/*! 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 1 Find the derivative. It may be t... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 3: Problem 1

Find the derivative. It may be to your advantage to simplify before differentiating. Assume \(a, b, c,\) and \(k\) are constants. $$f(t)=\ln \left(t^{2}+1\right)$$

Short Answer

Expert verified
The derivative is \(\frac{2t}{t^2 + 1}\).

Step by step solution

01

Understand the function

We start by considering the function given, which is \(f(t) = \ln(t^2 + 1)\). The function is the natural logarithm of the expression \(t^2 + 1\). We will need to differentiate this with respect to \(t\).
02

Apply the Chain Rule

To differentiate \(f(t) = \ln(t^2 + 1)\), we use the chain rule. The chain rule states that if we have a composite function \(f(g(t))\), then its derivative is \(f'(g(t)) \cdot g'(t)\). Here, \(g(t) = t^2 + 1\) and \(f(u) = \ln(u)\).
03

Differentiate the outer function

Differentiate the outer function \(f(u) = \ln(u)\). The derivative of \(\ln(u)\) with respect to \(u\) is \(\frac{1}{u}\). So, \(f'(g(t)) = \frac{1}{g(t)} = \frac{1}{t^2 + 1}\).
04

Differentiate the inner function

The inner function is \(g(t) = t^2 + 1\). Its derivative is \(g'(t) = 2t\).
05

Combine using the chain rule

By the chain rule, the derivative of \(f(t) = \ln(t^2 + 1)\) is \(f'(t) = \frac{1}{t^2 + 1} \cdot (2t) = \frac{2t}{t^2 + 1}\).

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 91影视!

Key Concepts

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

Chain Rule
The chain rule is an essential tool in calculus used when differentiating composite functions. A composite function is one where one function is nested inside another, such as \( f(g(t)) \). To apply the chain rule, you differentiate the outer function and then multiply by the derivative of the inner function. This is often expressed as \( f'(g(t)) \cdot g'(t) \).
  • Identify the inner function \( g(t) \) and the outer function \( f(u) \). In this exercise, \( g(t) = t^2 + 1 \) and \( f(u) = \ln(u) \).
  • Differentiating the outer function gives \( f'(u) = \frac{1}{u} \), while differentiating the inner function gives \( g'(t) = 2t \).
  • Combine these results to find the derivative of the entire function: \( f'(t) = \frac{1}{g(t)} \cdot g'(t) = \frac{2t}{t^2 + 1} \).
The chain rule provides a structured method for dealing with composite functions, making differentiation manageable even for more complex expressions.
Natural Logarithm
The natural logarithm, denoted as \( \ln \), is a logarithm to the base \( e \), where \( e \approx 2.71828 \). This function is particularly significant in calculus because of its simple derivative. The derivative of \( \ln(u) \) with respect to \( u \) is \( \frac{1}{u} \). This property makes the natural logarithm very straightforward to differentiate compared to logarithms of other bases.
  • Comprehending the natural logarithm's properties is crucial, especially when it appears as part of a larger function in differentiation problems.
  • In our exercise, the function \( f(t) = \ln(t^2 + 1) \) centers around finding the derivative of a natural logarithm of another function.
  • This feature of the natural logarithm simplifies the derivative to involve only a fraction where the numerator is the derivative of the inner function.
Understanding and applying the natural logarithm's derivative rule can significantly speed up the process of solving calculus problems, particularly those involving the chain rule.
Differentiation
Differentiation is a fundamental concept in calculus, focusing on finding the rate at which a function changes at any given point. The derivative of a function, denoted \( f'(t) \) or \( \frac{df}{dt} \), gives us this rate of change.
  • In our given function \( f(t) = \ln(t^2 + 1) \), differentiation seeks to determine how quickly \( f(t) \) changes as \( t \) changes.
  • The processes we've explored, such as applying the chain rule, serve the aim of differentiation, helping us determine \( f'(t) \).
  • The final result from the original problem, \( \frac{2t}{t^2 + 1} \), reflects the derivative of the function \( f(t) \), indicating its slope at various points.
Acquiring a robust understanding of differentiation, including techniques like the chain rule, equips students to tackle a variety of problems that involve rates of change in real-world scenarios.

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

Are the statements true or false for a function \(f\) whose domain is all real numbers? If a statement is true, explain how you know. If a statement is false, give a counterexample. If \(f^{\prime}(x) \leq g^{\prime}(x)\) for all \(x,\) then \(f(x) \leq g(x)\) for all \(x\)

Imagine you are zooming in on the graph of each of the following functions near the origin: $$\begin{aligned} &y=x \quad y=\sqrt{x}\\\ &y=x^{2} \quad y=\sin x\\\ &y=x \sin x \quad y=\tan x\\\ &y=\sqrt{x /(x+1)} y=x^{3}\\\ &y=\ln (x+1) \quad y=\frac{1}{2} \ln \left(x^{2}+1\right)\\\ &y=1-\cos x \quad y=\sqrt{2 x-x^{2}} \end{aligned}$$ Which of them look the same? Group together those functions which become indistinguishable, and give the equations of the lines they look like.

Find a formula for the error \(E(x)\) in the tangent line approximation to the function near \(x=a\). Using a table of values for \(E(x) /(x-a)\) near \(x=a\), find a value of \(k\) such that \(E(x) /(x-a) \approx k(x-a) .\) Check that, approximately, \(k=f^{\prime \prime}(a) / 2\) and that \(E(x) \approx\left(f^{\prime \prime}(a) / 2\right)(x-a)^{2}\). $$f(x)=\sqrt{x}, a=1$$

Let \(P=f(t)\) give the US population \(^{6}\) in millions in year \(t.\) (a) What does the statement \(f(2005)=296\) tell you about the US population? (b) Find and interpret \(f^{-1}(296) .\) Give units. (c) What does the statement \(f^{\prime}(2005)=2.65\) tell you about the population? Give units. (d) Evaluate and interpret \(\left(f^{-1}\right)^{\prime}(296) .\) Give units.

Give an example of: A function that is equal to a constant multiple of its derivative but that is not equal to its derivative.
See all solutions

Recommended explanations on Math Textbooks

Pure Maths

Read Explanation

Calculus

Read Explanation

Probability and Statistics

Read Explanation

Logic and Functions

Read Explanation

Applied Mathematics

Read Explanation

Decision Maths

Read Explanation
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.