Another Deriv

helpm3pl3ase

(sec4x + 4arctanx^2)=

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

Office_Shredder

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

courtrigrad

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

helpm3pl3ase

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

or

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

helpm3pl3ase

or do i keep just 1 4?? like this..

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

Office_Shredder

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)