1. untuk fungsi fungsi f(x) dibawah ini tunjukan bahwa {f "(x)}² = 4 cos² 2x a. f(x) = sin²x b f(x) = cos²x
2, diketahui fungsi fungsi f(x) = sin x dan g(x) = cos x tunjukan bahwa a. f ' (x) . g(x) + f(x) . g'(x) = cos 2x b. f " (x) . g'(x) - f'(x) . g "(x) = 1
subebe
1a. f(x) = sin²x f(x) = 1/2 - 1/2 (cos2x) f' (x) = sin 2x f " (x) = 2 cos 2x (f "(x)² = (2 cos2x)² = 4 cos ² 2x (terbukti) b. f(x) = cos²x = 1/2 (cos 2x) - 1/2 f ' (x) = - sin 2x f " (x) = - 2 cos 2x (f"(x))² = ( -2 cos2x)² (f"(x))² = 4 cos² 2x (terbukti)
2. f (x) = sin x f'(x) = cos x f " (x) = - sin x g(x) = cos x g ' (x) = - sin x g " (x) = - cos x a. f ' (x) . g(x) + f(x) . g'(x) = cos x . cos x + sin x (- sin x) = cos² x - sin² x = cos 2x (terbukti) b. (-sin x) (-sin x) - (cos x) ( - cos x) = sin² x + cos² x = 1 (terbukti)
f(x) = 1/2 - 1/2 (cos2x)
f' (x) = sin 2x
f " (x) = 2 cos 2x
(f "(x)² = (2 cos2x)²
= 4 cos ² 2x (terbukti)
b. f(x) = cos²x
= 1/2 (cos 2x) - 1/2
f ' (x) = - sin 2x
f " (x) = - 2 cos 2x
(f"(x))² = ( -2 cos2x)²
(f"(x))² = 4 cos² 2x (terbukti)
2. f (x) = sin x
f'(x) = cos x
f " (x) = - sin x
g(x) = cos x
g ' (x) = - sin x
g " (x) = - cos x
a. f ' (x) . g(x) + f(x) . g'(x) = cos x . cos x + sin x (- sin x)
= cos² x - sin² x
= cos 2x (terbukti)
b. (-sin x) (-sin x) - (cos x) ( - cos x) = sin² x + cos² x = 1 (terbukti)
a.)
b.)
Nomor 2.
f(x) = sin x, maka f'(x) = cos x, dan f''(x) = -sin x
g(x) = cos x, maka g'(x) = -sin x dan g''(x) = -cos x
Sehingga,
a.)
cos x.cos x + sin x (-sin x)
= cos²x - sin²x
= cos 2x
b.)
(-sin x).(-sin x) - cos x.(-cos x)
= sin²x+cos²x
= 1