MACLAURIN'S EXPANSION 109 



66. We can also apply Maclaurin's Theorem to the case in 

 which the process of finding the successive differential coefficients 

 is rendered simple by working to a general rule. Functions 

 such as tan x, cot x, sec x, cosec x, can be expanded in this manner. 



Thus, to find the expansion for sec x, we have to find the suc- 

 cessive differential coefficients of sec x, and to do this we shall 

 require the differential coefficients of sec x tan n x. 



y = sec x tan" x 



= n sec x tan n-1 x sec 2 x + tan n x sec x tan x 

 dx 



- sec x{n tan"- 1 x(l + tan 2 x) + tan n+1 x} 

 = sec x{n tan"- 1 x + (n + 1) tan n+1 a;} 



dt/ 



Thus when n = 0, y = sec x and -2- = sec x tan x 



ux 



di/ 

 when n = 1, y = sec x tan x and -p = sec a;{l + 2 tan 2 x} 



when n = 2, y - sec x tan 2 x and -j- = sec x {2 tan x + 3 tan 3 # } 



when n= 3, y = sec x tan 3 x and -p = sec x {3 tan 2 a; +4 tan 4 a;} 



Using these results we can now successively differentiate sec x. 

 y = sec x 



du 



-r- = sec x tan x 

 dx 



= sec a:{l + 2 tan 2 a?} 

 dx 



-r4 = sec a; (tan x + 2(2 tan x + 3 tan 3 a;)} 

 dx 3 



= sec #{5 tan x + 6 tan 3 x} 

 -3^ = sec x {5(1 + 2 tan 2 a?) + 6(3 tan 2 x + 4 tan 4 x) } 



= sec a; (5 + 28 tan 2 x + 24 tan 4 a;} 

 -T-| = sec x {5 tan x + 28(2 tana; + 3 tan 3 x) + 24(4 tan 3 a; + 5 tan 5 a:)} 



= sec x (61 tan x + 180 tan 3 x + 120 tan 5 x} 

 -r^ + sec a? {61(1 + 2 tan 2 a?) + 180(3 tan 2 x + 4 tan 4 x) + 120(5 tan 4 x 

 + 6 tan 6 ar) } 



= sec a; {61 + 662 tan 2 x + 1320 tan 4 x + 720 tan 6 a;} 



