Problem 13 $11-14$ Find the differential ... [FREE SOLUTION]

Chapter 3: Problem 13

$11-14$ Find the differential of each function. (a) $$ y=\frac{u+1}{u-1} \quad \text { (b) } y=\left(1+r^{3}\right)^{-2} $$

Short Answer

Expert verified

(a) $ dy = \frac{-2}{(u-1)^2} du $; (b) $ dy = -6r^2(1 + r^3)^{-3} dr $.

Step by step solution

Differentiate the First Function (a)

The given function is $ y = \frac{u+1}{u-1} $. Use the quotient rule for differentiation, which states that if $ y = \frac{f(u)}{g(u)} $, then $ \frac{dy}{du} = \frac{f'(u)g(u) - f(u)g'(u)}{[g(u)]^2} $. Here, $ f(u) = u+1 $ and $ g(u) = u-1 $.First, compute the derivatives: $ f'(u) = 1 $ and $ g'(u) = 1 $.Apply the quotient rule: \[\frac{dy}{du} = \frac{(1)(u-1) - (u+1)(1)}{(u-1)^2} = \frac{u - 1 - u - 1}{(u-1)^2} = \frac{-2}{(u-1)^2}\].Thus, the differential is $ dy = \frac{-2}{(u-1)^2} du $.

Differentiate the Second Function (b)

The given function is $ y = (1 + r^3)^{-2} $. Use the chain rule for differentiation, which states that if $ y = [u(r)]^n $, then $ \frac{dy}{dr} = nu'(r)u(r)^{n-1} $ where $ u(r) = 1 + r^3 $ and $ n = -2$.First, compute $ u'(r) $: $ u'(r) = 3r^2 $.Apply the chain rule: \[\frac{dy}{dr} = (-2)(3r^2)(1 + r^3)^{-3} = -6r^2(1 + r^3)^{-3}\].Thus, the differential is $ dy = -6r^2(1 + r^3)^{-3} dr $.

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.

Understanding the Quotient Rule

The quotient rule is a handy tool in differential calculus for finding derivatives of functions that are ratios of two other functions. Specifically, if you have a function given by the ratio of two functions such as $ y = \frac{f(u)}{g(u)} $, the quotient rule helps you find its derivative easily.
To apply the quotient rule:

Identify the functions in the numerator $ f(u) $ and denominator $ g(u) $.
Differentiating both: Find $ f'(u) $ and $ g'(u) $.
Apply the rule: $ \frac{dy}{du} = \frac{f'(u)g(u) - f(u)g'(u)}{[g(u)]^2} $.

This formula helps manage complex derivatives by breaking them down into simpler parts. For example, in the solution for $ y = \frac{u+1}{u-1} $, by using the quotient rule, we found $ \frac{dy}{du} $ to be $ \frac{-2}{(u-1)^2} $, simplifying a potentially tricky computation.

Exploring the Chain Rule

The chain rule allows you to differentiate compositions of functions. Imagine a function composed of two functions, like $ y = [u(r)]^n $. You might see this in expressions like powers of inner functions, where the inner function ($ u(r) $ in this case) plays a crucial role.
To employ the chain rule:

Differentiate the outer function first, treating the inner function as a constant.
Then, multiply by the derivative of the inner function.

This approach is vital for functions nested within one another, like in the example $ y = (1 + r^3)^{-2} $. By differentiating the outer and inner functions separately and multiplying, we find the derivative $ \frac{dy}{dr} = -6r^2(1 + r^3)^{-3} $. This technique simplifies finding derivatives that might seem challenging at first glance.

Basics of Differential Calculus

Differential calculus focuses on how functions change. It is a foundational element of mathematics for analyzing variable rates and understanding phenomena ranging from physics to economics.
Key aspects include:

Derivatives: Which represent the slope of the function at any point.
Rules: Such as the product, quotient, and chain, to ease computation.

The core purpose of differential calculus is to help us determine rates of change and understand how functions behave. It provides the tools鈥攍ike derivatives鈥攖o solve complex problems by breaking functions into manageable parts through various rules and techniques like the quotient and chain rules.

Embracing Derivatives

Understanding derivatives is crucial as they measure how a function changes as its input changes, providing insight into the function's behavior.
Derivatives represent:

The function's slope or rate of change at any point, crucial for understanding trends.
Changing dynamics, useful for fields like physics, where they describe velocity, acceleration, etc.

Derivations find their application across numerous disciplines. Whether it's optimizing a function or analyzing physical change, derivatives are fundamental. In our exercise, derivatives help us understand changes in functions represented by fractions and compositions. Utilizing rules like the quotient and chain rules simplifies the process of finding these derivatives, highlighting their broad utility in solving real-world problems.

One App. One Place for Learning.

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

Get started for free

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

91影视

\(11-14\) Find the differential of each function. (a) $$ y=\frac{u+1}{u-1} \quad \text { (b) } y=\left(1+r^{3}\right)^{-2} $$

Short Answer

Step by step solution

Differentiate the First Function (a)

Differentiate the Second Function (b)

Key Concepts

Understanding the Quotient Rule

Exploring the Chain Rule

Basics of Differential Calculus

Embracing Derivatives

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Math Textbooks

Decision Maths

Geometry

Statistics

Pure Maths

Discrete Mathematics

Logic and Functions

Study anywhere. Anytime. Across all devices.