求解(y=sin^4(X\/4)+cos^4(X\/4)的导数)
a2-18a+56=0 a=4,a=14 c=a-8>0 所以a=14 b=a-4=10 c=a-8=6
y=sin^4(x\/4)+cos^4(x\/4) 求导数
y'=[sin^4(x\/4)+cos^4(x\/4)]'=4cos^3(x\/4)-4sin^3(x\/4)这个是复合函数求导,里面的sin,cos也要求导,x\/4还要求导 正确结果是:y=sin^4(x\/4)+cos^4(x\/4)y'=[sin^4(x\/4)+cos^4(x\/4)]'=4sin^3(x\/4)·cos(x\/4)·1\/4+4cos^3(x\/4)·(-sin(x\/4)...
y=sin^4 x\/4 + cos^4 x\/4 求导数 ………
=(sinx\/2)\/2 * (-cosx\/2)=(-sinx)\/4 y=(sin^4 x\/4 + cos^4 x\/4)=(sin^2(x\/4)+cos^2(x\/4))²-2sin^2(x\/4)cos^2(x\/4)=1-2sin^2(x\/4)cos^2(x\/4)=1-(1\/2)(sinx\/2)²求导 y=-1\/2 * 2(sinx\/2) *cos(x\/2) *(1\/2)=(-1\/4)sinx ...
y=sin^4x\/4+cos^4x的导数怎么求
原式=分母为(sin^4x)'(4+cos^4x)-sin^4X(4+COS^4X)'分子为(4+cos^4x)^2 其中sin^4x的导数为4sin^3xcosx,而4+cos^4x的导数为-4cos^3xsinx
求y = (sinx\/4)^4 + (cosx\/4)^4导数,请列解题过程,多谢
=1\/2*sin(x\/2)*(-cos(x\/2))=-1\/4*sinx 或者 y=(sin(x\/4))^4 + (cos(x\/4))^4 =(sin(x\/4))^4 + (cos(x\/4))^4+2(sin(x\/4)*cos(x\/4))^2-2(sin(x\/4)*cos(x\/4))^2 =((sin(x\/4))^2+(cos(x\/4))^2)^2-1\/2*(sin(x\/2))^2 =1-1\/4*(1-cos...
y=sin4(x\/4) +cos4(x\/4)化简
这样化简:y=sin4(x\/4) +cos4(x\/4)=[sin2(x\/4) +cos2(x\/4)]的平方—2sin2(x\/4)*cos2(x\/4);y=1—1\/2sin2(x\/2);y=1—1\/2(1\/2—cosx\/2);y=3\/4+1\/4cosx
函数y=sinx的4次方+cosx的4次方的值域是?
y=[(sinx)^2+(cosx)^2]^2-2(sinx)^2+(cosx)^2 =1^2-1\/2*(2sinxcosx)^2 =-1\/2*(sin2x)^2+1 =-1\/2*(1-cos4x)\/2+1 =1\/2*cos4x+1\/2 -1<=cos4x<=1 0<=1\/2cos4x+1\/2<=1 所以值域[0,1]
函数y=sin4(4次方)x+cos4(4次方)x的值域是
{sin(x)}^4={1\/2*(1-cos2x)}^2 ,{cos(x)}^4={1\/2*(1+cos2x)}^2 ,因此,两者相加后的结果是1\/4*(2+2(cos(2x))^2),继续化简使用二倍角公式,原式=1\/4*(3+cos(4x)),这样就可以得知,-1<=cos(4x)<=1,2<=3+cos(4x)<=4,所以值域就是(1\/2,1)...
y=sin^4x+cos^4x的n阶导数如何求?谁能给出详细步骤,感谢
y =(sinx)^4+(cosx)^4 =[ (sinx)^2 + (cosx)^2 ] ^2 - 2(sinx.cosx)^2 =1 - 2(sinx.cosx)^2 =1 - (1\/2)(sin2x)^2 y'=-(sin2x). ( sin2x)'=-(sin2x). ( cos2x) (2x)'=-2(sin2x). ( cos2x)=-cos4x y^(n)= -4^(n-1) . cos4x ; n= 1,5,...
求∫sinxcosx\/(sinx^4+cosx^4 )dx
sinx^4+cosx^4]=1\/2sin2x\/[(sinx^2+cosx^2)^2-2sinx^2cosx^2]=1\/2sin2x\/(1-(sin2x)^2\/2)=sin2x\/(2-(sin2x)^2)=sin2x\/(1+(cos2x)^2)∫[sinxcosx]dx\/[sinx^4+cosx^4]=∫sin2xdx\/[1+(cos2x)^2]=-1\/2∫dcos2x\/(1+(cos2x)^2)= -1\/2arctan(cos2x)...