Problem 22 - University Calculus: Early Transcendentals

Question

Use implicit differentiation to find \(d y / d x\) and then \(d^{2} y / d x^{2}\) $$x^{2 / 3}+y^{2 / 3}=1$$

Step-by-Step Solution

Answer

\(\frac{dy}{dx} = -\left(\frac{x}{y}\right)^{1/3}\), \(\frac{d^2y}{dx^2} = \frac{x^{1/3}}{3y^{5/3}} + \frac{2x^{4/3}}{3y^{8/3}}\)."

Step 1: Differentiate Both Sides with Respect to x

Given the equation \(x^{2/3} + y^{2/3} = 1\), implicitly differentiate both sides with respect to \(x\). The derivative of \(x^{2/3}\) with respect to \(x\) is \(\frac{2}{3} x^{-1/3}\). For \(y^{2/3}\), use the chain rule: \(\frac{2}{3} y^{-1/3} \cdot \frac{dy}{dx}\). The result is: \(\frac{2}{3} x^{-1/3} + \frac{2}{3} y^{-1/3} \frac{dy}{dx} = 0\).

Step 2: Solve for \(\frac{dy}{dx}\)

Rearrange the equation \(\frac{2}{3} x^{-1/3} + \frac{2}{3} y^{-1/3} \frac{dy}{dx} = 0\) to solve for \(\frac{dy}{dx}\). First, isolate the term involving \(\frac{dy}{dx}\): \(\frac{2}{3} y^{-1/3} \frac{dy}{dx} = -\frac{2}{3} x^{-1/3}\). Divide both sides by \(\frac{2}{3} y^{-1/3}\) to find \(\frac{dy}{dx} = -\left(\frac{x}{y}\right)^{1/3}\).

Step 3: Differentiate \(\frac{dy}{dx}\) to Find \(\frac{d^2y}{dx^2}\)

Now that \(\frac{dy}{dx} = -\left(\frac{x}{y}\right)^{1/3}\), differentiate \(\frac{dy}{dx}\) with respect to \(x\). Applying the quotient rule and chain rule, we have:1. \(\frac{d}{dx}\left(-\left(\frac{x}{y}\right)^{1/3}\right) = -\frac{1}{3}\left(\frac{x}{y}\right)^{-2/3}\left(\frac{y(1) - x\frac{dy}{dx}}{y^2}\right)\).2. Simplifying, use \(\frac{dy}{dx} = -\left(\frac{x}{y}\right)^{1/3}\) in the derivative to find: \(\frac{d^2y}{dx^2} = \frac{1}{3}\left(\frac{x}{y}\right)^{-2/3} \left(\frac{y + x\left(\frac{x}{y}\right)^{1/3}}{y^2}\right)\).

Step 4: Simplify the Expression for \(\frac{d^2y}{dx^2}\)

Simplify the expression for \(\frac{d^2y}{dx^2}\). Substitute \(\frac{dy}{dx} = -\left(\frac{x}{y}\right)^{1/3}\) into the formula and simplify each term. After simplification, the expression for \(\frac{d^2y}{dx^2}\) is: \(\frac{x^{1/3}}{3y^{5/3}} + \frac{2x^{4/3}}{3y^{8/3}}\).

Key Concepts

Chain Rule

The chain rule is a fundamental concept in calculus used to differentiate composite functions. When you have a function within another function, the chain rule helps you find the derivative of the overall composition. Imagine you have a function \(y=f(u)\) and \(u=g(x)\). To find the derivative \(\frac{dy}{dx}\), you use the chain rule formula:

• \(\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}\).

In the context of implicit differentiation, particularly with equations involving both \(x\) and \(y\), the chain rule simplifies finding \(\frac{dy}{dx}\) when the relationship between \(x\) and \(y\) is not given explicitly. For example, in the original exercise given (\(x^{2/3} + y^{2/3} = 1\)), the chain rule was applied to differentiate \(y^{2/3}\), treating \(y\) as a function of \(x\). By applying the rule, we find

• \(\frac{d}{dx}[y^{2/3}] = \frac{2}{3} y^{-1/3} \cdot \frac{dy}{dx}\).

This approach is crucial because it allows you to differentiate complex implicit functions by breaking them into smaller, more manageable parts.

Quotient Rule

The quotient rule is essential for differentiating functions that are represented as the quotient of two other functions. If you have a function \(h(x) = \frac{f(x)}{g(x)}\), the quotient rule provides you with a way to find its derivative:

• \(h'(x) = \frac{f'(x)g(x) - f(x)g'(x)}{(g(x))^2}\)

In the exercise solution, we first calculate \(\frac{dy}{dx}\) implicitly and then need to find the second derivative, \(\frac{d^2y}{dx^2}\). During this step, the derivative of \(-\left(\frac{x}{y}\right)^{1/3}\) is calculated using the quotient rule in combination with the chain rule. Here's a simplified breakdown of this step:

• Define \(u = x\) and \(v = y\), with \(\frac{u}{v} = \frac{x}{y}\).
• Apply the quotient rule, noting that both \(u\) and \(v\) also need implicit differentiation.

This complex derivative calculation is made manageable by systematically applying these rules, highlighting how interconnected differentiation techniques can be effectively used together.

Second Derivative

Finding the second derivative involves differentiating the first derivative again. In many calculus problems, especially those involving implicit differentiation, this step can become complex. In the original problem, after finding that \(\frac{dy}{dx} = -\left(\frac{x}{y}\right)^{1/3}\), we differentiate again to find \(\frac{d^2y}{dx^2}\).This process is meticulous because it involves using the results of the previous step, applying both the chain rule and the quotient rule simultaneously. The second derivative \(\frac{d^2y}{dx^2}\) can reveal more about the curvature and the concavity of the function represented by the initial implicit equation. In our specific case, after finding \(\frac{dy}{dx}\), we used'the earlier chain and quotient rules into the expression for the second derivative. It's about constructing the derivative correctly:

• The presence of composite functions within our derivative requires a careful reapplication of differentiation rules.
• This process leads to the formula \(\frac{d^2y}{dx^2} = \frac{x^{1/3}}{3y^{5/3}} + \frac{2x^{4/3}}{3y^{8/3}}\).

This expression helps understand how changes in \(x\) and \(y\) quartic evolve. Thus, mastering these second derivative techniques unlocks deeper insights into the behavior of implicit functions.