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Two derivations of the identity sec¬1(-x) = # – sec-lx a. (Geometric) Here is a pictorial proof that sec-'(-x) = T - sec-lx. See if you can tell what is going on. y = sec-lx х b. (Algebraic) Derive the identity sec¬'(-x) = T – sec-lx by combining the following two equations from the text: cos(-x) = T – cosx Eq. (4), Section 1.6 sec-lx = cos'(1/x) Eq. (1)

Two derivations of the identity sec¬1(-x) = # – sec-lx a. (Geometric) Here is a pictorial proof that sec-'(-x) = T - sec-lx. See if you can tell what is going on. y = sec-lx х b. (Algebraic) Derive the identity sec¬'(-x) = T – sec-lx by combining the following two equations from the text: cos(-x) = T – cosx Eq. (4), Section 1.6 sec-lx = cos'(1/x) Eq. (1)

Calculus: Early Transcendentals
Calculus: Early Transcendentals
8th Edition
ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
Section: Chapter Questions
Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
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Two derivations of the identity sec¬1(-x) = # – sec-lx a. (Geometric) Here is a pictorial proof that sec-'(-x) = T - sec-lx. See if you can tell what is going on. y = sec-lx х b. (Algebraic) Derive the identity sec¬'(-x) = T – sec-lx by combining the following two equations from the text: cos(-x) = T – cosx Eq. (4), Section 1.6 sec-lx = cos'(1/x) Eq. (1)
Transcribed Image Text:Two derivations of the identity sec¬1(-x) = # – sec-lx a. (Geometric) Here is a pictorial proof that sec-'(-x) = T - sec-lx. See if you can tell what is going on. y = sec-lx х b. (Algebraic) Derive the identity sec¬'(-x) = T – sec-lx by combining the following two equations from the text: cos(-x) = T – cosx Eq. (4), Section 1.6 sec-lx = cos'(1/x) Eq. (1)
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