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Math
CalculusSingle Variable Calculus: Early Transcendentals, Volume I8th Edition

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Chapter 3.5, Problem 61E
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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 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
Problem 65E
Problem 66E
Problem 67E
Problem 68E
Problem 69E
Problem 70E
Problem 71E
Problem 73E
Problem 74E
Problem 75E
Problem 76E
Problem 77E
Problem 78E
Problem 79E
Problem 80E
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Single Variable Calculus: Early Tr...

8th Edition
James Stewart
ISBN: 9781305270343
Textbook Problem

Find f′(x). Check that your answer is reasonable by comparing the graphs of f and f′.

61. f ( x ) = 1 − x 2 arcsin x

To determine

To find: The derivative of the function to check the derivative f(x), f′(x) is reasonable by comparing the graph of f(x) and f′(x).

Explanation

Given:

The function is f(x)=1−x2arcsinx.

Derivative rules:

(1) Chain rule: dydx=dydu⋅dudx

(2) Derivative of the inverse trigonometric function: ddx(tan−1(x))=11+x2.

(3) Quotient rule: ddx(fg)=gddx(f)−fddx(g)g2

Calculation:

Obtain the derivative of the function.

Consider the function f(x)=1−x2arcsinx

Differentiate the function implicitly with respect to t,

f′(x)=ddx(1−x2arcsinx)

Apply the product rule (3) and simplify the problem,

f′(x)=1−x2ddx(arcsinx)+arcsinxddx(1−x2)=1−x2(11−x2)+arcsinxddx(1−x2)=1+arcsinxddx(1−x2)

Let v=1−x2.

f′(x)=1+arcsinxddx(v)

Apply the chain rule (1) and simplify the terms,

f′(x)=1+arcsinxddv(v)dvdx=1+arcsinx[12v−12]dvdx

Substitute v=1−x2,

f′(x)=1+arcsinx[12(1

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