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 + cos x)(x - tan x)`

Answer 1

Let f(x) = (x + cos x) (x − tan x)

By product rule,

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

= `(x + cos x) [d/dx (x) -d/dx (tan x)] + (x - tan x) (1-sin x)`

= `(x + cos x) + [1 - d/dx tan x] + (x - tan x) (1 - sin x)`    ...(i)

Let g(x) = tan x. Accordingly, g(x + h) = tan(x + h)

By first principle,

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

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

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

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

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

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

= `1/cos x .1 . 1/(cos (x + 0))`

= `1/cos^2 x`

= sec²x                ...(ii)

Therefore, from (i) and (ii), we obtain

f'(x) = (x + cos x) (1 - sec² x) + (x - tan x) (1 - sin x)

= (x + cos x)(- tan² x) + (x tan x) (1 - sin x)

= tan² x(x + cos x) + (x - tan x) (1 - sin x)