错在第一步.大意失荆州.
f(x)=sin(x/2)
f(x+⊿x)=sin(x/2+⊿x/2)
⊿y=sin(x/2+⊿x/2)-sin(x/2)=2cos(x/2+⊿x/4)sin(⊿x/4)
⊿y/⊿x=2cos(x/2+⊿x/4)sin(⊿x/4)/⊿x=cos(x/2+⊿x/4)sin(⊿x/4)/[2(⊿x/4)]
所以(⊿x→0)lim⊿y/⊿x=lim[cos(x/2+⊿x/4)]•1/2*[ sin(⊿x/4)/(⊿x/4)]=1/2*cos(x/2)
错在第一步.大意失荆州.
f(x)=sin(x/2)
f(x+⊿x)=sin(x/2+⊿x/2)
⊿y=sin(x/2+⊿x/2)-sin(x/2)=2cos(x/2+⊿x/4)sin(⊿x/4)
⊿y/⊿x=2cos(x/2+⊿x/4)sin(⊿x/4)/⊿x=cos(x/2+⊿x/4)sin(⊿x/4)/[2(⊿x/4)]
所以(⊿x→0)lim⊿y/⊿x=lim[cos(x/2+⊿x/4)]•1/2*[ sin(⊿x/4)/(⊿x/4)]=1/2*cos(x/2)