If ` x^2 +y^2 = 1 ` then

Options

A. ` y y' -2 (y')^2 +1 =0`
B. `y y '' +(y' )^2 +1 =0`
C. `Yy ' -(y ')^2 -1=0`
D. `y Y' +2 (y')^2 +1=0`

Answer

To solve the problem given the equation \( x^2 + y^2 = 1 \), we will differentiate it step by step to find the second derivative \( y'' \).

### Step-by-Step Solution

1. **Differentiate the equation with respect to \( x \)**:
\[ \frac{d}{dx}(x^2 + y^2) = \frac{d}{dx}(1) \]
This gives us:
\[ 2x + 2y \frac{dy}{dx} = 0 \]

2. **Rearranging the equation**:
Divide the entire equation by 2:
\[ x + y \frac{dy}{dx} = 0 \]
Now, isolate \( \frac{dy}{dx} \):
\[ y \frac{dy}{dx} = -x \implies \frac{dy}{dx} = -\frac{x}{y} \]

3. **Differentiate again to find the second derivative**:
Differentiate \( \frac{dy}{dx} = -\frac{x}{y} \) with respect to \( x \):
\[ \frac{d^2y}{dx^2} = \frac{d}{dx}\left(-\frac{x}{y}\right) \]
Using the quotient rule:
\[ \frac{d^2y}{dx^2} = -\frac{y \cdot \frac{d}{dx}(x) - x \cdot \frac{d}{dx}(y)}{y^2} \]
This simplifies to:
\[ \frac{d^2y}{dx^2} = -\frac{y \cdot 1 - x \cdot \frac{dy}{dx}}{y^2} \]

4. **Substituting \( \frac{dy}{dx} \)**:
Substitute \( \frac{dy}{dx} = -\frac{x}{y} \):
\[ \frac{d^2y}{dx^2} = -\frac{y + x \cdot \left(-\frac{x}{y}\right)}{y^2} \]
This simplifies to:
\[ \frac{d^2y}{dx^2} = -\frac{y + \frac{x^2}{y}}{y^2} \]
Combine the terms in the numerator:
\[ \frac{d^2y}{dx^2} = -\frac{y^2 + x^2}{y^3} \]

5. **Using the original equation**:
From the original equation \( x^2 + y^2 = 1 \), we can substitute:
\[ \frac{d^2y}{dx^2} = -\frac{1}{y^3} \]

### Final Result
Thus, we have:
\[ y'' = -\frac{1}{y^3} \]