Problem 13 - Calculus Early Transcendentals: Pearson New International Edition

Question

In Problems 1-20, find \(D_{x} y\). $$ y=\left(\frac{x+1}{x-1}\right)^{3} $$

Step-by-Step Solution

Answer

The derivative is \( \frac{dy}{dx} = \frac{-6(x+1)^2}{(x-1)^4} \).

Step 1: Identify the Function Form

The given function is \( y = \left(\frac{x+1}{x-1}\right)^{3} \). This is a composition of functions where \( u = \frac{x+1}{x-1} \) and then \( y = u^3 \).

Step 2: Apply the Chain Rule

To find \( \frac{dy}{dx} \), use the chain rule. The chain rule states that \( \frac{dy}{dx} = \frac{d}{du} (u^3) \cdot \frac{du}{dx} \).

Step 3: Differentiate the Outer Function

Differentiating \( y = u^3 \) with respect to \( u \), we have \( \frac{d}{du} (u^3) = 3u^2 \).

Step 4: Differentiate the Inner Function

Now, differentiate \( u = \frac{x+1}{x-1} \). Use the quotient rule: \( \frac{du}{dx} = \frac{(x-1)(1) - (x+1)(1)}{(x-1)^2} = \frac{x-1-x-1}{(x-1)^2} = \frac{-2}{(x-1)^2} \).

Step 5: Combine the Results

Multiply the derivatives from Steps 3 and 4: \( \frac{dy}{dx} = 3u^2 \cdot \frac{-2}{(x-1)^2} = \frac{-6u^2}{(x-1)^2} \).

Step 6: Substitute Back for \( u \)

Substitute \( u = \frac{x+1}{x-1} \) back into \( \frac{dy}{dx} \) to express the derivative in terms of \( x \): \( \frac{dy}{dx} = \frac{-6\left(\frac{x+1}{x-1}\right)^2}{(x-1)^2} = \frac{-6(x+1)^2}{(x-1)^4} \).

Key Concepts

Chain Rule

The chain rule is a fundamental tool in calculus for differentiating composite functions, which are functions composed of two or more functions. Imagine you have an outer function and an inner function. The chain rule helps us find the derivative of the outer function with respect to the inner function, and then multiply this by the derivative of the inner function with respect to the original variable. This method is like peeling an onion layer by layer. For our exercise, the outer function is \(y = u^3\) and the inner function is \(u = \frac{x+1}{x-1}\).

The chain rule states:

• Find the derivative of the outer function with respect to the inner function.
• Multiply it by the derivative of the inner function with respect to the variable, in this case, \(x\).

Applying this, we started by differentiating \(y = u^3\), giving us \(3u^2\). Then, using the quotient rule, we found the derivative of the inner function. The result is multiplying these two derivatives to form the overall derivative of the composite function.

Quotient Rule

The quotient rule is essential when differentiating a function that is the division of two other functions. It's crucial when you have expressions in a fraction, like \(\frac{x+1}{x-1}\) in our exercise. This rule states that if you have a function \(\frac{f(x)}{g(x)}\), then its derivative is:

• \(\frac{d}{dx}\left(\frac{f(x)}{g(x)}\right) = \frac{g(x)\cdot f'(x) - f(x)\cdot g'(x)}{[g(x)]^2}\)

In the exercise, \(f(x) = x+1\) and \(g(x) = x-1\). Differentiating these, we got \(f'(x) = 1\) and \(g'(x) = 1\). Plugging into the quotient rule:

• \(\frac{(x-1)\cdot 1 - (x+1)\cdot 1}{(x-1)^2} = \frac{-2}{(x-1)^2}\)

This derivative of the quotient helps us, along with the previously found \(3u^2\), use the chain rule effectively to differentiate the entire function.

Composite Functions

Composite functions involve applying one function to the result of another function, typically denoted as \(f(g(x))\). In our example, we have the composite function \(y = \left(\frac{x+1}{x-1}\right)^3\), where \(g(x) = \frac{x+1}{x-1}\) and \(f(u) = u^3\).

Composite functions are common in calculus, and understanding them is vital because they form the basis for using the chain rule. You can think of them as a process where you first transform \(x\) into \(u\) using the inner function, and then apply the outer function on \(u\) to find \(y\).

Recognizing and working with composite functions allows us to easily utilize rules such as the chain rule to differentiate them efficiently. In our exercise, identifying \(u = \frac{x+1}{x-1}\) was a key step in simplifying the problem and properly applying calculus techniques.