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Let's solve the given equation \( 3x^2 + 2y^2 = 9 \) using implicit differentiation.

### Step 1: Finding \(\frac{dy}{dx}\) (First Derivative)

Differentiate both sides of the equation with respect to \(x\):

\[
\frac{d}{dx}(3x^2 + 2y^2) = \frac{d}{dx}(9)
\]

Now, applying the chain rule for the \(y^2\) term (since \(y\) is a function of \(x\)):

\[
6x + 4y \frac{dy}{dx} = 0
\]

Solve for \(\frac{dy}{dx}\):

\[
4y \frac{dy}{dx} = -6x
\]
\[
\frac{dy}{dx} = \frac{-6x}{4y} = \frac{-3x}{2y}
\]

### Step 2: Finding \(\frac{d^2y}{dx^2}\) (Second Derivative)

To find the second derivative, we differentiate \(\frac{dy}{dx} = \frac{-3x}{2y}\) with respect to \(x\), using the quotient rule:

\[
\frac{d}{dx}\left(\frac{-3x}{2y}\right) = \frac{(2y)(-3) - (-3x)(2 \frac{dy}{dx})}{(2y)^2}
\]

Substitute \(\frac{dy}{dx} = \frac{-3x}{2y}\) into the equation:

\[
= \frac{-6y - (-3x)(2) \left(\frac{-3x}{2y}\right)}{4y^2}
\]
\[
= \frac{-6y + 9x^2/y}{4y^2}
\]
\[
= \frac{-6y^2 + 9x^2}{4y^3}
\]

Thus, the second derivative is:

\[
\frac{d^2y}{dx^2} = \frac{9x^2 - 6y^2}{4y^3}
\]

### Summary of Results:
1. \(\frac{dy}{dx} = \frac{-3x}{2y}\)
2. \(\frac{d^2y}{dx^2} = \frac{9x^2 - 6y^2}{4y^3}\)

Would you like further clarification on any of these steps?

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Here are 5 relative questions to explore further:
1. How does implicit differentiation handle terms involving both \(x\) and \(y\)?
2. What is the quotient rule for derivatives and when is it used?
3. What is the chain rule, and how does it apply to implicit differentiation?
4. How does knowing the first derivative help in understanding the curve of the equation?
5. Why do we need to find the second derivative in some problems?

**Tip**: When using implicit differentiation, always remember to apply the chain rule whenever differentiating a term with \(y\), since \(y\) is a function of \(x\).

Use implicit differentiation to find dy/dx and d^2y/dx^2 for the equation 3x^2 + 2y^2 = 9.

Learn how to use implicit differentiation to find dy/dx and d^2y/dx^2 for the equation 3x^2 + 2y^2 = 9. This step-by-step solution covers core calculus concepts like the chain rule and quotient rule, suitable for high school students in grades 10-12.

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