Another Deriv


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

(sec4x)(tan4x) + (4)(1/1+x^4)(2x).. Did I derive this correctly??
Phys.Org News Partner Science news on Phys.org
NASA's space station Robonaut finally getting legs
Free the seed: OSSI nurtures growing plants without patent barriers
Going nuts? Turkey looks to pistachios to heat new eco-city
Office_Shredder
Office_Shredder is offline
#2
Nov1-06, 02:33 PM
Mentor
P: 4,499
Did you use the chain rule on the sec(4x) term?
courtrigrad
courtrigrad is offline
#3
Nov1-06, 02:45 PM
P: 1,239
you forgot to multiply by [tex] du = 4 [/tex] in the first term

helpm3pl3ase
helpm3pl3ase is offline
#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
helpm3pl3ase is offline
#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
Office_Shredder is offline
#6
Nov1-06, 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)


Register to reply

Related Discussions
Calculating a deriv using logarithmic differentiation Calculus & Beyond Homework 1
Taylor series / 2nd deriv test Calculus & Beyond Homework 2
deriv of trig function Calculus & Beyond Homework 4
unsure on how to approach sin^4 Calculus 5
Deriv of xtanx Calculus 3