102 PRACTICAL MATHEMATICS 



As a further application of the use of this method, let us find 

 the first four differential coefficients of e? sec x. 



To differentiate e* sec x successively, we require to be able to 

 differentiate e* sec x tan n x. 



y = c? sec x tan" x 

 log y = x + log sec x + n log tan x 

 1 dy sec x tan x sec 2 x 



- -r- = 1 + h n 



?/ cte sec a; tan x 



= {tan x + tan 2 x + n(l + tan 2 x) } 



tan x l 



-^ = e* sec a? tan"" 1 x{n + tan + (n + 1) tan 2 a?} 

 dzr 



= e* sec #{n tan"- 1 x + tan" a; + (n + 1) tan w+1 a;} 



du 



when w = 0. w = e* sec a;. -?- = er sec a? (1 + tan a?) 



cte 



when w = 1, y = 6 s sec a; tan a;, "T" = ^ sec ^ (1 + tsJ 1 ^ 



+ 2 tan 2 a;) 



y7^y 



when n =2, v = & sec a? tan 2 , TT ^^ sec x (2 tan a? + tan 2 x 



ax 



when w = 3, i/ = e* sec a) tan 3 a;, ~ = e 6 ?>&c x (3 tan 2 # + tan 3 a; 



+ 3 tan 3 x) 

 <? sec a; (3 1 

 + 4 tan 4 a;) 



Thus if y = e* sec a? 



flu 



j^ = e* sec a? (1 + tan x) 



dhi 



-r^ = (? sec {(! + tan as) + (1 + tan x + 2 tan 2 #) } 



= e* sec a? (2 + 2 tan a; + 2 tan 2 a;) 



d 3 w 



-T-| = * sec a; (2(1 + tan or) + 2(1 + tan x + 2 tan 2 a;) 



+ 2(2 tan x + tan 2 a? + 3 tan 3 a;) } 



= # sec x (4 + 8 tan x + 6 tan 2 # + 6 tan 3 a?) 



d*u 



-T-| = c* sec a; (4(1 + tan a?) + 8(1 + tan a? + 2 tan 2 a?) 



+ 6(2 tan a? + tan 2 a? + 3 tan 3 a;) + 6 (3 tan 2 x + tan 3 x 

 + 4 tan 4 a?) } 



= * sec a? (12 + 24 tan a; + 40 tan 2 x + 24 tan 3 x + 24 tan 4 #} 



