If `x^(2)+y^(2)+siny=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 \(|\frac{d^2y}{dx^2}|\) at the point \((-2, 0)\) given the equation:

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

### Step 1: Differentiate the equation with respect to \(x\)

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: Rearrange the equation

We can rearrange the equation to isolate \(\frac{dy}{dx}\):

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

This leads to:

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

Thus,

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

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

Now we substitute \(x = -2\) and \(y = 0\):

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

### Step 4: Differentiate again to find \(\frac{d^2y}{dx^2}\)

Next, we differentiate the equation \(2x + (2y + \cos y) \frac{dy}{dx} = 0\) again with respect to \(x\):

Using the product rule on \((2y + \cos y) \frac{dy}{dx}\):

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

This gives:

\[ 2 + \left(2 \frac{dy}{dx} + (2y + \cos y) \frac{d^2y}{dx^2} - \sin y \left(\frac{dy}{dx}\right)^2\right) = 0 \]

### Step 5: Substitute the values at the point \((-2, 0)\)

Now substituting \(x = -2\), \(y = 0\), and \(\frac{dy}{dx} = 4\):

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

This simplifies to:

\[ 2 + (8 + 1 \cdot \frac{d^2y}{dx^2} - 0) = 0 \]

So we have:

\[ 2 + 8 + \frac{d^2y}{dx^2} = 0 \]

### Step 6: Solve for \(\frac{d^2y}{dx^2}\)

This leads to:

\[ \frac{d^2y}{dx^2} = -10 \]

### Step 7: Find the absolute value

Finally, we need the absolute value:

\[ |\frac{d^2y}{dx^2}| = |-10| = 10 \]

### Final Answer

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

\[ \boxed{10} \]