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how did the first derivative turn into the second?

To compute the second derivative \( \frac{d^2y}{dx^2} \) from the first derivative \( \frac{dy}{dx} \), you follow the process of implicit differentiation with respect to \( x \).

### First Derivative:
The first derivative is given as:
\[
\frac{dy}{dx} = \frac{y^2 + 3xy + 2x^2}{x^2}.
\]

### Steps to Calculate the Second Derivative:
1. **Rewrite the First Derivative**:
   Simplify the given first derivative, if necessary:
   \[
   \frac{dy}{dx} = \frac{y^2}{x^2} + \frac{3y}{x} + 2.
   \]

2. **Implicit Differentiation**:
   Differentiate both sides of the equation with respect to \( x \). Remember to use the **product rule** and the **chain rule** when differentiating terms involving \( y \), since \( y \) is a function of \( x \).

   Let’s apply the differentiation step-by-step:
   - For \( \frac{y^2}{x^2} \), use the quotient rule:
     \[
     \frac{d}{dx} \left(\frac{y^2}{x^2}\right) = \frac{2y \frac{dy}{dx} \cdot x^2 - y^2 \cdot 2x}{x^4} = \frac{2y \frac{dy}{dx} x^2 - 2xy^2}{x^4}.
     \]

   - For \( \frac{3y}{x} \), use the product rule:
     \[
     \frac{d}{dx} \left(\frac{3y}{x}\right) = \frac{3 \frac{dy}{dx} \cdot x - 3y}{x^2}.
     \]

   - The derivative of \( 2 \) is simply \( 0 \).

   Combining all terms, we get:
   \[
   \frac{d^2y}{dx^2} = \frac{\text{Numerator (from differentiation)}}{x^3}.
   \]

3. **Express the Second Derivative in Terms of \( x \), \( y \), and \( \frac{dy}{dx} \)**:
   Substitute back \( \frac{dy}{dx} \) into the resulting expression to express the second derivative entirely in terms of \( x \), \( y \), and \( \frac{dy}{dx} \). This gives the form:
   \[
   \frac{d^2y}{dx^2} = \frac{(2y + 3x) \left( x\frac{dy}{dx} - y \right)}{x^3}.
   \]

### Why This Happens:
- The structure of the second derivative depends on the chain rule (because \( y \) is a function of \( x \)) and quotient/product rules when combining terms from the first derivative.
- The key insight here is substituting \( \frac{dy}{dx} \) from the first equation into the second derivative formula, ensuring the consistency of the expression.

Would you like a more detailed breakdown of any step?

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### Related Questions:
1. What is the quotient rule, and how is it applied in differentiation?
2. How can we simplify complex second derivative expressions?
3. Why does the chain rule appear when differentiating \( y^2 \)?
4. How can we verify the correctness of the derived second derivative?
5. Can this second derivative be simplified further for specific values of \( x \) and \( y \)?

### Tip:
When working with derivatives involving \( y \), always remember that \( y \) is implicitly a function of \( x \), so the chain rule applies when differentiating terms involving \( y \).

How did the first derivative turn into the second?

Understand the step-by-step process of finding the second derivative from the given first derivative dy/dx = (y^2 + 3xy + 2x^2)/x^2. Explore the use of implicit differentiation, the chain rule, and the quotient rule to solve this calculus problem.

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