y=(cos(1–2x))^3求导
cos (\u03c0/3-2x) \u6c42\u5bfcf(x) = cos(\u03c0/3-2x)
f ' (x) = - sin(\u03c0/3-2x) * (-2) = 2sin(\u03c0/3-2x)
\u7528\u5230\u7684\u516c\u5f0f\uff1a
f(x) = cosx \u5219 f ' (x) = - sinx
y=cos(2x²+1)
y'=-4xsin(2x²+1)
y'=3cos(1-2x)×［-sin(1-2x)］×(-2)
=-3sin(2-4x)cos(1-2x)
-48sin（1-2x）
绛旓細鍥炵瓟锛y'=3cos(1-2x)脳[-sin(1-2x)]脳(-2) =-3sin(2-4x)cos(1-2x)
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绛旓細瀹冧滑鏄嚎鎬х浉鍏崇殑銆備簨瀹炰笂 2(cosx)^2=1+cos锛2x锛.
绛旓細=1-cos(2x)+sin(2x)=1+鈭2(鈭2/2*sin(2x) -鈭2/2*cos(2x))=1+鈭2 sin(2x-蟺/4)鍑芥暟鐨勬渶澶у兼槸1+鈭2锛 鏈灏忓1-鈭2锛屽懆鏈熸槸蟺.2. y=3- cos x - sin ² x =3- cos x 鈥(1- cos ²x)= cos ²x- cos x+2 =(cosx-1/2) ²+7/4 Cosx...
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绛旓細y=f(x)鈫抷=Af(蠅x+蠁)鍏蜂綋鍙傜収涓夎鍑芥暟鐨勫浘璞″彉鎹 涓涓噸瑕佺粨璁:鑻(a-x)=f(a+x),鍒欏嚱鏁皔=f(x)鐨勫浘鍍忓叧浜庣洿绾縳=a瀵圭О; 濡: 鐨勫浘璞″鍥,浣滃嚭涓嬪垪鍑芥暟鍥捐薄: (1) ;(2) ; (3) ;(4) ; (5) ;(6) ; (7) ;(8) ; (9) 銆 浜斻佸弽鍑芥暟: (1)瀹氫箟: (2)鍑芥暟瀛樺湪鍙嶅嚱鏁扮殑鏉′欢: ;...
