91Ó°ÊÓ

The chain rule is a fundamental principle in calculus that helps us find the derivative of composite functions. When we apply the chain rule, we essentially differentiate a function within another function.

In the exercise, we have the expression \( y^{2/3} \). Here, \( y \) is a function of \( x \), so we consider it as a composite function. To differentiate \( y^{2/3} \) with respect to \( x \), the rule is to first differentiate \( u^{2/3} \) with respect to \( u \) (where \( u = y \)), giving us \( \frac{2}{3} u^{-1/3} \).

Then, we multiply by the derivative of \( y \) with respect to \( x \), i.e., \( \frac{dy}{dx} \). This step is crucial and reflects the essence of the chain rule, which helps capture how \( y \) changes with \( x \). By applying the chain rule, we accordingly obtain \( \frac{2}{3} y^{-1/3} \cdot \frac{dy}{dx} \).

In a consistent manner, we can use the rule to smoothly manage differentials in implicitly given equations.

The second derivative, represented as \( \frac{d^2y}{dx^2} \), provides information about the curvature of a function. It's essentially the derivative of the derivative, indicating how the rate of change of one quantity with respect to another is itself changing.

In the context of this exercise, once we have found \( \frac{dy}{dx} \), we differentiate it again with respect to \( x \) to find \( \frac{d^2y}{dx^2} \).

This process often requires applying the chain rule and product rule together, because our previously derived expression \( -\left(\frac{y}{x}\right)^{1/3} \) is not a simple function of \( x \). Instead, it involves both \( y \) and \( x \), which makes implicit differentiation necessary.

By carefully substituting \( \frac{dy}{dx} \) into our differentiated form and simplifying, we delve deeper into the function's behavior, unveiling how the slopes' gradient unfolds across various points.

Algebraic manipulation involves rearranging and simplifying equations to isolate terms or change forms, making it easier to solve or interpret them.

In this problem, algebraic manipulation is used critically in several steps. Once we have the differentiated forms, manipulating these equations allows us to express \( \frac{dy}{dx} \) and \( \frac{d^2y}{dx^2} \) in a comprehensible manner.

For example, to solve for \( \frac{dy}{dx} \), we had to rearrange the initial differential equation by moving terms around and dividing by coefficients like \( \frac{2}{3} y^{-1/3} \). This helped isolate the \( \frac{dy}{dx} \) term.

Similarly, for finding the second derivative, substituting back and managing fractions requires steady manipulation to maintain clarity and precision in results. The algebraic manipulation bridges the calculus process with interpretable solutions, demystifying the interrelations between the variables involved.