Question

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

sin (3x)

Answer 1

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