函数 (x)=asin x+dfrac (1)(3)sin 3x 在x=π/3 处有极值,则 a= __-|||-__

已知函数(x)=asin x+dfrac (1)(4)sin 4x-|||-__，则常数(x)=asin x+dfrac (1)(4)sin 4x-|||-__

[题目]试问:a为何值时,函数 (x)=asin x+dfrac (1)(3)sin 3x-|||-在 =dfrac (pi )(3) 处取得极值?它是极大值还

若函数=sin 3x，则=sin 3x_____.若函数，则_____.

(3) lim _(xarrow 0)dfrac (tan x-sin x)({sin )^3x}

(3) lim _(xarrow 0)dfrac (tan x-sin x)({sin )^3x};

15、填空 已知函数f(x)=asinx+(1)/(3)sin3x在点x=(pi)/(3)处取 得极值，则常数a=_____.15、填空 已知函数$f(x)=a

3.lim _(xarrow 0)dfrac (tan x-sin x)({sin )^3x}.3..

int dfrac (sin x)({cos )^3x}dx=______.______.

3.问a,b为何值时,有极限-|||-lim _(xarrow 0)(dfrac (sin 3x)({x)^3}+dfrac (a)({x)^2}+b)=0-|

求下列函数的导数 ( 1 ) =sin dfrac (1)(x)( 2 )=sin dfrac (1)(x)( 3 )=sin dfrac (1)(x)求下列函