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: $\frac{d^2y}{dx^2} = -\frac{1}{(y+1)^3}$

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!