   Chapter 2.9, Problem 39E

Chapter
Section
Textbook Problem

# Establish the following rules for working with differentials (where c denotes a constant and u and v are functions of x).(a) d c = 0 (b) d ( c u ) = c d u (c) d ( u + v ) = d u + d v (d) d ( u v ) = u d v + v d u (e) ( u v ) = v d u − u d v v 2 (f) d ( x n ) = n x n − 1 d x

To determine

To establish:

The given rules using differentials.

Explanation

Concept:

If a function y = f(x) then dy and dx are differentials and relation between them is,

dy = f’(x) dx

Calculation:

We shall verify each of the given statements.

(a)

dc=0

By using differentials,

dc= ddxc*dx =0……….since c denotes a constant

Thus, dc=0

(b)

d cu=c du

By using differentials,

d cu=  ddxcu*dx

By using constant multiple rule,

d cu=cddxu*dx = c  du

Thus, d cu=c  du

(c)

d u+v=du+dv

By using differentials,

d u+v =ddxu+v dx

By using sum rule,

d u+v= ddxu dx+ddxvdx= du+dv

Thus, d u+v =du+dv

(d)

d uv=u dv+v du

By using differentials,

d uv=ddxuv dx

By using product rule,

d uv=uddx v+vddxu dx=uddx<

### Still sussing out bartleby?

Check out a sample textbook solution.

See a sample solution

#### The Solution to Your Study Problems

Bartleby provides explanations to thousands of textbook problems written by our experts, many with advanced degrees!

Get Started

#### Calculate y'. 30. y=(x2+1)4(2x+1)3(3x1)5

Single Variable Calculus: Early Transcendentals, Volume I

#### In Exercises 49-62, find the indicated limit, if it exists. 62. limx24x22x2+x3

Applied Calculus for the Managerial, Life, and Social Sciences: A Brief Approach

#### Solve the differential equation. 1. dydx=3x2y2

Single Variable Calculus: Early Transcendentals

#### 2 1 0 does not exist

Study Guide for Stewart's Single Variable Calculus: Early Transcendentals, 8th

#### Which curve is simple but not closed?

Study Guide for Stewart's Multivariable Calculus, 8th 