Problem 83 - Calculus Early Transcendentals: Pearson New International Edition

Question

Give a second proof of the Quotient Rule. Write $$ D_{x}\left(\frac{f(x)}{g(x)}\right)=D_{x}\left(f(x) \frac{1}{g(x)}\right) $$ and use the Product Rule and the Chain Rule.

Step-by-Step Solution

Answer

The derivative is \( \frac{f'(x)g(x) - f(x)g'(x)}{g(x)^2} \), using the Product and Chain Rules.

Step 1: Set Up the Differentiation

Write the derivative of the quotient \( \frac{f(x)}{g(x)} \) as \( D_x \left( f(x) \cdot \frac{1}{g(x)} \right) \). This sets the stage for applying the Product Rule to differentiate the expression \( f(x) \cdot \frac{1}{g(x)} \).

Step 2: Apply the Product Rule

Recall the Product Rule: if you have two functions \( u(x) = f(x) \) and \( v(x) = \frac{1}{g(x)} \), then \( D_x(u \cdot v) = u'v + uv' \). Apply it: \( D_x(f(x) \cdot \frac{1}{g(x)}) = f'(x) \cdot \frac{1}{g(x)} + f(x) \cdot \frac{d}{dx} \left( \frac{1}{g(x)} \right) \).

Step 3: Differentiate Using the Chain Rule

Differentiate \( \frac{1}{g(x)} \) using the Chain Rule. Express \( \frac{1}{g(x)} \) as \( (g(x))^{-1} \). Thus, \( \frac{d}{dx} \left( \frac{1}{g(x)} \right) = \frac{d}{dx} \left( (g(x))^{-1} \right) = -g'(x)(g(x))^{-2} \).

Step 4: Substitute Back the Derivative

Substitute back the derivative found in Step 3 into the Product Rule application. This gives: \( D_x \left( f(x) \cdot \frac{1}{g(x)} \right) = f'(x) \cdot \frac{1}{g(x)} - f(x) \cdot \frac{g'(x)}{g(x)^2} \).

Step 5: Simplify the Expression

Combine the derivatives into a single fraction: \[ \frac{f'(x)g(x) - f(x)g'(x)}{g(x)^2} \]. This simplifies to the standard form of the quotient rule.

Key Concepts

Product Rule

The Product Rule is an essential tool in calculus when dealing with the derivative of the product of two differentiable functions. It tells us how to differentiate expressions like \( u(x) \cdot v(x) \). Here, \( u(x) \) and \( v(x) \) are functions of \( x \).

• First, identify the functions you are multiplying. In the context of the Quotient Rule exercise, set \( u(x) = f(x) \) and \( v(x) = \frac{1}{g(x)} \).
• Apply the rule: the derivative of the product \( u \cdot v \) can be expressed as \( u'v + uv' \).

This rule offers a methodical way to differentiate products by breaking them into smaller, manageable parts. This is crucial when proving or applying the Quotient Rule as seen in this exercise.

Chain Rule

The Chain Rule is another critical principle of calculus, especially useful when dealing with composite functions. A composite function occurs when one function nests within another, like \( h(x) = (g(x))^{-1} \).

• Determine the components of the composite. For \( (g(x))^{-1} \), think of \( h(u) = u^{-1} \) and \( u = g(x) \).
• The Chain Rule states that the derivative of \( h(g(x)) \) is \( h'(g(x)) \cdot g'(x) \).

Applying the Chain Rule allows us to correctly find the derivative of complex expressions. This exercise demonstrates its utility in differentiating \( \frac{1}{g(x)} \) to complete the evaluation of the Quotient Rule.

Differentiation Techniques

Differentiation techniques such as the Product and Chain Rules comprise the broader skill set necessary to tackle a variety of calculus problems. Mastery of these techniques enables the solution of complex derivatives systematically. Some key points about differentiation:

• The Product Rule applies when you're dealing with products of functions, as shown in the exercise.
• The Chain Rule comes into play for compositions of functions, a crucial step in the Quotient Rule exercise.

By combining these techniques, you can break down and solve almost any differentiable expression you encounter in calculus. Practice these regularly to deepen your understanding and skills in calculus.

Calculus

At its core, calculus is the study of change. It provides the tools to calculate rates of change and the accumulation of quantities. Differentiation, in particular, focuses on rates of change.

• It allows the determination of how a function's output changes as its input changes.
• Key differentiation rules like the Product and Chain Rules form the backbone of calculus problem-solving.
• The Quotient Rule, a derivative application, combines these rules seamlessly to handle ratios of functions.

Understanding calculus and its differentiation techniques opens the door to solving real-world problems. From engineering to economics, these principles help model and predict behavior across various fields.