Expand the function p (z + h) as follows. p (z + h) = V(w+h)²–1 Substitute the function p (z + h) in equation (1). (w+h)2 –1 p = limh¬0 h (w²_ /(w+h)² –1 limh¬0 = h/(w+h)²_1 (w² –1) (w²–1)-/(w+h)²–1)(/ (w²- (w+h)? – -1 = limh¬0 h/(w+h)²–1\/ (w²–1) (w²–1)-/(w+h)²–1 w²–1-(w²+2wh+h?–1) = lim½¬0 h/(w+h)²_1. (w²–1)( /(w²-1)-V(w+h)²=1 Simplify the Equation. w² –1-(w²+2wh+h²–1) z = limh¬0 h/(w+h²-1 /[w²-1)(/(w²-1)-/(w+h²-1 (w2_j = lim½¬0 -2w-h (w2 –1)( /(w2 –1 2-1)-/(w+h)²– (w²–1) 3/½ Thus, the derivative of the function vis (w²-1) ½
Expand the function p (z + h) as follows. p (z + h) = V(w+h)²–1 Substitute the function p (z + h) in equation (1). (w+h)2 –1 p = limh¬0 h (w²_ /(w+h)² –1 limh¬0 = h/(w+h)²_1 (w² –1) (w²–1)-/(w+h)²–1)(/ (w²- (w+h)? – -1 = limh¬0 h/(w+h)²–1\/ (w²–1) (w²–1)-/(w+h)²–1 w²–1-(w²+2wh+h?–1) = lim½¬0 h/(w+h)²_1. (w²–1)( /(w²-1)-V(w+h)²=1 Simplify the Equation. w² –1-(w²+2wh+h²–1) z = limh¬0 h/(w+h²-1 /[w²-1)(/(w²-1)-/(w+h²-1 (w2_j = lim½¬0 -2w-h (w2 –1)( /(w2 –1 2-1)-/(w+h)²– (w²–1) 3/½ Thus, the derivative of the function vis (w²-1) ½
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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