To find the second derivative of \( y \) with respect to \( x \) for the given equation \( x^2 + y^2 = a^2 \), follow these steps:

### Step 1: Differentiate the equation implicitly with respect to \( x \)
The given equation is:

\[
x^2 + y^2 = a^2
\]

Differentiate both sides with respect to \( x \):

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

1. Differentiate \( x^2 \) with respect to \( x \):
   \[
   \frac{d}{dx}(x^2) = 2x
   \]

2. Differentiate \( y^2 \) with respect to \( x \). Apply the chain rule:
   \[
   \frac{d}{dx}(y^2) = 2y \cdot \frac{dy}{dx} = 2yy'
   \]

3. Since \( a^2 \) is a constant, its derivative is zero:
   \[
   \frac{d}{dx}(a^2) = 0
   \]

So the first derivative gives:

\[
2x + 2yy' = 0
\]

Simplify this equation to solve for \( y' \):

\[
yy' = -x
\]

\[
y' = \frac{dy}{dx} = -\frac{x}{y}
\]

### Step 2: Differentiate the first derivative to find the second derivative
Now, differentiate \( y' = -\frac{x}{y} \) with respect to \( x \):

\[
\frac{d}{dx}\left( y' \right) = \frac{d}{dx}\left( -\frac{x}{y} \right)
\]

Use the quotient rule, where if \( u = -x \) and \( v = y \), then:

\[
\frac{d}{dx}\left( \frac{u}{v} \right) = \frac{v \cdot \frac{du}{dx} - u \cdot \frac{dv}{dx}}{v^2}
\]

1. \( u = -x \) and \( v = y \)
2. \( \frac{du}{dx} = -1 \)
3. \( \frac{dv}{dx} = y' \)

Substitute these into the quotient rule:

\[
y'' = \frac{y \cdot (-1) - (-x) \cdot y'}{y^2}
\]

Substitute \( y' = -\frac{x}{y} \) into the equation:

\[
y'' = \frac{-y + x \cdot \left( \frac{x}{y} \right)}{y^2}
\]

Simplify the expression:

\[
y'' = \frac{-y + \frac{x^2}{y}}{y^2}
\]

Combine the terms:

\[
y'' = \frac{-y^2 + x^2}{y^3}
\]

Thus, the second derivative of \( y \) with respect to \( x \) is:

\[
y'' = \frac{x^2 - y^2}{y^3}
\]

Would you like more details or have any questions? Here are 5 questions related to this problem:

1. Why is the quotient rule necessary when finding the second derivative?
2. How would the second derivative change if the equation was \( x^2 - y^2 = a^2 \)?
3. Can you interpret the geometric meaning of the second derivative for the curve \( x^2 + y^2 = a^2 \)?
4. How does the chain rule apply in the differentiation of \( y^2 \) with respect to \( x \)?
5. What does the second derivative tell us about the concavity of the curve?

**Tip:** When working with implicit differentiation, always carefully apply the chain rule to functions involving \( y \), as \( y \) is a function of \( x \).

Explore the detailed process of finding the second derivative of the equation x^2 + y^2 = a^2 using implicit differentiation and the quotient rule. This problem delves into advanced mathematical concepts including implicit differentiation, chain rule applications, and quotient rule usage.

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