Find a general expression for the derivative of f(x)= „ by using the limit definition of the derivative (do not use quotient/product rule if you know it) Remark: the derivative of g(x) "his is because g(æ) = f(x) – 1 and using our sum and constant rule we find g' (x) = f (x) – 1)' = f'(æ) – 1' = f'(x) – 0 = f'(x). The derivative of g(x) also is a lot easie o calculate! This is a cool application of using the fact that the derivative of a constant is O! is the same as the derivative that you just found. x+1 | |

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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Find a general expression for the derivative of f(x)
by using the limit definition of the
x+1
derivative (do not use quotient/product rule if you know it)
1
is the same as the derivative that you just found.
Remark: the derivative of g(x) :
This is because g(x) = f(x) – 1 and using our sum and constant rule we find g' (x) =
(f(x) – 1)' = f'(x) – 1' = f'(x) – 0 = f'(x). The derivative of g(x) also is a lot easier
%3D
x+1
to calculate! This is a cool application of using the fact that the derivative of a constant is O!
Transcribed Image Text:Find a general expression for the derivative of f(x) by using the limit definition of the x+1 derivative (do not use quotient/product rule if you know it) 1 is the same as the derivative that you just found. Remark: the derivative of g(x) : This is because g(x) = f(x) – 1 and using our sum and constant rule we find g' (x) = (f(x) – 1)' = f'(x) – 1' = f'(x) – 0 = f'(x). The derivative of g(x) also is a lot easier %3D x+1 to calculate! This is a cool application of using the fact that the derivative of a constant is O!
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