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Chapter 3: Problem 62

The Reciprocal Rule a. The Reciprocal Rule says that at any point where the function \(v(x)\) is differentiable and different from zero, $$\frac{d}{d x}\left(\frac{1}{v}\right)=-\frac{1}{v^{2}} \frac{d v}{d x}$$ Show that the Reciprocal Rule is a special case of the Derivative Quotient Rule. b. Show that the Reciprocal Rule and the Derivative Product Rule together imply the Derivative Quotient Rule.

Short Answer

Expert verified
The Reciprocal Rule is derived using the Quotient Rule by setting \(u=1\), and the Quotient Rule is derived using the Product and Reciprocal Rules.

Step by step solution

01

Understand the Reciprocal Rule

The Reciprocal Rule states that the derivative of a function \( \frac{1}{v(x)} \) with respect to \( x \) is given by \( \frac{d}{dx}\left(\frac{1}{v}\right) = -\frac{1}{v^2} \frac{dv}{dx} \), where \( v(x) \) is differentiable and not equal to zero.
02

Understand the Derivative Quotient Rule

The Quotient Rule for derivatives states that if you have two functions \( u(x) \) and \( v(x) \), where both are differentiable, the derivative of their quotient is \( \left(\frac{u}{v}\right)' = \frac{v(du/dx) - u(dv/dx)}{v^2} \).
03

Prove Reciprocal Rule using Quotient Rule

To prove that the Reciprocal Rule is a special case of the Quotient Rule, consider \( u(x) = 1 \) and \( v(x) = v(x) \). Then, using the Quotient Rule, \( \left(\frac{1}{v}\right)' = \frac{v(0) - 1(dv/dx)}{v^2} = -\frac{1}{v^2} \frac{dv}{dx} \), which matches the Reciprocal Rule.
04

Use Product and Reciprocal Rule to Derive Quotient Rule

Consider \( f(x) = \frac{u(x)}{v(x)} = u(x) \cdot \frac{1}{v(x)} \). Using the Product Rule, \( \left(\frac{u}{v}\right)' = u'(\frac{1}{v}) + u\left(-\frac{1}{v^2} \frac{dv}{dx}\right) \). Simplifying, we get \( = u'\frac{1}{v} - u\frac{v'}{v^2} = \frac{u'v - uv'}{v^2} \), which is the Quotient Rule.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Reciprocal Rule
The Reciprocal Rule is a handy tool when you are differentiating the reciprocal of a function. When you have a function that can be written as \( \frac{1}{v(x)} \), you're dealing with the reciprocal of \( v(x) \). To find its derivative with respect to \( x \), the Reciprocal Rule states:\[\frac{d}{d x}\left(\frac{1}{v}\right)=-\frac{1}{v^{2}} \frac{d v}{d x}\]This rule is essential in calculus because it simplifies taking derivatives of reciprocal functions. Here's why:
  • First, this derivative is negative, reflecting the opposite behavior between \( v(x) \) and its reciprocal.
  • The \( \frac{1}{v^2} \) term helps account for the rate of change in the denominator \( v(x) \).
So, whenever you differentiate a reciprocal function, this rule makes it quick and easy, saving you time in your calculations while reducing mistakes.
Quotient Rule
The Quotient Rule is fundamental when differentiating a ratio of two functions, say \( \frac{u(x)}{v(x)} \). It helps to determine how the two functions interact as their values change. The rule is expressed as:\[\left(\frac{u}{v}\right)' = \frac{v(du/dx) - u(dv/dx)}{v^2}\]Key points to remember:
  • The numerator \( v(du/dx) - u(dv/dx) \) calculates the difference in the influence of each function's rate of change.
  • The denominator \( v^2 \) normalizes this difference by the square of the denominator \( v(x) \).
By using the Quotient Rule, we can navigate the challenge of changing ratios effectively, keeping every term's transformation in check. This rule is particularly useful when functions interact in complex ways, ensuring mathematically accurate results.
Product Rule
The Product Rule comes into play when differentiating the product of two functions, \( f(x) = u(x) \cdot v(x) \). It's essential when both functions have changing contributions to the larger product. The Product Rule is written as:\[(f\cdot g)' = f'g + fg'\]Let's break it down:
  • This formula accounts for changes both functions introduce in the product.
  • \( f'g \) considers the effect of the derivative of \( f(x) \) while keeping \( g(x) \) constant to see its impact.
  • Similarly, \( fg' \) captures the influence of \( g(x) \)'s derivative.
The Product Rule allows us to systematically handle the complexity of products in calculus. By treating each function's contribution separately, it simplifies understanding how changes propagate through mathematical expressions. Combining it with the Reciprocal Rule can derive the Quotient Rule, showcasing the harmony in differentiation rules.

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Most popular questions from this chapter

The radius \(r\) of a circle is measured with an error of at most 2\(\%\) . What is the maximum corresponding percentage error in computing the circle's $$\text{a. circumference?} \quad \text{b. area?}$$

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