Derivative: sec4x tan4x + 8x/(1+x^4)

[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 $$du = 4$$ 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)