We are tasked with finding the first derivative $\frac{dy}{dx}$ and the second derivative $\frac{d^2y}{dx^2}$ implicitly for the equation:

$x^2 + y^2 = 6$

Step 1: First derivative $\frac{dy}{dx}$

Differentiate both sides of the equation with respect to $x$, using implicit differentiation for $y$ (since $y$ is a function of $x$):

$\frac{d}{dx}(x^2 + y^2) = \frac{d}{dx}(6)$

This gives:

$2x + 2y \frac{dy}{dx} = 0$

Now, solve for $\frac{dy}{dx}$:

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

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

Step 2: Second derivative $\frac{d^2y}{dx^2}$

Differentiate $\frac{dy}{dx} = \frac{-x}{y}$ with respect to $x$ again, using the quotient rule:

$\frac{d}{dx}\left(\frac{-x}{y}\right) = \frac{y\frac{d}{dx}(-x) - (-x)\frac{d}{dx}(y)}{y^2}$

This becomes:

$\frac{d^2y}{dx^2} = \frac{y(-1) - (-x)\left(\frac{dy}{dx}\right)}{y^2}$

Substitute $\frac{dy}{dx} = \frac{-x}{y}$ into the equation:

$\frac{d^2y}{dx^2} = \frac{-y + x\left(\frac{-x}{y}\right)}{y^2}$

Simplify the numerator:

$\frac{d^2y}{dx^2} = \frac{-y + \frac{-x^2}{y}}{y^2} = \frac{-y^2 - x^2}{y^3}$

Since $x^2 + y^2 = 6$, we can substitute $x^2 + y^2 = 6$ into the numerator:

$\frac{d^2y}{dx^2} = \frac{-6}{y^3}$

Final answers:

• $\frac{dy}{dx} = \frac{-x}{y}$
• $\frac{d^2y}{dx^2} = \frac{-6}{y^3}$

Let me know if you need further clarification or more detailed steps!

Follow-up questions:

1. How do you apply implicit differentiation to functions of multiple variables?
2. Why is the quotient rule needed for the second derivative in this problem?
3. What are the general conditions for implicit differentiation?
4. How would the result change if the right-hand side of the equation were something other than a constant?
5. Can you explain how implicit differentiation differs from explicit differentiation?

Tip: When differentiating implicitly, remember to apply the chain rule carefully whenever you encounter a function of $y$.