关于y=sin(x/2)的求导,我不知道我的推导哪里错了⊿y=sin(x/2+⊿x)-sin(x/2)=2cos(x/2+⊿x/2)sin(⊿x/2) ⊿y/⊿x=2cos(x/2+⊿x/2)sin(⊿x/2)/⊿x=cos(x/2+⊿x/2)sin(⊿x/2)/(⊿x/2) 所以lim⊿x→0⊿y/⊿x=lim⊿x→0cos(x/2+⊿

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关于y=sin(x/2)的求导,我不知道我的推导哪里错了⊿y=sin(x/2+⊿x)-sin(x/2)=2cos(x/2+⊿x/2)sin(⊿x/2) ⊿y/⊿x=2cos(x/2+⊿x/2)sin(⊿x/2)/⊿x=cos(x/2+⊿x/2)sin(⊿x/2)/(⊿x/2) 所以lim⊿x→0⊿y/⊿x=lim⊿x→0cos(x/2+⊿

关于y=sin(x/2)的求导,我不知道我的推导哪里错了⊿y=sin(x/2+⊿x)-sin(x/2)=2cos(x/2+⊿x/2)sin(⊿x/2) ⊿y/⊿x=2cos(x/2+⊿x/2)sin(⊿x/2)/⊿x=cos(x/2+⊿x/2)sin(⊿x/2)/(⊿x/2) 所以lim⊿x→0⊿y/⊿x=lim⊿x→0cos(x/2+⊿
关于y=sin(x/2)的求导,我不知道我的推导哪里错了
⊿y=sin(x/2+⊿x)-sin(x/2)=2cos(x/2+⊿x/2)sin(⊿x/2)
⊿y/⊿x=2cos(x/2+⊿x/2)sin(⊿x/2)/⊿x=cos(x/2+⊿x/2)sin(⊿x/2)/(⊿x/2)
所以lim⊿x→0⊿y/⊿x=lim⊿x→0cos(x/2+⊿x/2)•lim⊿x→0sin(⊿x/2)/(⊿x/2)=cos(x/2)

关于y=sin(x/2)的求导,我不知道我的推导哪里错了⊿y=sin(x/2+⊿x)-sin(x/2)=2cos(x/2+⊿x/2)sin(⊿x/2) ⊿y/⊿x=2cos(x/2+⊿x/2)sin(⊿x/2)/⊿x=cos(x/2+⊿x/2)sin(⊿x/2)/(⊿x/2) 所以lim⊿x→0⊿y/⊿x=lim⊿x→0cos(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)

你这里的⊿x只是把x/2看成了整体,而不是x,需要最后再对x/2求导。