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We are tasked with finding the second derivative \( \frac{d^2 y}{dx^2} \), given the equation:

\[
\sqrt{y} + 4xy = 4
\]

### Step 1: Differentiate the equation implicitly
First, let's rewrite the equation:
\[
\sqrt{y} + 4xy = 4
\]
This simplifies to:
\[
y^{1/2} + 4xy = 4
\]

Now, differentiate both sides with respect to \(x\), using implicit differentiation.

For \(y^{1/2}\), use the chain rule:
\[
\frac{d}{dx}(y^{1/2}) = \frac{1}{2}y^{-1/2} \cdot \frac{dy}{dx} = \frac{1}{2\sqrt{y}} \cdot y'
\]

For \(4xy\), use the product rule:
\[
\frac{d}{dx}(4xy) = 4 \left( y + x \cdot y' \right)
\]

On the right-hand side, the derivative of the constant 4 is zero:
\[
\frac{d}{dx}(4) = 0
\]

Thus, the derivative of the entire equation becomes:
\[
\frac{1}{2\sqrt{y}} \cdot y' + 4 \left( y + x \cdot y' \right) = 0
\]

### Step 2: Solve for \( y' \)
This is a first-order differential equation. Let's isolate \( y' \) to solve for the first derivative.

\[
\frac{1}{2\sqrt{y}} y' + 4y + 4x y' = 0
\]
Combine the terms with \( y' \):
\[
\left( \frac{1}{2\sqrt{y}} + 4x \right) y' = -4y
\]
Now solve for \( y' \):
\[
y' = \frac{-4y}{\frac{1}{2\sqrt{y}} + 4x}
\]

### Step 3: Differentiate again to find \( \frac{d^2 y}{dx^2} \)
Now we need to differentiate \( y' \) with respect to \(x\) to get \( y'' \). This is a bit more complex since \(y\) is a function of \(x\), so we apply both the quotient rule and chain rule.

Let me calculate the second derivative \( y'' \) for you.The second derivative \( \frac{d^2 y}{dx^2} \) is:

\[
\frac{d^2 y}{dx^2} = \frac{64y^2}{(8x\sqrt{y} + 1)^2}
\]

This represents the curvature or concavity of the function \( y(x) \) based on the given implicit equation.

Do you want further details or have any questions on this? Here are some related questions to deepen your understanding:

1. How does implicit differentiation differ from explicit differentiation?
2. What does the second derivative \( \frac{d^2y}{dx^2} \) tell us about the behavior of \( y(x) \)?
3. Can you find specific values for \( y' \) and \( y'' \) if you are given specific values for \( x \) and \( y \)?
4. How does the product rule apply in the context of \( 4xy \)?
5. Why is it important to apply the chain rule when differentiating \( \sqrt{y} \)?

**Tip:** Always double-check the structure of the problem to ensure correct application of differentiation rules, especially in implicit differentiation.

Learn how to find the second derivative (d²y/dx²) of an implicit equation involving √y + 4xy = 4 using implicit differentiation, the chain rule, and the product rule. Suitable for advanced calculus students in Grades 11-12.

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