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

Explanation

To find the value of \(|\frac{d^2y}{dx^2}|\) at the point \((-2, 0)\) given the equation \(x^2 + y^2 + \sin y = 4\), we will follow these steps:

### Step 1: Differentiate the given equation
We start with the equation:
\[ x^2 + y^2 + \sin y = 4 \]
Differentiate both sides 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
Rearranging the equation, we have:
\[ 2x + (2y + \cos y) \frac{dy}{dx} = 0 \]
From this, we can express \(\frac{dy}{dx}\):
\[ \frac{dy}{dx} = -\frac{2x}{2y + \cos y} \]

### 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:
\[ \frac{d^2y}{dx^2} = \frac{d}{dx}\left(-\frac{2x}{2y + \cos y}\right) \]
Using the quotient rule:
\[ \frac{d^2y}{dx^2} = \frac{(2y + \cos y)(-2) - (-2x)(\frac{d}{dx}(2y + \cos y))}{(2y + \cos y)^2} \]

### Step 5: Compute \(\frac{d}{dx}(2y + \cos y)\)
Using the chain rule:
\[ \frac{d}{dx}(2y + \cos y) = 2\frac{dy}{dx} - \sin y \frac{dy}{dx} \]
Substituting \(\frac{dy}{dx} = 4\) and \(y = 0\):
\[ \frac{d}{dx}(2y + \cos y) = 2(4) - \sin(0)(4) = 8 \]

### Step 6: Substitute back into the second derivative
Now substituting back:
\[ \frac{d^2y}{dx^2} = \frac{(1)(-2) - (-2)(8)}{(1)^2} = -2 + 16 = 14 \]

### Step 7: Find the absolute value
Finally, we need the absolute value:
\[ |\frac{d^2y}{dx^2}| = |14| = 14 \]

### Final Answer
The value of \(|\frac{d^2y}{dx^2}|\) at the point \((-2, 0)\) is:
\[ \boxed{14} \]