value of \( \left| \frac{d^2y}{dx^2} \right| \) at the point \( (-2, 0) \) given the equation:

\[ x^2 + y^2 + \sin y = 4 \]

### Step 1: Differentiate the equation implicitly
We start by differentiating both sides of the equation with respect to \( x \):

\[ \frac{d}{dx}(x^2) + \frac{d}{dx}(y^2) + \frac{d}{dx}(\sin y) = \frac{d}{dx}(4) \]

This gives us:

\[ 2x + 2y \frac{dy}{dx} + \cos y \frac{dy}{dx} = 0 \]

### Step 2: Solve for \( \frac{dy}{dx} \)
Now, we can factor out \( \frac{dy}{dx} \):

\[ 2x + \left(2y + \cos y\right) \frac{dy}{dx} = 0 \]

Rearranging gives:

\[ \frac{dy}{dx} = -\frac{2x}{2y + \cos y} \]

### Step 3: Evaluate \( \frac{dy}{dx} \) at the point \( (-2, 0) \)
Substituting \( x = -2 \) and \( y = 0 \):

\[ \frac{dy}{dx} = -\frac{2(-2)}{2(0) + \cos(0)} = -\frac{-4}{1} = 4 \]

### Step 4: Differentiate again to find \( \frac{d^2y}{dx^2} \)
Now we differentiate the expression for \( \frac{dy}{dx} \) again:

Using the quotient rule:

\[ \frac{d^2y}{dx^2} = \frac{(2y + \cos y)(-2) - (-2x)(\frac{d}{dx}(2y + \cos y)}{(2y + \cos y)^2} \]

We need to compute \( \frac{d}{dx}(2y + \cos y) \):

\[ \frac{d}{dx}(2y + \cos y) = 2\frac{dy}{dx} - \sin y \frac{dy}{dx} \]

Substituting \( \frac{dy}{dx} = 4 \) and \( y = 0 \):

\[ \frac{d}{dx}(2y + \cos y) = 2(4) - \sin(0)(4) = 8 - 0 = 8 \]

### Step 5: Substitute back to find \( \frac{d^2y}{dx^2} \)
Now substituting back into the formula for \( \frac{d^2y}{dx^2} \):

\[ \frac{d^2y}{dx^2} = \frac{(2(0) + \cos(0))(-2) - (-2(-2))(8)}{(2(0) + \cos(0))^2} \]

This simplifies to:

\[ \frac{d^2y}{dx^2} = \frac{(1)(-2) - (4)(8)}{(1)^2} = -2 - 32 = -34 \]

### Step 6: Find the absolute value
Finally, we find the absolute value:

\[ \left| \frac{d^2y}{dx^2} \right| = | -34 | = 34 \]

### Final Answer
Thus, the value of \( \left| \frac{d^2y}{dx^2} \right| \) at the point \( (-2, 0) \) is:

\[ \boxed{34} \]