Find the derivative of y = arctan(1 + e²x). О 1 yl= 1+(1+e²)² О 1+e2x yl= ○ yl= 1+(1+e²x)² e2x 2 1+(1+e²*)² 2 О yl= 2e2x 2x 2 ,

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 the derivative of y = arctan(1 + e²x).
О
1
yl=
1+(1+e²)²
О
1+e2x
yl=
○ yl=
1+(1+e²x)²
e2x
2
1+(1+e²*)²
2
О
yl=
2e2x
2x 2
,
Transcribed Image Text:Find the derivative of y = arctan(1 + e²x). О 1 yl= 1+(1+e²)² О 1+e2x yl= ○ yl= 1+(1+e²x)² e2x 2 1+(1+e²*)² 2 О yl= 2e2x 2x 2 ,
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