/*! 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 111 Differentiate. $$ f(t)=\ln \... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Differentiate. $$ f(t)=\ln \left|\frac{1-t}{1+t}\right| $$

Short Answer

Expert verified
The derivative is \( f'(t) = \frac{-2}{1-t^2} \).

Step by step solution

01

Understand the Problem

We need to differentiate the function \( f(t) = \ln \left| \frac{1-t}{1+t} \right| \). This involves using the chain rule and the derivative of logarithmic functions.
02

Rewrite the Function

Notice that \( \ln \left| \frac{1-t}{1+t} \right| \) can be expressed as \( \ln \left( \frac{1-t}{1+t} \right) \) since we are differentiating with respect to \( t \), assuming \( 1-t eq 0 \) and \( 1+t eq 0 \).
03

Use the Chain Rule

Let \( g(t) = \frac{1-t}{1+t} \). Then \( f(t) = \ln |g(t)| \), and we apply the chain rule to find \( f'(t) = \frac{1}{g(t)} \cdot g'(t) \).
04

Differentiate \( g(t) \)

We need \( g'(t) \). Use the quotient rule: \( g'(t) = \frac{(1+t)(-1) - (1-t)(1)}{(1+t)^2} = \frac{-1-t - (1-t)}{(1+t)^2} = \frac{-2}{(1+t)^2} \).
05

Combine Results

Substitute \( g(t) \) and \( g'(t) \) in the expression from Step 3: \( f'(t) = \frac{1}{\frac{1-t}{1+t}} \cdot \frac{-2}{(1+t)^2} = \frac{-2(1+t)}{(1-t)(1+t)^2} = \frac{-2}{(1-t)(1+t)} \).
06

Simplify the Expression

Since the expression simplifies to \( \frac{-2}{1-t^2} \), we have found the derivative of the function given. Note \( f'(t) = \frac{-2}{1-t^2} \).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Chain Rule
The Chain Rule is an essential technique in calculus used for differentiating composite functions. In simple terms, when you have a function composed of two other functions, you apply the Chain Rule to differentiate.
For example, if you have a function like \( f(x) = h(g(x)) \), the derivative \( f'(x) \) is found by multiplying the derivative of the outer function \( h \) evaluated at the inner function \( g \), with the derivative of the inner function \( g \). This gives us \( f'(x) = h'(g(x)) \cdot g'(x) \).
In the given problem, we apply the Chain Rule when finding the derivative of \( f(t) = \ln |g(t)| \), where \( g(t) = \frac{1-t}{1+t} \).
  • The derivative of \( \ln |x| \) is \( \frac{1}{x} \).
  • Using the chain rule, the derivative \( f'(t) = \frac{1}{g(t)} \cdot g'(t) \).
This rule simplifies the process and ensures precision in evaluating derivatives of complex, composite functions.
Quotient Rule
The Quotient Rule is utilized when differentiating functions that are ratios of two expressions. Knowing how to apply this rule is crucial whenever you face a division situation in functions.
For a function expressed as a quotient, such as \( g(x) = \frac{u(x)}{v(x)} \), the derivative is:
\[ g'(x) = \frac{v(x)u'(x) - u(x)v'(x)}{(v(x))^2} \]
Applying this rule ensures you accurately account for changes in both the numerator and the denominator functions.
In our example, \( g(t) = \frac{1-t}{1+t} \). Following the Quotient Rule:
  • \( u(t) = 1-t \) and \( u'(t) = -1 \).
  • \( v(t) = 1+t \) and \( v'(t) = 1 \).
  • Thus, \( g'(t) = \frac{(1+t)(-1) - (1-t)(1)}{(1+t)^2} = \frac{-2}{(1+t)^2} \).
This provides an efficient way to handle differentiations involving fractions.
Logarithmic Differentiation
Logarithmic Differentiation is a powerful technique, especially useful for differentiating functions with complex structures or products and quotients of exponential expressions. By leveraging the properties of logarithms, complex multiplication and division are transformed into addition and subtraction, thereby easing the differentiation process.
Consider a function like \( y = \ln \left| \frac{1-t}{1+t} \right| \) from our problem. Using logarithmic differentiation:
  • First, simplify by rewriting as \( y = \ln \left( \frac{1-t}{1+t} \right) \), assuming that \( 1-t eq 0 \) and \( 1+t eq 0 \).
  • Next, differentiate the expression using the Chain Rule: \( y' = \frac{1}{g(t)} \cdot g'(t) \).
This method capitalizes on the simpler properties of logarithms to differentiate, particularly useful in our exercise where the natural log function is involved. The result \( f'(t) = \frac{-2}{1-t^2} \) elegantly captures the derivative of the original function using this technique.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

The revenue of Red Rocks, Inc., in millions of dollars, is given by the function $$ R(t)=\frac{4000}{1+1999 e^{-0.5 t}} $$ where \(t\) is measured in years a) What is \(R(0),\) and what does it represent? b) Find \(\lim _{t \rightarrow \infty} R(t) .\) Call this value \(R_{\max },\) and explain what it means c) Find the value of \(t\) (to the nearest integer) for which \(R(t)=0.99 R_{\max }\)

Find an expression relating the exponential growth rate \(k\) and the tripling time \(T_{3}\).

As part of a study, students in a psychology class took a final exam. They took equivalent forms of the exam at monthly intervals thereafter. After \(t\) months, the average score \(S(t),\) as a percentage, was found to be given by $$ S(t)=78-15 \ln (t+1), \quad t \geq 0 $$. a) What was the average score when they initially took the test, \(t=0 ?\) b) What was the average score after 4 months? c) What was the average score after 24 months? d) What percentage of their original answers did the students retain after 2 years ( 24 months)? e) Find \(S^{\prime}(t)\). f) Find the maximum and minimum values, if they exist. g) Find \(\lim _{t \rightarrow \infty} S(t)\) and discuss its meaning.

Differentiate. $$ f(x)=\ln \frac{1+\sqrt{x}}{1-\sqrt{x}} $$

The tortoise population, \(P(t),\) in a square mile of the Mojave Desert after \(t\) years can be approximated by the logistic equation $$ P(t)=\frac{3000}{20+130 e^{-0.214 t}} $$ a) Find the tortoise population after \(0 \mathrm{yr}, 5 \mathrm{yr}, 15 \mathrm{yr},\) and 25 yr. b) Find the rate of change in the population, \(P^{\prime}(t)\). c) Find the rate of change in the population after \(0 \mathrm{yr}\) \(5 \mathrm{yr}, 15, \mathrm{yr},\) and \(25 \mathrm{yr}\) d) What is the limiting value (see Exercise 42 ) for the population of tortoises in a square mile of the Mojave Desert?

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.