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20.11.2011, 10:08
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это ответ
Здравствуйте, Alejandro!
Пусть f(x) = [$8730$](1 - 0,5 [$183$] sin2 x) = (1 - (sin x)2/2)1/2. Тогда
f'(x) = 1/2 [$183$] (1 - (sin x)2/2)-1/2 [$183$] (1 - (sin x)2/2)' = 1/2 [$183$] (1 - (sin x)2/2)-1/2 [$183$] (-1/2) [$183$] 2 [$183$] sin x [$183$] cos x = (-1/4) [$183$] sin 2x [$183$] (1 - (sin x)2/2)-1/2,
f"(x) = ((-1/4) [$183$] sin 2x [$183$] (1 - (sin x)2/2)-1/2)' = -1/4 [$183$] ((sin 2x)' [$183$] (1 - (sin x)2/2)-1/2 + sin 2x [$183$] ((1 - (sin x)2/2)-1/2)') =
= -1/4 [$183$] (2 [$183$] cos 2x [$183$] (1 - (sin x)2/2)-1/2 + sin 2x [$183$] (-1/2) [$183$] (1 - (sin x)2/2)-3/2 [$183$] (-1/2) [$183$] 2 [$183$] sin x [$183$] cos x) =
= -1/4 [$183$] (2 [$183$] cos 2x [$183$] (1 - (sin x)2/2)-1/2 + 1/4 [$183$] (sin 2x)2 [$183$] (1 - (sin x)2/2)-3/2) =
= -1/4 [$183$] (1 - (sin x)2/2)-1/2 [$183$] (2 [$183$] cos 2x + 1/4 [$183$] (sin 2x)2 [$183$] (1 - (sin x)2/2)-1).
С уважением.
Об авторе:
Facta loquuntur.