Let h(x) = tan(2x + 5). h'(x) h"(x) = = 2 sec (2x+5)(2) OF ४

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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Now find the derivative of the given function.

h'x=dhxdx=ddxtan2x+5=sec22x+5·ddx2x+5  Apply the Chain rule=sec22x+5ddx2x+ddx5=sec22x+52·ddxx+5·ddx1=sec22x+521x1-1+50  Apply the Power law=sec22x+521+0=sec22x+52=2sec22x+5

Now find the second derivative of the given function.

h''x=d2hxdx2=ddxdhxdx=ddx2sec22x+5=2·ddxsec22x+5=2ddxsec2x+52=22sec2x+52-1·ddxsec2x+5   Apply the Chain rule=22sec2x+51sec2x+5tan2x+5·ddx2x+5   Again apply the Chain rule=22sec22x+5tan2x+5ddx2x+ddx5=22sec22x+5tan2x+52ddxx+5ddx1=22sec22x+5tan2x+521x1-1+50  Apply the Power law=22sec22x+5tan2x+52=8sec22x+5tan2x+5

Therefore,

h'x=2sec2(2x+5)h''x=8sec2(2x+5)tan(2x+5)

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Let h(x) = tan(2x + 5).
h'(x)
h"(x)
=
=
2
sec (2x+5)(2)
OF
४
Transcribed Image Text:Let h(x) = tan(2x + 5). h'(x) h"(x) = = 2 sec (2x+5)(2) OF ४
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