Question

18th one pls
Differentiate wrt x
Pls urgent.

Answer 1

XViii) differentiation 

given f(x) = (sec x + tan x) / (sec x - tan x)

Multiply by (sec x + tan x) on Nr and Dr.

$f(x)=\frac{(sec \: x+ tan \:x)^2}{sec^2x-tan^2x}= sec^2x +tan^2x + 2 \: Sec \: x \: tan \: x\\\\=1 + 2 tan^2 x+2 sec \: x \: tan \: x$

So differentiate now wrt x. (apply chain rule).

f '(x) = (1)' + 2 tan x * (tan x)' + 2 * Sec x * (tan x)' + 2 (sec x)' * tan x 
         = 2 tan x * sec² x + 2 sec x * (sec²x) + 2 (sec x tan x) * tan x
        =  2 * sec² x [ tan x + 2 sec x  ] - 2 sec x 

Answer 2

Here is your answer✌️✌️

f(X)=y

$y=sex+ tanx /secx-tanx$

multiply by secx +tanx on NR and Dr-

$(secx+tanx)^2/(sec^2x-tan^2x)$

$sec^2x+tan^2x+2secx.tanx$

$=1+2tan^2x+2secx.tanx$

d.w.r.t.x

$=2tanx*sec^2x+2secx*sec^2x+$

$2(secx.tanx)*tanx$

$=2sec^2x(tanx+2secx)-2secx$