anyone know how to get the derivative of cos 2x - 5cos^2 X?
Here are some choices.
1. -6cos2x
2. 3sin2x
3.6cos2x
4.6sin2x
5.3cos2x
2007-10-08 15:53:23 · 4 answers · asked by Lil j 1 in Science & Mathematics ➔ Mathematics
f(x)=cos2x-5cos^2x
f'(x)=-2sin2x+-5(2cosx)(-sinx)
=10cosxsinx-2sin2x
=5sin2x-2sin2x
=3sin2x
2007-10-08 16:00:18 · answer #1 · answered by slick_licker88 3 · 0⤊ 0⤋
f (x) = cos 2 x - 5 cos ² x
f `(x) = - 2 sin 2 x - 10 cos x (- sin x)
f `(x) = - 2 sin 2 x + 10 sin x cos x
f `(x) = - 2 sin 2 x + 5 (2 sin x cos x)
f `(x) = - 2 sin 2 x + 5 sin 2 x
f `(x) = 3 sin 2 x
2007-10-09 04:09:52 · answer #2 · answered by Como 7 · 0⤊ 0⤋
d ( cos 2x - 5cos² x )/dx
. = d ( cos 2x )/dx + d ( -5cos² x )/dx
. = -sin 2x * 2 - 5*2cos x * -sin x
. = -2sin 2x + 10sin x * cos x
. = -2sin 2x + 5sin 2x
. = 3sin 2x . . (answer 2)
2007-10-08 23:01:57 · answer #3 · answered by Ben 3 · 0⤊ 0⤋
-sin2x *2 - 10 cosx (-sinx)
-2sin2x + 10sinxcox
-2sin2x +5(sin2x)
3sin(2x)
2007-10-08 22:59:02 · answer #4 · answered by Marcus C 3 · 0⤊ 0⤋