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The image presents a question asking to find the second derivative \(\frac{d^2y}{dx^2}\) of the following functions:

### a) \( x^2 + y^2 = 1 \)
This equation represents a circle. To find the second derivative, we will use implicit differentiation.

1. Differentiate both sides of the equation with respect to \(x\) to get the first derivative:
   \[
   2x + 2y \frac{dy}{dx} = 0
   \]
   Simplify for \(\frac{dy}{dx}\):
   \[
   \frac{dy}{dx} = -\frac{x}{y}
   \]

2. Differentiate again to get the second derivative:
   \[
   \frac{d^2y}{dx^2} = \frac{d}{dx}\left(-\frac{x}{y}\right)
   \]
   Using the quotient rule:
   \[
   \frac{d^2y}{dx^2} = -\frac{y \cdot 1 - x \cdot \frac{dy}{dx}}{y^2}
   \]
   Substitute \(\frac{dy}{dx} = -\frac{x}{y}\):
   \[
   \frac{d^2y}{dx^2} = -\frac{y + \frac{x^2}{y}}{y^2}
   \]
   Simplify:
   \[
   \frac{d^2y}{dx^2} = -\frac{y^2 + x^2}{y^3}
   \]
   Since \(x^2 + y^2 = 1\), the second derivative simplifies to:
   \[
   \frac{d^2y}{dx^2} = -\frac{1}{y^3}
   \]

### b) \( y^2 - 2x = 1 - 2y \)
For this equation, implicit differentiation is again needed.

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

2. Differentiate again to get the second derivative:
   \[
   \frac{d^2y}{dx^2} = \frac{d}{dx}\left(\frac{1}{y+1}\right)
   \]
   Using the chain rule:
   \[
   \frac{d^2y}{dx^2} = -\frac{1}{(y+1)^2} \cdot \frac{dy}{dx}
   \]
   Substitute \(\frac{dy}{dx} = \frac{1}{y+1}\):
   \[
   \frac{d^2y}{dx^2} = -\frac{1}{(y+1)^2} \cdot \frac{1}{y+1}
   \]
   Simplify:
   \[
   \frac{d^2y}{dx^2} = -\frac{1}{(y+1)^3}
   \]

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Would you like further details or clarifications? Here are five related questions:

1. What is implicit differentiation, and when is it used?
2. How do you use the quotient rule in derivatives?
3. What is the geometrical meaning of \(\frac{d^2y}{dx^2}\) for a circle?
4. How does the chain rule apply when differentiating functions implicitly?
5. How does the second derivative relate to the curvature of a function?

**Tip**: Always remember to substitute any known values (like \(x^2 + y^2 = 1\)) at the end to simplify your second derivative!

Carilah d^2y/dx^2 dari fungsi-fungsi berikut: a) x^2 + y^2 = 1, b) y^2 - 2x = 1 - 2y.

Learn how to find the second derivative d^2y/dx^2 using implicit differentiation for the equations x^2 + y^2 = 1 and y^2 - 2x = 1 - 2y. This solution involves key concepts such as the quotient rule and chain rule, suitable for college-level calculus students.

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