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Math
CalculusSingle Variable Calculus: Early Transcendentals, Volume I8th Edition

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Chapter 3.2, Problem 2E
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Single Variable Calculus: Early Tr...

8th Edition
James Stewart
ISBN: 9781305270343

Solutions

Chapter
Section
Problem 1E
Problem 2E
Problem 3E
Problem 4E
Problem 5E
Problem 6E
Problem 7E
Problem 8E
Problem 9E
Problem 10E
Problem 11E
Problem 12E
Problem 13E
Problem 14E
Problem 15E
Problem 16E
Problem 17E
Problem 18E
Problem 19E
Problem 20E
Problem 21E
Problem 22E
Problem 23E
Problem 24E
Problem 25E
Problem 26E
Problem 27E
Problem 28E
Problem 29E
Problem 30E
Problem 31E
Problem 32E
Problem 33E
Problem 34E
Problem 35E
Problem 36E
Problem 37E
Problem 38E
Problem 39E
Problem 40E
Problem 41E
Problem 42E
Problem 43E
Problem 44E
Problem 45E
Problem 46E
Problem 47E
Problem 48E
Problem 49E
Problem 50E
Problem 51E
Problem 52E
Problem 53E
Problem 54E
Problem 55E
Problem 56E
Problem 57E
Problem 58E
Problem 59E
Problem 60E
Problem 61E
Problem 62E
Problem 63E
Problem 64E
BuyFindarrow_forward

Single Variable Calculus: Early Tr...

8th Edition
James Stewart
ISBN: 9781305270343
Textbook Problem

Find the derivative o f the function

F ( x ) = x 4 − 5 x 3 + x x 2

in two ways: by using the Quotient Rule and by simplifying first. Show that your answers are equivalent. Which method do you prefer?

To determine

To find: The derivative of the function F(x)=x4−5x3+xx2.

Explanation

Given:

The function F(x)=x4−5x3+xx2.

Derivative rule:

(1) Quotient Rule: If f1(x) and f2(x) are both differentiable, then

ddx[f1(x)f2(x)]=f2(x)ddx[f1(x)]−f1(x)ddx[f2(x)][f2(x)]2

(2) Power Rule: ddx(xn)=nxn−1

(3). Sum rule: ddx(f+g)=ddx(f)+ddx(g)

(4) Constant multiple rule: ddx(cf)=cddx(f)

(5) Difference rule: ddx(f−g)=ddx(f)−ddx(g)

Calculation:

Method 1:

Obtain the derivative of F(x) by using Quotient Rule.

The derivative of the function F(x) is F′(x), which is obtained as follows.

F′(x)=ddx(F(x))=ddx(x4−5x3+xx2)

Substitute x4−5x3+x for f1(x) and x2 for f2(x) in the quotient rule (1),

F′(x)=(x2)ddx(x4−5x3+x)−(x4−5x3+x)ddx(x2)(x2)2

Apply the derivative rules (3), (4), and (5),

F′(x)=(x2)[ddx(x4)−ddx(5x3)+ddx(x12)]−[(x4−5x3+x)ddx(x2)](x2)2=x2[ddx(x4)−5ddx(x3)+ddx(x12)]−[(x4−5x3+x12)ddx(x2)]x4

Apply the power rule (2) and simplify the terms,

F′(x)=x2[4x3−5(3x3−1)+(12x12−1)]−(x4−</

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