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प्रश्न
Find the derivative of the following w. r. t. x by using method of first principle:
sec (5x − 2)
उत्तर
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)
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