设Z=X/2,则Z(x)'=1/2,
设U=cos(X/2)=cos Z,
U(x)'=U(z)'·Z(x)'=(-sin Z)·(1/2)=-(1/2)sin (X/2)
Y=[COS(x/2)]^5=U^5,所以
Y(x)'=Y(u)'·U(x)'=(5U^4)·[-(1/2)Sin (X/2)]
=[5cos^4 (X/2)]·[-(1/2)Sin (X/2)]
=-(5/2)Sin (X/2)cos^4 (X/2)
设Z=X/2,则Z(x)'=1/2,
设U=cos(X/2)=cos Z,
U(x)'=U(z)'·Z(x)'=(-sin Z)·(1/2)=-(1/2)sin (X/2)
Y=[COS(x/2)]^5=U^5,所以
Y(x)'=Y(u)'·U(x)'=(5U^4)·[-(1/2)Sin (X/2)]
=[5cos^4 (X/2)]·[-(1/2)Sin (X/2)]
=-(5/2)Sin (X/2)cos^4 (X/2)