/*! 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 40 Find (a) \(\frac{d}{d x}\left[... [FREE SOLUTION] | 91影视

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Chapter 3: Problem 40

Find (a) \(\frac{d}{d x}\left[\log _{(1 / x)} e\right]\) (b) \(\frac{d}{d x}\left[\log _{(\ln x)} e\right]\)

Short Answer

Expert verified
(a) \( \frac{1}{x(\ln x)^2} \); (b) \(-\frac{1}{x \ln x (\ln(\ln x))^2} \).

Step by step solution

01

Express Derivative Problem

Given the function \( y = \log_{(1/x)} e \), express it in terms of natural logarithms using the change of base formula:\[ y = \frac{\ln e}{\ln(1/x)} = \frac{1}{-\ln x} \]
02

Differentiate with Respect to x

Now we differentiate \( y = \frac{1}{-\ln x} \) with respect to \( x \) using the quotient rule. The derivative \( \frac{d}{dx} \left( \frac{1}{-\ln x} \right) \) is:\[ \frac{d}{dx} \left( \frac{1}{-\ln x} \right) = \frac{0 \cdot (-\ln x) - 1 \cdot \left(-\frac{1}{x}\right)}{(-\ln x)^2} = \frac{1/x}{(\ln x)^2} \]Simplifying this yields:\[ \frac{1}{x(\ln x)^2} \]
03

Express second Derivative Problem

For the function \( y = \log_{(\ln x)} e \), use the change of base formula to express it as:\[ y = \frac{\ln e}{\ln(\ln x)} = \frac{1}{\ln(\ln x)} \]
04

Differentiate second Function

Differentiate \( y = \frac{1}{\ln(\ln x)} \) with respect to \( x \). The derivative \( \frac{d}{dx} \left( \frac{1}{\ln(\ln x)} \right) \) can be found using the chain rule and the quotient rule:\[ \frac{d}{dx} \left( \frac{1}{\ln(\ln x)} \right) = \frac{0 \cdot \ln(\ln x) - 1 \cdot \frac{1}{x \ln x}}{(\ln(\ln x))^2} = -\frac{1}{x \ln x (\ln(\ln x))^2} \]

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Key Concepts

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

Change of Base Formula
The Change of Base Formula is a useful tool in calculus, especially when dealing with logarithms of arbitrary bases. If you have a logarithm such as \( \log_b a \), you can convert it into a fraction involving natural logarithms, which are more common in calculus problems. The formula is:
  • \( \log_b a = \frac{\ln a}{\ln b} \)
In the exercise provided, this formula helps express logarithms of different bases in terms of natural logarithms. For instance, \( \log_{(1/x)} e \) can be expressed as \( \frac{\ln e}{\ln(1/x)} \), reducing it to \( \frac{1}{-\ln x} \) since \( \ln e = 1 \). This transformation simplifies the problem and makes it more manageable when you need to take the derivative.
Quotient Rule
The Quotient Rule is essential when differentiating a function that is the ratio of two other functions. If you have a function \( f(x) = \frac{u(x)}{v(x)} \), the Quotient Rule gives you the derivative:
  • \( f'(x) = \frac{u'(x)v(x) - u(x)v'(x)}{(v(x))^2} \)
This rule is particularly helpful in our exercise. For the function \( y = \frac{1}{-\ln x} \), you can see it as \( y = \frac{u(x)}{v(x)} \) with \( u(x) = 1 \) and \( v(x) = -\ln x \). This application allows calculation of the derivative as \( \frac{1/x}{(\ln x)^2} \), by identifying how both numerator and denominator change with respect to \( x \).
Natural Logarithms
Natural Logarithms, denoted as \( \ln \), are logarithms with the base \( e \), where \( e \approx 2.71828 \). These logarithms are fundamentally important because their calculus derivatives and integrals simplify significantly compared to other bases.
  • The differentiator enables expressions and functions to be simplified using properties like \( \ln e = 1 \), and transformations such as\( \ln(ab) = \ln a + \ln b \).
  • In our step-by-step solution, natural logarithms allow reformulation of \( \log_{(1/x)} e \) and \( \log_{(\ln x)} e \) in a more calculus-friendly form, making computation of derivatives straightforward.
Chain Rule
The Chain Rule is a powerful technique for differentiating composite functions. If a function \( y \) can be expressed as \( g(f(x)) \), its derivative is given by:
  • \( \frac{dy}{dx} = g'(f(x)) \times f'(x) \)
This rule is particularly used in differentiating the second function in our exercise, \( y = \frac{1}{\ln(\ln x)} \). Using the Chain Rule, you first differentiate \( \ln(\ln x) \) with respect to \( x \) yielding \( \frac{1}{x \ln x} \). The derivative of the outer function \( \frac{1}{u} \) in terms of \( u \) is \( -\frac{1}{u^2} \). Multiplying these results yields the final derivative \( -\frac{1}{x \ln x \cdot (\ln(\ln x))^2} \). Using chain rule ensures all layers of the function are addressed.

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Most popular questions from this chapter

Find the limits. $$\lim _{x \rightarrow \pi / 2^{-}} \sec 3 x \cos 5 x$$

Find the differential \(d y\). (a) \(y=\frac{1}{x^{3}-1}\) (b) \(y=\frac{1-x^{3}}{2-x}\)

The side of a square is measured to be \(10 \mathrm{ft}\), with a possible error of \(\pm 0.1 \mathrm{ft}\). (a) Use differentials to estimate the error in the calculated area. (b) Estimate the percentage errors in the side and the area.

A long high-voltage power line is 18 feet above the ground. The electric current in the line generates a magnetic field whose magnitude \(F\) (in microtesla) is given by \(F=36 / L\) where \(L\) is the (perpendicular) distance to the line in feet. Suppose that a new regulation requires the power line to be raised everywhere by \(0.9 \mathrm{ft} .\) Use a differential to estimate the decrease in the value of \(F\) at a point on the ground directly beneath the line.

Determine whether the statement is true or false. Explain your answer. A local linear approximation to a nonconstant function can never be constant.
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