Question

Find the derivative of the following function:

2tan x – 7sec x

Answer 1

Let f (x) = 2 tan x – 7 sec x Accordingly, from the first principle,

f'(x) = `lim_(h->0) (f(x + h) - f(x))/h`

= `lim_(h->0)1/h [2tan (x + h) - 7 sec (x + h) -2 tan x + 7sec x]`

= `2 lim_(h->0)1/h[2{tan (x + h) - tan x} - 7{sec (x + h) -sec x}]`

= `2 lim_(h->0)1/h[tan (x + h) - tan x] - 7lim_(h->0)1/h[sec (x + h) -sec x]`

= `2 lim_(h->0) 1/h [sin (x + h)/(cos (x + h))-(sinx)/(cos x)] -7 lim_(h->0)1/h [1/(cos (x + h)) - 1/(cos x)]`

= `2 lim_(h->0)1/h [(sin(x + h) cos x - sin x cos (x + h))/(cos x cos (x + h))] -7lim_(h->0)1/h[(cos x - cos (x + h))/(cos x cos (x + h))]`

= `2 lim_(h->0) [(sin (x + h - x))/(cos x cos (x + h))] -7 lim_(h->0)[(-2 sin ((x + x + h)/2) sin ((x - x - h)/(2)))/(cos x cos (x + h))]`

= `2 lim_(h->0) [((sin h)/h) 1/(cos x cos (x +h))] - 7 lim_(h->0)1/h[(-2 sin ((2x + h)/h) sin (-h/2))/(cos x cos (x + h))]`

= `2 (lim_(h->0)(sin h)/h) (lim_(h->0) 1/(cos x cos (x + h)))-7 (lim_(h->0) (sin  h/2)/(h/2)) (lim_(h ->0) (sin ((2x + h)/2))/(cos x cos (x + h)))`

= 2.1 `1/(cos x cos x) - 7.1 (sin x/(cos x cos x))`

= 2 sec² x - 7 sec x tan x