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?

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 $y$.