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TOPIC 1 Higher Derivatives 1-x 1 + X Q. #1 Given f(x)=- , please find derivatives below. Hint: Always simplify between derivatives ! a) f(x) = G=0 f23=(1-x)(x+1)-(1-x)(1+ (1+x) fl= -(x+1)-(1-x)(1) (1+x)³ b) f(x)= x()](1+x)²-(+) (-3) [(1 + x)²³) ³ c) f"(x) = 0 - (1+x) (2) (- (1 + x)² = 4 (1+x) (1+x) Final Answer -2 f'(x = (1+x²² Ofw=-x-1-1+X (1+x) P'(x) = (1+x) Final Answer Final Answer

TOPIC 1 Higher Derivatives 1-x 1 + X Q. #1 Given f(x)=- , please find derivatives below. Hint: Always simplify between derivatives ! a) f(x) = G=0 f23=(1-x)(x+1)-(1-x)(1+ (1+x) fl= -(x+1)-(1-x)(1) (1+x)³ b) f(x)= x()](1+x)²-(+) (-3) [(1 + x)²³) ³ c) f"(x) = 0 - (1+x) (2) (- (1 + x)² = 4 (1+x) (1+x) Final Answer -2 f'(x = (1+x²² Ofw=-x-1-1+X (1+x) P'(x) = (1+x) Final Answer Final Answer

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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TOPIC 1 Higher Derivatives 1-.x Q. #1 Given f(x)= , please find derivatives below. Hint: Always simplify between derivatives ! 1 + X a) f(x) = G=0 f23=(1-x)(x+1)-(1-x)(1+x) (1+x) f² = -(x+1)-(1-x)(1) tr b) ƒ"(x) = x (-)](1+x)ª- (+ (-3) [(1 + x)²³) ³ c) f"(x) 2 = 0-(1+x) (2) (-2 (1 + x)² = 4 (1+x) (1+x)" Final Answer -2 f'(x) = (²+²x³²² Ofw=-x-1-1+X (1+x) P'(x) = (1+x) Final Answer Final Answer
Transcribed Image Text:TOPIC 1 Higher Derivatives 1-.x Q. #1 Given f(x)= , please find derivatives below. Hint: Always simplify between derivatives ! 1 + X a) f(x) = G=0 f23=(1-x)(x+1)-(1-x)(1+x) (1+x) f² = -(x+1)-(1-x)(1) tr b) ƒ"(x) = x (-)](1+x)ª- (+ (-3) [(1 + x)²³) ³ c) f"(x) 2 = 0-(1+x) (2) (-2 (1 + x)² = 4 (1+x) (1+x)" Final Answer -2 f'(x) = (²+²x³²² Ofw=-x-1-1+X (1+x) P'(x) = (1+x) Final Answer Final Answer
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