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The task is to find the second derivative \(\frac{d^2y}{dx^2}\) for the given functions in the image:

### a) \(x^2 + y^2 = 1\)
This is the equation of a circle. To find the second derivative \(\frac{d^2y}{dx^2}\), we proceed step by step using implicit differentiation.

1. Differentiate both sides with respect to \(x\):
\[
2x + 2y \frac{dy}{dx} = 0
\]
Solve for \(\frac{dy}{dx}\):
\[
\frac{dy}{dx} = -\frac{x}{y}
\]

2. Now differentiate \(\frac{dy}{dx} = -\frac{x}{y}\) again with respect to \(x\) (using the quotient rule):
\[
\frac{d^2y}{dx^2} = -\frac{y - x \frac{dy}{dx}}{y^2}
\]
Substitute \(\frac{dy}{dx} = -\frac{x}{y}\) from the first derivative:
\[
\frac{d^2y}{dx^2} = -\frac{y - x \left(-\frac{x}{y}\right)}{y^2} = -\frac{y + \frac{x^2}{y}}{y^2}
\]
Simplify:
\[
\frac{d^2y}{dx^2} = -\frac{y^2 + x^2}{y^3}
\]
Since \(x^2 + y^2 = 1\) from the original equation:
\[
\frac{d^2y}{dx^2} = -\frac{1}{y^3}
\]

### b) \(y^2 - 2x = 1 - 2y\)
For this equation, we will follow the same steps to find \(\frac{d^2y}{dx^2}\).

1. Differentiate both sides with respect to \(x\):
\[
2y \frac{dy}{dx} - 2 = -2 \frac{dy}{dx}
\]
Simplify:
\[
2y \frac{dy}{dx} + 2 \frac{dy}{dx} = 2
\]
Factor \(\frac{dy}{dx}\):
\[
\frac{dy}{dx}(2y + 2) = 2
\]
Solve for \(\frac{dy}{dx}\):
\[
\frac{dy}{dx} = \frac{2}{2(y + 1)} = \frac{1}{y + 1}
\]

2. Now differentiate \(\frac{dy}{dx} = \frac{1}{y + 1}\) again with respect to \(x\):
\[
\frac{d^2y}{dx^2} = -\frac{1}{(y + 1)^2} \cdot \frac{dy}{dx}
\]
Substitute \(\frac{dy}{dx} = \frac{1}{y + 1}\):
\[
\frac{d^2y}{dx^2} = -\frac{1}{(y + 1)^2} \cdot \frac{1}{y + 1} = -\frac{1}{(y + 1)^3}
\]

### Summary of Second Derivatives:
- For \(x^2 + y^2 = 1\): \(\frac{d^2y}{dx^2} = -\frac{1}{y^3}\)
- For \(y^2 - 2x = 1 - 2y\): \(\frac{d^2y}{dx^2} = -\frac{1}{(y+1)^3}\)

Would you like more details or have any further questions?

Here are five questions for deeper exploration:
1. Why is implicit differentiation used for these equations?
2. How do we apply the quotient rule in implicit differentiation?
3. What is the geometric meaning of the second derivative in these cases?
4. Can you find the second derivative of other conic sections like ellipses and hyperbolas?
5. How does the second derivative help us understand the curvature of the graph?

**Tip:** Always check if you can substitute the original equation into your derivatives to simplify the expression!

Find the second derivative \(\frac{d^2y}{dx^2}\) from the following functions: a) \(x^2 + y^2 = 1\), b) \(y^2 - 2x = 1 - 2y\).

Learn how to find the second derivative \(\frac{d^2y}{dx^2}\) for the functions \(x^2 + y^2 = 1\) and \(y^2 - 2x = 1 - 2y\). This problem involves using implicit differentiation and applying the quotient rule. Ideal for advanced high school or college-level calculus students.

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