4道高数求导求解：y=arccos(1-2x) y=lncot(x/2) y=e的负3分之x次方×sin3x y=cos

利用复合函数求导法,很简单的.
1、y'=-1/√[1-(1-2x)^2]*(-2)=2/√(4x-4x^2)=1/√(x-x^2)
2、y'=1/cot(x/2)*[-csc^2(x/2)]*1/2=sin(x/2)/cos(x/2)*[-1/sin^2(x/2)]*1/2
=-1/[2sin(x/2)cos(x/2)]=1/sin(x/2)=csc(x/2)
3、y'=e^(-x/3)*(-1/3)*sin(3x)+e^(-x/3)*cos(3x)*3=e^(-x/3)*[3cos(3x)-1/3*sin(3x)]
4、y'=2cosx*(-sinx)*cos(x^2)+cos^2(x)*[-sin(x^2)]*2x=-sin(2x)cos(x^2)-2xsin(x^2)cos^2(x)