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By applying the Product Rule twice, one can prove that if f, g, and h are differentiable, then (fgh)' = f'gh+fg'h+fgh'. Now, in the above result, letting f = g = h yields d -[f(x)]³ = 3[f(x)]²ƒ'(x). dx Use this last formula to differentiate y = e³¸ y' = B By applying the Product Rule twice, one can prove that if f, g, and h are differentiable, then (fgh)' = f'gh+fg'h+fgh'. Now, in the above result, letting f = g = h yields d -[f(x)]³ = 3[f(x)]²ƒ'(x). dx Use this last formula to differentiate y = e³¸ y' = B

By applying the Product Rule twice, one can prove that if f, g, and h are differentiable, then (fgh)' = f'gh+fg'h+fgh'. Now, in the above result, letting f = g = h yields d -[f(x)]³ = 3[f(x)]²ƒ'(x). dx Use this last formula to differentiate y = e³¸ y' = B By applying the Product Rule twice, one can prove that if f, g, and h are differentiable, then (fgh)' = f'gh+fg'h+fgh'. Now, in the above result, letting f = g = h yields d -[f(x)]³ = 3[f(x)]²ƒ'(x). dx Use this last formula to differentiate y = e³¸ y' = B

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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By applying the Product Rule twice, one can prove that if f, g, and h are differentiable, then (fgh)' = f'gh+fg'h+fgh'. Now, in the above result, letting f = g = h yields d -[f(x)]³ = 3[f(x)]²ƒ'(x). dx Use this last formula to differentiate y = e³¸ y' = B
Transcribed Image Text:By applying the Product Rule twice, one can prove that if f, g, and h are differentiable, then (fgh)' = f'gh+fg'h+fgh'. Now, in the above result, letting f = g = h yields d -[f(x)]³ = 3[f(x)]²ƒ'(x). dx Use this last formula to differentiate y = e³¸ y' = B
By applying the Product Rule twice, one can prove that if f, g, and h are differentiable, then (fgh)' = f'gh+fg'h+fgh'. Now, in the above result, letting f = g = h yields d -[f(x)]³ = 3[f(x)]²ƒ'(x). dx Use this last formula to differentiate y = e³¸ y' = B
Transcribed Image Text:By applying the Product Rule twice, one can prove that if f, g, and h are differentiable, then (fgh)' = f'gh+fg'h+fgh'. Now, in the above result, letting f = g = h yields d -[f(x)]³ = 3[f(x)]²ƒ'(x). dx Use this last formula to differentiate y = e³¸ y' = B
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