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प्रश्न
Find the derivative of the following w. r. t. x by using method of first principle:
sin (3x)
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उत्तर
Let f(x) = sin 3x
∴ f(x + h) = sin [3(x + h)] = sin (3x + 3h)
f(x + h) – f(x) = sin(3x + 3h) – sin 3x
= `2cos((3x + 3"h" + 3x)/2)*sin((3x + 3"h" - 3x)/2)`
= `2cos((6x + 3"h")/2)*sin ((3"h")/2)`
By definition,
f'(x) = `lim_("h"-> 0) ("f"(x + "h") - "f"(x))/"h"`
= `lim_("h" -> 0) (2cos((6x + 3"h")/2)*sin((3"h")/2))/ "h"`
= `lim_("h" -> 0) 2[cos ((6x + 3"h")/2)]*[(sin ((3"h")/2))/(((3"h")/2))] xx 3/2`
= `3[lim_("h" -> 0) cos((6x + 3"h")/2)] xx [lim_("h" -> 0) (sin((3"h")/2))/(((3"h")/2))]`
= `3[cos ((6x + 0)/2) ] xx 1 ...[because "h" -> 0"," (3"h")/2 -> 0 "and" lim_(theta-> 0) sintheta/theta = 1]`
3 cos 3x
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