How can i find f''(x) of the function f(x)= 2(cosx)(sec^2x)

Answer 1

I would first simplify it.
f(x)= 2(cosx)(sec^2x)
= 2cosx(1/cos^2x) = 2/cosx = 2secx
f'(x) = 2tanxsecx

Answer 2

y = 2 cos x sec^2x

y' = 2[ cos x d/dx(sec^2x) + sec^2x (-sinx)]

= 2[ cos x{ sec x d/dx sec x +sec xd/dx sec x} -sin x sec^2 x]

= 2 [cos x {2 sec x d/dx sec x} -sin x sec^2 x ]
(but cos x sec x = 1)
= 2 [2 sec x tan x -sin x sec^2 x]

= 2 [2 sin x/(cos^2 x) -sin x/cos^2 x]

= 2 sin x/cos^2 x = 2 tan x sec x

Answer 3

LOL, well more fun to do it the hard way.

Answer 4

LOL, what else to do on a nice warm Sunday afternoon.

Damon, btw, can you take a peak at

http://www.jiskha.com/display.cgi?id=1270812113

Can you see a better way?

Answer 5

I do not see a better way.