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Another Deriv

by helpm3pl3ase
Tags: deriv
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helpm3pl3ase
#1
Nov1-06, 02:22 PM
P: 79
(sec4x + 4arctanx^2)=

(sec4x)(tan4x) + (4)(1/1+x^4)(2x).. Did I derive this correctly??
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Office_Shredder
#2
Nov1-06, 02:33 PM
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Did you use the chain rule on the sec(4x) term?
courtrigrad
#3
Nov1-06, 02:45 PM
P: 1,236
you forgot to multiply by [tex] du = 4 [/tex] in the first term

helpm3pl3ase
#4
Nov1-06, 03:04 PM
P: 79
Another Deriv

(sec4x)(4)(tan4x)(4) + (4)(1/1+x^4)(2x)

or

16(sec4x)(tan4x) + (4)(1/1+x^4)(2x)??
helpm3pl3ase
#5
Nov1-06, 03:04 PM
P: 79
or do i keep just 1 4?? like this..

(sec4x)(tan4x)(4) + (4)(1/1+x^4)(2x)
Office_Shredder
#6
Nov1-06, 04:19 PM
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It would only be one four. Let sec(4x)=sec(u).

Then d(sec4x)/dx = d(secu)/dx = secu*tanu*du/dx

du/dx = 4, so

sec(4x)' = 4sec(4x)tan(4x)


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