/*! 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 31 Find the derivative of each of t... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 23: Problem 31

Find the derivative of each of the given functions. $$f(R)=\sqrt{\frac{2 R+1}{4 R+1}}$$

Short Answer

Expert verified
The derivative is \\( f'(R) = \frac{1}{2}\sqrt{\frac{4R+1}{2R+1}} \cdot \frac{-2}{(4R+1)^2} \\).

Step by step solution

01

Identify the Function Type

The given function is a composition of functions involving a square root and a rational expression. To differentiate it, we'll use the chain rule and the quotient rule.
02

Apply Chain Rule First

The outer function is the square root. Let \(u(R) = \frac{2R+1}{4R+1}\). Then the original function becomes \(f(R) = \sqrt{u(R)}\). The derivative of \(\sqrt{u}\) with respect to \(u\) is \(\frac{1}{2\sqrt{u}}\). Apply the chain rule: \(f'(R) = \frac{1}{2\sqrt{u(R)}} \cdot u'(R)\).

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
When determining the derivative of a composition of functions, the chain rule is a fundamental tool. The chain rule enables us to differentiate complex functions that are composed of two or more functions. For instance, if you have a function that is applied to the output of another function, you'll need to utilize the chain rule. This rule states that if you have a composite function, say \(f(g(x))\), then its derivative is \(f'(g(x)) \, g'(x)\).
Let's break it down:
  • Determine the outer and inner functions. The outer function is the last operation performed, and the inner function is the operation within.
  • Differentiating the outer function with respect to the inner function gives \(f'(g(x))\).
  • Differentiating the inner function with respect to \(x\) gives \(g'(x)\).
By multiplying these derivatives, you obtain the derivative of the composite function. In the original exercise, the square root function is the outer function, and the rational expression is the inner function.
Quotient Rule
The quotient rule is utilized when differentiating a function that is the division of two functions, \(f(x) = \frac{u(x)}{v(x)}\). It's essential when dealing with rational functions to find their derivatives correctly.
The formula for the quotient rule is:\[(f/g)' = \frac{(g(x) \cdot u'(x) - u(x) \cdot g'(x))}{(g(x))^2}\]Here's how to apply it:
  • Identify the top function, \(u(x)\), and the bottom function, \(v(x)\).
  • Find the derivative of the numerator, \(u'(x)\), and the derivative of the denominator, \(v'(x)\).
  • Apply the quotient rule by plugging these derivatives into the formula.
Returning to our original problem: \(u(x) = 2R + 1\) and \(v(x) = 4R + 1\). You would first compute these derivatives before proceeding with the rest of the differentiation process.
Composition of Functions
Understanding a composition of functions is crucial when finding derivatives of complex functions. A composite function is created when a function is plugged into another function, like \(f(g(x))\). It's important to identify which parts of a complex expression correspond to different functions.
For example:
  • In our exercise, the overall function is a composition involving a square root and a rational expression. \(f(R) = \sqrt{u(R)}\) where \(u(R) = \frac{2R+1}{4R+1}\).
  • Recognizing this structure allows us to use the chain rule effectively, as discussed earlier.
  • This methodical approach not only simplifies the process but also ensures accuracy in differentiation.
By understanding the essence of function composition, you can untangle even the most intricate functions for successful differentiation.

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

Solve the given problems by using implicit differentiation. A computer is programmed to draw the graph of the implicit function \(\left(x^{2}+y^{2}\right)^{3}=64 x^{2} y^{2}\) (see Fig. 23.45 and Example 7 on page 607 ). Find the slope of a line tangent to this curve at (2.00,0.56) and at (2.00,3.07)

Solve the given problems by finding the appropriate derivatives.If the population of a city is \(P(t)=8000\left(1+0.02 t+0.005 t^{2}\right)\) (t is in years from 2010 ), what is the acceleration in the size of the population?

Find the acceleration of an object for which the displacement \(s\) (in \(\mathrm{m}\) ) is given as a function of the time \(t\) (in s) for the given value of \(t\). $$s=3(1+2 t)^{4}, t=0.500 \mathrm{s}$$

Evaluate the indicated limits by direct evaluation as in Examples \(10-14 .\) Change the form of the function where necessary. $$\lim _{x \rightarrow 8} \frac{x-8}{\sqrt[3]{x}-2}$$

\(\lim _{x \rightarrow a^{-}} f(x)\) means to find the limit as x approaches a from the left only, and \(\lim _{x \rightarrow a^{+}} f(x)\) means to find the limit as \(x\) approaches a from the right only. These are called one-sided limits. Solve the following problems. In Einstein's theory of relativity, the length \(L\) of an object moving at a velocity \(v\) is \(L=L_{0} \sqrt{1-\frac{v^{2}}{c^{2}}},\) where \(c\) is the speed of light and \(L_{0}\) is the length of the object at rest. Find lim \(L\) and explain why a limit from the left is used.
See all solutions

Recommended explanations on Math Textbooks

Geometry

Read Explanation

Theoretical and Mathematical Physics

Read Explanation

Pure Maths

Read Explanation

Probability and Statistics

Read Explanation

Mechanics Maths

Read Explanation

Discrete Mathematics

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.