   Chapter 3.5, Problem 64E ### Single Variable Calculus: Early Tr...

8th Edition
James Stewart
ISBN: 9781305270343

#### Solutions

Chapter
Section ### Single Variable Calculus: Early Tr...

8th Edition
James Stewart
ISBN: 9781305270343
Textbook Problem

# (a) One way of defining sec–1x is to say that y = sec − 1 x ⇔ sec y = x   and   0 ≤ y < π / 2 or   π ≤ y < 3 π / 2 Show that, with this definition, d d x ( sec − 1 x ) = 1 x 2 − 1 x (b) Another way of defining sec–1x that is sometimes used is to say that y = sec − 1 x ⇔ sec y = x and 0 ≤ y ≤π, y ≠ π/2. Show that, with this definition, d d x ( sec − 1 x ) = 1 | x | x 2 − 1

(a)

To determine

To show: The formula ddx(sec1(x))=1xx21 when 0yπ2.

Explanation

Given:

The function is y=sec1(x).

Derivative rules:

Chain rule: dydx=dydududx

Proof:

The given function can be expressed as follows,

y=sec1(x)secy=x

Differentiate the function implicitly with respect to x,

ddx(secy)=ddx(x)

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

ddy(secy)dydx=ddx(x)secytanydydx=1dydx=1secytany

Squaring on both sides,

[dydx]2=1sec2ytan2y=1sec2ytan2y        [Qtan2y=sec2y1]=1sec2y(sec2y1)

Take square root on both sides,

[dydx]2=1sec2y(sec2y1)[dydx]2=1sec2y(

(b)

To determine

To show: The formula ddx(sec1(x))=1|x|x21 when 0yπ and yπ2

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