OK, can someone solve this diferencial: d/dx (Sec x Tan x - Ln | Sec x + Tan x |)?

Answer 1

d/dx(Sec x Tan x - ln |Sec x + Tan x|)
= tanx secx tanx + secx sec^2 x - {(secx tanx + sec^2 x)/(sec x + tan x)}
= tan^2 x sec x + sec^3 x - {(secx tanx + sec^2 x)/(sec x + tan x)}

Answer 2

First We remember that
Sec² x = tan² x + 1

we have F(x) = sec x.tan x - ln | sec x + tan x |
F' (x) = sec x .sec² x + tan x.sec x .tan x -
(1 *(secx.tanx + sec² x) / (sec x + tan x)

= sec³ x + sec x (sec² x -1) -
[ (sec x (tan x + sec x ) / ( sec x + tan x) ]

= sec ³ x + sec ³ x - sec x + sec x

= 2 sec ³ x
*******************
Notice that sec² x = sec^2 x & sec³ x = sec^3 x
I hope to have the ten points
Thanx in advance

Answer 3

i'm not really good at it.. but use this funda.. the whole func is just one func.. it can't be two bcoz of the minus sign in between.

use chain rule to apply the differentials:
secx > secx.tanx
tanx > secx*sqaure
ln > 1/secx+tanx (and then the differential of secx+tanx) again..

Hope u get ur answer.. this is what i got..
[secx.tanx.tan^x + secx.sex^x] - [1/(secx+tanx)][secx.tanx+sec^x]

Answer 4

it goes like this:
Secx.Sec^2 x + Tan^2 x.Secx - 1/(Secx+Tanx).(Secx.Tanx+Sec^2 x)

Answer 5

dude... that is EASY! go look up CALCULUS textbook:


product rule
substraction rule
d/dx of ln, sec, tan
chain rule

OK THAT IS IT!!!

Answer 6

(sec x)^3+(sec x)(tan x)^2-sec x