/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 65 Logarithmic differentials Let \(... [FREE SOLUTION] | 91影视

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Chapter 13: Problem 65

Logarithmic differentials Let \(f\) be a differentiable function of one or more variables that is positive on its domain. a. Show that \(d(\ln f)=\frac{d f}{f}\) b. Use part (a) to explain the statement that the absolute change in \(\ln f\) is approximately equal to the relative change in \(f\) c. Let \(f(x, y)=x y,\) note that \(\ln f=\ln x+\ln y,\) and show that relative changes add; that is, \(d f / f=d x / x+d y / y\) d. Let \(f(x, y)=x / y,\) note that \(\ln f=\ln x-\ln y,\) and show that relative changes subtract; that is \(d f / f=d x / x-d y / y\) e. Show that in a product of \(n\) numbers, \(f=x_{1} x_{2} \cdots x_{n},\) the relative change in \(f\) is approximately equal to the sum of the relative changes in the variables.

Short Answer

Expert verified
Question: Prove that for the function \(f(x) = x^{\ 2}\), the relative change in \(f\) is approximately equal to twice the relative change in \(x\). Solution: To find the relative change in \(f(x) = x^{\ 2}\), we first need to find its differential: \(d(f) = 2x\ dx\) Now, let's find the relative change in \(f(x)\), which is given by the ratio of the absolute change in \(f(x)\) to the function's value: \(\frac{d(f)}{f} = \frac{2x\ dx}{x^{\ 2}} = \frac{2\ dx}{x}\) To prove that the relative change in \(f\) is approximately equal to twice the relative change in \(x\), we need to find the differential of \(\ln{x}\): \(d(\ln{x}) = \frac{1}{x}\ dx\) Using the result from part a, the relative change in \(x\) is given by: \(\frac{dx}{x} = d(\ln x) = \frac{1}{x}\ dx\) Now, let's multiply the expression for the relative change in \(x\) by 2: \(2\cdot\frac{dx}{x} = 2\cdot\frac{1}{x}\ dx = \frac{2\ dx}{x}\) Comparing the expressions for the relative change in \(f\) and twice the relative change in \(x\), we find that they are equal: \(\frac{d(f)}{f} = \frac{2\ dx}{x} = 2\cdot\frac{dx}{x}\) Thus, the relative change in \(f(x) = x^{\ 2}\) is approximately equal to twice the relative change in \(x\).

Step by step solution

01

Part a: Proving \(d(\ln f) = \frac{df}{f}\)

First, let's start by applying the chain rule to find the differential of \(\ln f\). If \(f\) is a function, \(g(x) = \ln x\), then \(g(f(x)) = \ln f(x)\). Now we can apply the chain rule: \(\frac{d(\ln f)}{dx} = \frac{d(\ln f)}{df} \cdot \frac{df}{dx}\) Since \(\frac{d(\ln x)}{dx} = \frac{1}{x}\), we get: \(\frac{d(\ln f)}{dx} = \frac{1}{f} \cdot \frac{df}{dx}\) Multiplying both sides by \(dx\) gives us the desired result: \(d(\ln f) = \frac{df}{f}\)
02

Part b: Absolute change in \(\ln f\) and relative change in \(f\)

The absolute change in a function can be defined as the absolute value of the differential of the function, whereas the relative change in a function is the ratio of the absolute change to the value of the function. We are given: \(d(\ln f) \approx \frac{df}{f}\) This states that the absolute change in \(\ln f\) is approximately equal to the ratio of the absolute change in \(f\) to the value of \(f\). This is the definition of the relative change in \(f\). Therefore, the absolute change in \(\ln f\) is approximately equal to the relative change in \(f\).
03

Part c: Relative changes add in \(f(x, y) = xy\)

We are given that \(f(x, y) = xy\) and \(\ln f = \ln x + \ln y\). To show that relative changes add, we need to find \(d(\ln f)\): \(d(\ln f) = d(\ln x + \ln y) = d(\ln x) + d(\ln y)\) Using the result from part a, we can now substitute: \(\frac{df}{f} = \frac{dx}{x} + \frac{dy}{y}\) We are given that \(f = xy\), so let's differentiate \(f\) with respect to \(x\) and \(y\): \(df = x\ dy + y\ dx\) Dividing by \(f = xy\) to express the equation in terms of relative changes: \(\frac{df}{f} = \frac{x\ dy + y\ dx}{xy} = \frac{dx}{x} + \frac{dy}{y}\) This shows that relative changes add for \(f(x, y) = xy\).
04

Part d: Relative changes subtract in \(f(x, y) = \frac{x}{y}\)

We are given that \(f(x,y)=\frac{x}{y}\) and \(\ln f = \ln x - \ln y\). To show that relative changes subtract, we need to find \(d(\ln f)\): \(d(\ln f) = d(\ln x - \ln y) = d(\ln x) - d(\ln y)\) Using the result from part a, we can now substitute: \(\frac{df}{f} = \frac{dx}{x} - \frac{dy}{y}\) We are given that \(f = \frac{x}{y}\), so let's differentiate \(f\) with respect to \(x\) and \(y\): \(df = \frac{y\ dx - x\ dy}{y^2}\) Dividing by \(f = \frac{x}{y}\) to express the equation in terms of relative changes: \(\frac{df}{f} = \frac{y\ dx - x\ dy}{xy} = \frac{dx}{x} - \frac{dy}{y}\) This shows that relative changes subtract for \(f(x, y) = \frac{x}{y}\).
05

Part e: Sum of relative changes in a product of \(n\) numbers

We are given a product of \(n\) numbers, \(f = x_1 x_2 \cdots x_n\). Taking the natural logarithm of both sides, we get: \(\ln f = \ln x_1 + \ln x_2 + \cdots + \ln x_n\) Now, finding the differential of \(\ln f\): \(d(\ln f) = d(\ln x_1) + d(\ln x_2) + \cdots + d(\ln x_n)\) Using the result from part a, we substitute for each term: \(\frac{df}{f} = \frac{dx_1}{x_1} + \frac{dx_2}{x_2} + \cdots + \frac{dx_n}{x_n}\) This is the sum of the relative changes in the variables, which shows that for a product of \(n\) numbers, the relative change in \(f\) is approximately equal to the sum of the relative changes in the variables.

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Key Concepts

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

Chain Rule
The chain rule is a key method in calculus used to differentiate compositions of functions. It helps us find the derivative of a composite function by breaking it down into simpler parts. A composite function is essentially a function inside another function. For example, if we have a function like \( g(f(x)) \), this forms a composite where "\( f(x) \)" is nested within "\( g \)".
To apply the chain rule, we differentiate the outer function (\( g(x) \)) with respect to its argument and then multiply it by the derivative of the inner function (\( f(x) \)). Mathematically, for \( h(x) = g(f(x)) \), the chain rule states, \[ \frac{dh}{dx} = \frac{dg}{df} \cdot \frac{df}{dx} \]
In logarithmic differentiation, this rule allows us to break down the natural logarithm of a function, leading to the relation \( d(\ln f) = \frac{df}{f} \), which simplifies handling of complicated functions.
Relative Change
Relative change is a concept used to compare the change in a variable in relation to its original value. It provides a way to express change in percentage terms, which can help us understand the significance of the change.
In mathematical terms, the relative change in a function \( f \) is given by \( \frac{df}{f} \). This represents the differential \( df \) (the change in \( f \)) divided by its original value \( f \).
  • For example, if a stock price changes from \\(100\ to \\)110\, the relative change would be \( \frac{10}{100} = 0.1 \) or 10%.
  • Relative change helps to easily compare changes across different scales or units.
This concept is directly connected to logarithmic differentials because logarithms naturally transform multiplication into addition, and the differential form \( d(\ln f) \) directly relates to the relative change \( \frac{df}{f} \), making it an effective tool in many areas of mathematics and applied sciences.
Differentiable Functions
Differentiable functions are functions that can be differentiated, meaning they have a derivative. This implies that the function is smooth and continuous over its domain without any jumps, cusps, or discontinuities. Differentiability is a crucial aspect of calculus because it allows us to examine the rates of change and slopes of curves on graphs.
  • A function \( f(x) \) is said to be differentiable at a point \( x = a \) if the derivative \( f'(a) \) exists.
  • If a function is differentiable over a range, it is also continuous over that range.
In the context of logarithmic differentials, we assume the function \( f \), where \( \ln f \) is being differentiated, is positive and differentiable over its domain. This allows us to leverage calculus techniques like the chain rule and find meaningful relationships between functions and their logs.
Natural Logarithm
The natural logarithm, denoted by \( \ln \), is a logarithm to the base \( e \), where \( e \approx 2.718 \). It's a fundamental constant in mathematics, playing a significant role in calculus, complex analysis, and number theory. Natural logarithms are used extensively due to their unique properties and relationship with exponential functions.
  • The natural logarithm of a number is the power to which \( e \) must be raised to equal that number.
  • It simplifies the appearance and behavior of many mathematical formulas.
  • For example, \( \ln(e^x) = x \), and \( e^{\ln x} = x \), showing a close relationship with exponential growth.
When differentiating functions using natural logarithms, the derivative of \( \ln x \) becomes \( \frac{1}{x} \), which ties directly into the chain rule and the simplification of composite functions. This makes natural logarithms a powerful tool in simplifying complex mathematical expressions.

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