Problem 38 - Thomas Calculus

Question

In Exercises \(37-42,\) 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

\(\frac{dy}{dx} = -\frac{y^{1/3}}{x^{1/3}}\), \(\frac{d^2y}{dx^2} = \frac{-x^{1/3}\cdot y^{-1/3} + y^{2/3}}{3x^{2/3}y^{2/3}}\).

Step 1: Differentiate Implicitly with Respect to x

We start by differentiating the given equation \(x^{2/3} + y^{2/3} = 1\) implicitly with respect to \(x\). The derivative of \(x^{2/3}\) with respect to \(x\) is \(\frac{2}{3}x^{-1/3}\), and for \(y^{2/3}\) with respect to \(y\) is \(\frac{2}{3}y^{-1/3}\cdot \frac{dy}{dx}\) by the chain rule. Thus, we have:\[\frac{2}{3}x^{-1/3} + \frac{2}{3}y^{-1/3}\cdot \frac{dy}{dx} = 0\]

Step 2: Solve for dy/dx

From the result in Step 1, isolate \(\frac{dy}{dx}\):\[\frac{2}{3}y^{-1/3}\cdot \frac{dy}{dx} = -\frac{2}{3}x^{-1/3}\]Divide both sides by \(\frac{2}{3}y^{-1/3}\):\[\frac{dy}{dx} = -\frac{x^{-1/3}}{y^{-1/3}} = -\frac{y^{1/3}}{x^{1/3}}\]

Step 3: Differentiate dy/dx Implicitly to Find d²y/dx²

Now, differentiate \(\frac{dy}{dx} = -\frac{y^{1/3}}{x^{1/3}}\) implicitly with respect to \(x\):Apply the quotient rule: If \(u = y^{1/3}\) and \(v = x^{1/3}\), then\[\frac{d^2y}{dx^2} = \frac{v \cdot \frac{du}{dx} - u \cdot \frac{dv}{dx}}{v^2}\]First, find \(\frac{du}{dx} = \frac{1}{3}y^{-2/3}\cdot \frac{dy}{dx}\) and \(\frac{dv}{dx} = \frac{1}{3}x^{-2/3}\).Substitute these into the quotient formula and use \(\frac{dy}{dx}\) from Step 2 for simplification:\[\frac{d^2y}{dx^2} = \frac{x^{1/3} \cdot \frac{1}{3}y^{-2/3}\left(-\frac{y^{1/3}}{x^{1/3}}\right) - y^{1/3} \cdot \frac{1}{3}x^{-2/3}}{x^{2/3}}\]This simplifies to:\[\frac{d^2y}{dx^2} = \frac{-x^{1/3}\cdot y^{-1/3} + y^{2/3}}{3 \cdot x^{2/3} \cdot y^{2/3}}\]

Step 4: Simplify the Expression for d²y/dx²

Combine the terms within the numerator and simplify using the initial equation \(x^{2/3} + y^{2/3} = 1\):\[\frac{d^2y}{dx^2} = \frac{-x^{1/3} \cdot y^{-1/3} + 1 - x^{2/3}}{3 \cdot x^{2/3} \cdot y^{2/3}}\]Use substitutions and identities to further simplify if required, or express in terms of \(x\) and \(y\) as:\[\frac{d^2y}{dx^2} = \text{retain form depending on given constants/variables.}\]

Key Concepts

Chain Rule

The chain rule is a fundamental concept in calculus, especially when dealing with composite functions. It's an essential tool when we want to differentiate a composite function, that is, a function of another function. For instance, if you have a function expressed as \(f(g(x))\), and you need to find its derivative, you would apply the chain rule. The chain rule states that the derivative of \(f(g(x))\) with respect to \(x\) is \(f'(g(x)) \, g'(x)\).

In the exercise given, the chain rule is applied when differentiating the term \(y^{2/3}\) with respect to \(x\). Here, you don't differentiate \(y\) directly. Instead, you treat \(y\) as a function of \(x\), so you multiply the derivative of \(y^{2/3}\) with respect to \(y\) by \(\frac{dy}{dx}\), which gives us \(\frac{2}{3}y^{-1/3}\cdot \frac{dy}{dx}\). This step allows us to handle equations where \(y\) is not isolated, which is often the case in implicit differentiation.

• Always identify your inner function and outer function when using the chain rule.
• Multiply the derivative of the outer function by the derivative of the inner function.
• In problems involving \(y\), remember to apply \(\frac{dy}{dx}\) to show that \(y\) is also a function of \(x\).

Quotient Rule

When you encounter a mathematical expression that is a ratio or a division of two functions, the quotient rule becomes incredibly useful. The quotient rule helps to differentiate expressions of the form \(\frac{u}{v}\), where both \(u\) and \(v\) are functions of \(x\).

The formula for the quotient rule is given by \(\frac{d}{dx}\left( \frac{u}{v} \right) = \frac{v \cdot \frac{du}{dx} - u \cdot \frac{dv}{dx}}{v^2}\). When differentiating such expressions, be careful to follow these steps:

• Identify the numerator function \(u\) and the denominator function \(v\).
• Find the derivative of each: \(\frac{du}{dx}\) and \(\frac{dv}{dx}\).
• Apply the quotient rule formula. Substitute and simplify as needed.

In the solution provided, the quotient rule is applied to differentiate \(\frac{dy}{dx} = -\frac{y^{1/3}}{x^{1/3}}\). This step is especially necessary when dealing with higher-order derivatives, as it keeps the structure of the function clear and allows us to methodically work through the differentiation process.

Higher-order Derivatives

Higher-order derivatives involve taking the derivative of a derivative. In problems involving implicit differentiation, once the first derivative, \(\frac{dy}{dx}\), is found, often the next task is to find the second derivative, or \(\frac{d^2y}{dx^2}\).

To find a higher-order derivative, you first compute the first derivative, which might involve using rules such as the chain rule or the quotient rule. Once you have \(\frac{dy}{dx}\), differentiate it again with respect to \(x\).

• Begin with a clean first derivative: reduce fractions and simplify as much as possible.
• Apply the necessary differentiation rules again (e.g., chain rule, product rule, or quotient rule).
• Simplify the expression for clarity and easier computation.

In this exercise, once \(\frac{dy}{dx}\) is determined, the second derivative \(\frac{d^2y}{dx^2}\) is found by applying the quotient rule to \(\frac{dy}{dx}\). The process of finding \(\frac{d^2y}{dx^2}\) can be seen as taking the problem one level deeper, and thus may require careful management of algebraic manipulation and prior identities like the original equation.