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We are asked to find the second derivative \( \frac{d^2 y}{dx^2} \) in terms of \(x\) and \(y\), given the equation \( y^3 = x^4 \).

### Step 1: Implicit Differentiation for the First Derivative
We start by differentiating both sides of the equation \( y^3 = x^4 \) with respect to \(x\), using implicit differentiation.

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

Using the chain rule for \(y^3\), we get:

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

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

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

### Step 2: Implicit Differentiation for the Second Derivative
Now, to find \( \frac{d^2y}{dx^2} \), we differentiate \( \frac{dy}{dx} = \frac{4x^3}{3y^2} \) again with respect to \(x\):

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

This requires using the quotient rule:

\[
\frac{d}{dx} \left( \frac{u}{v} \right) = \frac{v \frac{du}{dx} - u \frac{dv}{dx}}{v^2}
\]

Here, \( u = 4x^3 \) and \( v = 3y^2 \). Now, compute the derivatives:

- \( \frac{du}{dx} = 12x^2 \)
- \( \frac{dv}{dx} = 6y \frac{dy}{dx} \), and we substitute \( \frac{dy}{dx} = \frac{4x^3}{3y^2} \)

Thus:

\[
\frac{dv}{dx} = 6y \cdot \frac{4x^3}{3y^2} = \frac{24x^3}{y}
\]

Now, apply the quotient rule:

\[
\frac{d^2y}{dx^2} = \frac{3y^2 \cdot 12x^2 - 4x^3 \cdot \frac{24x^3}{y}}{(3y^2)^2}
\]

Simplify the numerator:

\[
= \frac{36y^2x^2 - 96x^6 / y}{9y^4}
\]

### Final Expression:

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

This is the second derivative of \(y\) with respect to \(x\) in terms of \(x\) and \(y\).

Would you like further details on any steps, or have any questions?

### Related questions:
1. How is implicit differentiation applied in multi-variable calculus?
2. What are the advantages of using the quotient rule in differentiation?
3. Can you explain how the chain rule works for implicit differentiation?
4. How would the result change if the equation was \( y^4 = x^5 \) instead?
5. What physical phenomena might be modeled by an equation like \( y^3 = x^4 \)?

### Tip:
Always simplify the expression as much as possible after differentiating to make interpretation and further manipulation easier.

Find d²y/dx² in terms of x and y, given y³ = x⁴.

Find the second derivative d²y/dx² in terms of x and y for the equation y³ = x⁴. This solution uses implicit differentiation, the chain rule, and the quotient rule to solve for the second derivative.

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