
#1
Nov106, 02:22 PM

P: 79

(sec4x + 4arctanx^2)=
(sec4x)(tan4x) + (4)(1/1+x^4)(2x).. Did I derive this correctly?? 



#2
Nov106, 02:33 PM

Mentor
P: 4,499

Did you use the chain rule on the sec(4x) term?




#3
Nov106, 02:45 PM

P: 1,239

you forgot to multiply by [tex] du = 4 [/tex] in the first term




#4
Nov106, 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)?? 



#5
Nov106, 03:04 PM

P: 79

or do i keep just 1 4?? like this..
(sec4x)(tan4x)(4) + (4)(1/1+x^4)(2x) 



#6
Nov106, 04:19 PM

Mentor
P: 4,499

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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