(1) y=sinβcos²β=sinβ(1-sin²β)=sinβ-sin³β.
dy/dβ=cosβ-3sin²βcosβ. Chain rule for differentiating sin³β. Let x=sinβ and z=x³. dx/dβ=cosβ, dz/dx=3x², dz/dβ=(dz/dx)(dx/dβ)=3x²cosβ=3sin²βcosβ.
(2) Let u=cot(x)=cos(x)/sin(x). Let p=cos(x), q=sin(x)
du/dx=(qdp/dx-pdq/dx)/q²=(-sin(x)sin(x)-cos(x)cos(x))/sin²(x)=-(sin²(x)+cos²(x))csc²(x)=-csc²(x).
Therefore d(cot(x/4))/dx=-csc²(x/4)/4.
Let v=cot(x/4) and y=v², then dy/dx=(dy/dv)(dv/dx)=2v(-csc²(x/4)/4)=-½cot(x/4)csc²(x/4).