This one's not easy... help me out!!!

Here's the first one.

First some derivatives

d secx / dx= sec x tan x
d tanx / dx = (sec x)^2
d f(x)^n = n f(x)^(n-1) df/dx

Now to your equation: f (x) = (sec x + tan x ) ^n , n E R, n is greater than zero

f'(x) = n (sec x + tan x)^(n-1) d(sec x + tan x)/dx
= n (sec x + tan x)^(n-1) ( sec x tan x + (sec x)^2)
= n (sec x + tan x)^(n-1) (sec x)(tan x + sec x)
= n (sec x + tan x)^n (sec x)
= n f(x) (sec x)

Hope that gets you started.

Second part:

Reminder: d f(x).g(x) = g(x).df(x) + f(x).dg(x)

f'' = {n f(x) (sec x)}'
= n{ df(x) sec x + f(x) d(sec x) }

We calculated df(x) above and we have d sec(x) from the link I posted.

= n { n f(x) (sec x) (sec x) + f(x) (sec x tan x)}

Factoring out f(x) sec x...

= n f(x) sec x (n sec x + tan x)

but n f(x) sec x = f'(x) so...

= f'(x) (n sec x + tan x)

Thanks, engineer, May God bless you in this life and the hereafter...