Question

Find the derivative of the following function:

5 sec x + 4 cos x

Answer 1

Let f (x) = 5 sec x + 4 cos x. Accordingly, from the first principle,

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

= `lim_(h->0) (5sec(x + h) + 4cos (x + h)- [5sec x + 4 cos x])/h`

= `5 lim_(h->0)[[sec(x + h) - sec x]]/h + 4 lim_(h->0)[[cos (x + h) - cos x]]/h`

= `5 lim_(h->0)[1/(cos(x + h)) - 1/(cos x)] + 4 lim_(h->0)1/h[cos(x + h) - cosx]`

= `5lim_(h->0)1/h[(cos x - cos (x + h))/(cos x cos (x + h))] + 4 lim_(h->0)1/h[cosx cos h - sin x sin h - cosx]`

= `5/(cosx) lim_(h->0)1/h[(-2sin  ((x + x + h))/2 sin  ((x - x - h))/2)/cos (x + h)] + 4lim_(h->0) 1/h[-cosx (1 - cos h) - sinx sin h]`

= `5/(cosx) lim_(h->0)1/h[(-2sin  ((2x + h)/2)  sin  (-h/2))/cos (x + h)] + 4 [-cos x lim_(h->0)((1 - cos h))/h - sinx lim_(h->0)(sin h)/h]`

= `5/cosx lim_(h->0)[(sin((2x + h)/2).sin(h/2)/(h/2))/(cos (x + h))] + 4[(-cosx).(0)-(sinx).1]`

= `5/(cos x)[lim_(h->0)sin((2x + h)/2)/(cos(x + h)). lim_(h->0)sin(h/2)/(h/2)]-4sinx`

= `5/(cos x).(sin x)/(cos x). 1 - 4 sin x`

= 5 sec x. tan x - 4 sin x