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

Options

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

Explanation

To solve the problem given by the equation \( x^2 + y^2 = 1 \), we need to differentiate this equation with respect to \( x \) twice to find \( y'' \) in terms of \( y \) and \( y' \).

### Step 1: Differentiate the equation once
Start with the equation:
\[ x^2 + y^2 = 1 \]
Now, differentiate both sides with respect to \( x \):
\[ \frac{d}{dx}(x^2) + \frac{d}{dx}(y^2) = \frac{d}{dx}(1) \]
This gives:
\[ 2x + 2y \frac{dy}{dx} = 0 \]
Let \( y' = \frac{dy}{dx} \). The equation simplifies to:
\[ 2x + 2y y' = 0 \]
Dividing by 2:
\[ x + y y' = 0 \]
Rearranging gives:
\[ y' = -\frac{x}{y} \]

### Step 2: Differentiate again
Now we differentiate \( y' \) with respect to \( x \):
\[ \frac{d}{dx}(y') = \frac{d}{dx}\left(-\frac{x}{y}\right) \]
Using the quotient rule:
\[ y'' = \frac{y \cdot \frac{d}{dx}(-x) - (-x) \cdot \frac{dy}{dx}}{y^2} \]
This simplifies to:
\[ y'' = \frac{-y - x y'}{y^2} \]
Substituting \( y' = -\frac{x}{y} \) into the equation:
\[ y'' = \frac{-y - x\left(-\frac{x}{y}\right)}{y^2} \]
This becomes:
\[ y'' = \frac{-y + \frac{x^2}{y}}{y^2} \]
Combining the terms:
\[ y'' = \frac{-y^2 + x^2}{y^3} \]

### Step 3: Substitute \( x^2 + y^2 = 1 \)
From the original equation \( x^2 + y^2 = 1 \), we can express \( x^2 \) as:
\[ x^2 = 1 - y^2 \]
Substituting this into the expression for \( y'' \):
\[ y'' = \frac{-y^2 + (1 - y^2)}{y^3} \]
This simplifies to:
\[ y'' = \frac{1 - 2y^2}{y^3} \]

### Conclusion
Thus, the final expression for \( y'' \) in terms of \( y \) is:
\[ y'' = \frac{1 - 2y^2}{y^3} \]