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Question
Find the derivative of the following w. r. t. x by using method of first principle:
tan (2x + 3)
Solution
Let f(x) = tan (2x + 3)
∴ f(x + h) = tan [2(x + h) + 3] = tan [(2x + 3) + 2h]
f(x + h) – f(x) = tan [(2x + 3) + 2h] – tan (2x + 3)
= `(sin[(2x + 3) + 2"h"])/(cos[(2x + 3) + 2"h"]) - (sin(2x + 3))/(cos(2x + 3))`
= `(sin[(2x + 3) + 2"h"] cos[2x + 3] - cos[(2x + 3) + 2"h"] sin[2x + 3])/(cos[2x + 3 + 2"h"]*cos[2x + 3])`
= `(sin[(2x + 3) + 2"h" - (2x + 3)])/(cos[2x + 3 + 2"h"]*cos[2x + 3])`
= `(sin2"h")/(cos[2x + 3 + 2"h"]* cos[2x + 3])`
By definition,
f'(x) = `lim_("h" -> 0) ("f"(x + "h") - "f"(x))/"h"`
= `lim_("h" -> 0) (sin 2"h")/("h" cos [2x + 3 + 2"h"]*cos[2x + 3])`
= `2/(cos(2x + 3)) lim_("h" -> 0) [((sin2"h")/(2"h"))* 1/(cos [2x + 3 + 2"h"])]`
= `2/(cos(2x + 3)) [lim_("h" -> 0) (sin2"h")/(2"h")] xx 1/(lim_("h" -> 0) [cos (2x + 2"h" + 3)]`
= `2/(cos(2x + 3)) xx 1 xx 1/(cos(2x + 0 + 3)) ...[because "h" -> 0, 2"h" -> 0 "and" lim_(theta -> 0) sintheta/theta = 1]`
= 2 sec2 (2x + 3)
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