Question

Find the derivative of the following w. r. t. x by using method of first principle:

sec (5x − 2)

Answer 1

Let f(x) = sec (5x − 2)

∴ f(x + h) = sec [5 (x + h) − 2]

= sec (5x + 5h − 2)

By first principle, we get

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

= `lim_("h" -> 0) (sec(5x + 5"h" - 2) - sec(5x - 2))/"h"`

= `lim_("h" -> 0) [(1/(cos(5x + 5"h" - 2)) - 1/(cos(5x - 2)))/"h"]`

= `lim_("h" -> 0) [(cos(5x - 2) - cos(5x + 5"h" - 2))/("h" cos(5x - 2) cos(5x + 5"h" - 2))]`

= `lim_("h" -> 0) [(2sin ((5x - 2 + 5x + 5"h" - 2)/2) sin((5x + 5"h" - 2 - 5x + 2)/2))/("h" cos(5x - 2) cos(5x + 5"h" - 2))]    ...[because cos "C" - cos "D" = 2sin (("C" + "D")/2) sin (("D" - "C")/2)]`

= `1/(cos(5x - 2)) lim_("h" -> 0) (2sin(5x - 2 + (5"h")/2) sin  (5"h")/2)/("h" cos (5x + 5"h" - 2)`

= `5/(cos (5x - 2)) lim_("h" -> 0) [(sin(5x - 2 + (5"h")/2))/(cos (5x + 5"h" - 2)) * (sin((5"h")/2))/((5"h")/2)]`

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

= `5/(cos (5x - 2)) * (sin(5x - 2))/(cos(5x - 2)) (1)  ...[lim_(theta -> 0) (sin"p"theta)/("p"theta) = 1]`

= 5 sec (5x – 2) tan (5x – 2)