Let's find the derivative of the equation

\[
3y(x^2 + 2x + 8) = 4x^2 + 5
\]

with respect to \(x\). 

### Step 1: Differentiate both sides with respect to \(x\)

We need to apply implicit differentiation since \(y\) is a function of \(x\). Differentiating both sides of the equation with respect to \(x\):

\[
\frac{d}{dx} \left[ 3y(x^2 + 2x + 8) \right] = \frac{d}{dx} \left[ 4x^2 + 5 \right]
\]

### Step 2: Apply the product rule on the left-hand side

The left-hand side involves the product of \(3y\) and \((x^2 + 2x + 8)\), so we apply the product rule:

\[
\frac{d}{dx} \left[ 3y(x^2 + 2x + 8) \right] = 3\left( \frac{dy}{dx}(x^2 + 2x + 8) + y \cdot \frac{d}{dx} \left( x^2 + 2x + 8 \right) \right)
\]

The derivative of \(x^2 + 2x + 8\) with respect to \(x\) is:

\[
\frac{d}{dx} \left( x^2 + 2x + 8 \right) = 2x + 2
\]

So the derivative of the left-hand side becomes:

\[
3 \left( \frac{dy}{dx} (x^2 + 2x + 8) + y(2x + 2) \right)
\]

### Step 3: Differentiate the right-hand side

The derivative of the right-hand side is straightforward:

\[
\frac{d}{dx} \left[ 4x^2 + 5 \right] = 8x
\]

### Step 4: Combine and solve for \(\frac{dy}{dx}\)

Now, equate the derivatives:

\[
3 \left( \frac{dy}{dx} (x^2 + 2x + 8) + y(2x + 2) \right) = 8x
\]

Distribute the 3 on the left-hand side:

\[
3(x^2 + 2x + 8) \frac{dy}{dx} + 3y(2x + 2) = 8x
\]

To solve for \(\frac{dy}{dx}\):

\[
3(x^2 + 2x + 8) \frac{dy}{dx} = 8x - 3y(2x + 2)
\]

Finally, divide both sides by \(3(x^2 + 2x + 8)\) to isolate \(\frac{dy}{dx}\):

\[
\frac{dy}{dx} = \frac{8x - 3y(2x + 2)}{3(x^2 + 2x + 8)}
\]

This is the derivative \(\frac{dy}{dx}\) of the given equation.

Do you have any questions or need further details?

Here are five follow-up questions to deepen your understanding:
1. What happens to the derivative \(\frac{dy}{dx}\) if \(y\) is a constant?
2. How would the derivative change if the equation involved \(y^2\) instead of \(y\)?
3. Can you interpret the physical meaning of this derivative in the context of a real-world problem?
4. What is the significance of implicit differentiation in calculus?
5. How would the process change if there were a trigonometric function of \(x\) involved?

**Tip:** When applying the product rule in implicit differentiation, always remember to differentiate both the dependent variable \(y\) and the independent variable \(x\) separately before combining them.

Learn how to find the derivative of the implicit equation 3yx^2 + 2x + 8 = 4x^2 + 5 using implicit differentiation and the product rule. Understand step-by-step with detailed explanations and solve for dy/dx. Suitable for college-level calculus students.

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