2006-09-24 03:38:49 · 5 answers · asked by babis p 1 in Science & Mathematics ➔ Mathematics
i)f(x)=cos x - sin x
2006-09-24 03:49:34 · answer #1 · answered by Anonymous · 0⤊ 0⤋
if f(x)=sinx + cosx
(i)f'(x)=cos x - sin x
(ii)f(x)=f'(x)
sinx + cosx = cos x - sin x
2sin x =0
sin x =0 it means either x= 0 or pi
2006-09-24 11:12:19 · answer #2 · answered by Amar Soni 7 · 0⤊ 0⤋
The derivative of sine is cosine, the derivative of cosine is -sine. The differential of (a+b) is (differential of a) + (differential of b). So, f'(x) = cos x - sin x. f(x)=f'(x) means cos x - sin x = sin x + cos x, which simplifies to sin x = 0, which means x = 0 or pi. Easier than you thought? Just be methodical.
2006-09-24 10:52:05 · answer #3 · answered by Sangmo 5 · 0⤊ 0⤋
f(x) = sin x + cos x
so you can differentiate term by term,
that is,
f'(x) = cos x - sin x
for f(x) = f'(x)
then,
sin x + cos x = cos x -sin x
so sin x = -sin x
then sin x = sin (-x)
so x is either 0 or pi in the given interval.
2006-09-24 10:56:50 · answer #4 · answered by yasiru89 6 · 0⤊ 0⤋
as said earlier ...
(i) f'(x) = cox-sinx
(ii) between [0,pi], sin(x) = -sinx, so f(x) = cosx-sinx = f'(x)
2006-09-24 10:45:04 · answer #5 · answered by sebourban 4 · 0⤊ 1⤋