The problem provides the equation:

$x^2 - 2y^2 = 1$

and asks to find $\frac{dy}{dx}$ using implicit differentiation.

Solution:

1. Differentiate both sides of the equation with respect to $x$, treating $y$ as a function of $x$: $\frac{d}{dx}(x^2) - \frac{d}{dx}(2y^2) = \frac{d}{dx}(1)$

2. Apply the differentiation:

   • For $x^2$, the derivative is $2x$.
   • For $2y^2$, use the chain rule. The derivative of $y^2$ with respect to $x$ is $2y \cdot \frac{dy}{dx}$, so the derivative of $2y^2$ is $2 \cdot 2y \cdot \frac{dy}{dx} = 4y \frac{dy}{dx}$.
   • The derivative of $1$ is $0$.

   Putting it all together: $2x - 4y \frac{dy}{dx} = 0$

3. Solve for $\frac{dy}{dx}$: $4y \frac{dy}{dx} = 2x$ $\frac{dy}{dx} = \frac{2x}{4y} = \frac{x}{2y}$

Final Answer:

$\frac{dy}{dx} = \frac{x}{2y}$

Would you like a more detailed breakdown of each step, or do you have any specific questions?

Here are 5 related questions to deepen your understanding:

1. How would you find $\frac{d^2y}{dx^2}$ for this equation using implicit differentiation?
2. If $x = 3$ and $y = 1$, what is the value of $\frac{dy}{dx}$?
3. How would the solution change if the equation were $x^2 + 2y^2 = 1$?
4. What does the expression for $\frac{dy}{dx}$ tell us about the slope of the curve at any point $(x, y)$?
5. Can implicit differentiation be used to find higher-order derivatives in this case?

Tip: Remember to apply the chain rule whenever differentiating terms with $y$ with respect to $x$.