Question

Find the derivative of the following function (it is to be understood that a, b, c, d, p, q, r and s are fixed non-zero constants and m and n are integers):

(x + sec x) (x – tan x)

Answer 1

Let f(x) = (x + sec x) (x  tan x)

By product rule,

f'(x) = `(x + sec x) d/dx (x - tan x) + (x - tan x) d/dx (x + sec x)`

= `(x + sec x) [dx/d (x) d/dx tanx] + (x − tan x) [d/dx (x) + d/dx sec x]`

= `(x + secx) [1– d/dx tan x] + (x - tan x) [1 + d/dx sec x]`     ...(i)

Let f₁ (x) = tan x, f₂(x) = sec x

Accordingly, f₁(x + h) = tan (x + h) and f₂(x + h) = sec (x + h)

f'(x) = `lim_(h->0) ((f_1(x + h) - f_1(x))/(h))`

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

= `lim_(h->0) [(tan(x + h) - tan x)/(h)]`

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

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

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

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

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

= `1 xx 1/cos^2 x = sec^2 x`

= `d/dx tan x = sec^2 x`     ...(ii)

f'₂(x) = `lim_(h->0) ((f_2(x + h) - f_2(x))/(h))`

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

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

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

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

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

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

= `sec x. (lim_(h->0)sin((2x + h)/2) {lim_(h->0) sin(h/2)/(h/2)})/(lim_(h->0) cos (x + h))`

= `sec x. (sin x.1)/(cos x)`

`=> d/dx sec x = sec x tan x`   ... (iii)

From (i), (ii), and (iii), we obtain

f'(x) = (x + sec x)(1 − sec² x) + (x − tan x) (1 + sec x tan x)