1. second derivitive ..HELP!

f(x)=3ln[sec(x)+tan(x)]
f(x)=

I got (((3sec(x))(tan(x))^2)-(3(sec(x))^2)-(2(sec(x))^2)(tan(x))(sec(x)tan(x))+(-3(sec(x)tan(x)+(sec(x))^2)(-sec(x)tan(x)+(sec(x))^2)))/((sec(x)+tan(x))^2)

2. Originally Posted by lmao
f(x)=3ln[sec(x)+tan(x)]
f(x)=

I got (((3sec(x))(tan(x))^2)-(3(sec(x))^2)-(2(sec(x))^2)(tan(x))(sec(x)tan(x))+(-3(sec(x)tan(x)+(sec(x))^2)(-sec(x)tan(x)+(sec(x))^2)))/((sec(x)+tan(x))^2)

you're making your life WWWAAAYYYY too complicated. simplify intermediate steps before moving on

well, by definition, the derivative of ln|sec(x) + tan(x)| = sec(x), but let's pretend we didn't know that.

we have to use the chain rule:

f(x) = 3ln|sec(x) + tan(x)|
=> f ' (x) = 3[1/(sec(x) + tan(x)) * (sec(x)tan(x) + sec^2(x))]
.............= 3[{sec(x)tan(x) + sec^2(x)}/{sec(x) + tan(x)}]
.............= 3[sec(x){tan(x) + sec(x)}/{sec(x) + tan(x)}]
.............= 3sec(x)

=> f '' (x) = 3sec(x)tan(x)

3. Originally Posted by lmao
f(x)=3ln[sec(x)+tan(x)]
f(x)=

I got (((3sec(x))(tan(x))^2)-(3(sec(x))^2)-(2(sec(x))^2)(tan(x))(sec(x)tan(x))+(-3(sec(x)tan(x)+(sec(x))^2)(-sec(x)tan(x)+(sec(x))^2)))/((sec(x)+tan(x))^2)

Well, let's make your life a bit easier:

f(x) = 3ln[secx+tanx]
f'(x) = 3*(secxtanx + sec^2x)/(secx + tanx)

We can simplify the first derivative.
Factor out secx from the numerator:

f'(x) = 3secx*(tanx + secx)/(secx + tanx)
f'(x) = 3sec(x)