y=arcsinx+arcosx 求其导数

y=arcsinx+arcosx 求其导数

定义域：-1<=x<=1
1)
-1<=x<=0：-pi/2<=arcsinx<=0pi/2<=arccosx<=pi
令arcsinx=t,x=sint, -pi/2<=t<=0
令arccosx=z，x=cosz，pi/2<=z<=pi
这时容易得到，pi/2<=t+pi<=pi sin(t+pi)=-sint=-xx=-sin(t+pi)
即cosz=-sin(t+pi)且pi/2<=z<=pi，pi/2<=t+pi<=pi
即sin(z-pi/2)=sin(t+pi) 0<=z-pi/2<=pi/2 pi/2<=t+pi<=pi
那么只有z-pi/2+t+pi=pi即z+t=pi/2
即arcsinx+arccosx=pi/2

2)
0<=x<=1：0<=arcsinx<=pi/20<=arccosx<=pi
令arcsinx=t,x=sint, 0<=t<=pi/2
令arccosx=z，x=cosz，0<=z<=pi/2
即x=sint=cos(pi/2-t)=cosz 0<=pi/2-t<=pi/20<=z<=pi/2
故只能pi/2-t=z即z+t=pi/2
即arcsinx+arccosx=pi/2

3）上面折腾那么多，只是为了证明arcsinx+arccosx=pi/2。
那么导数显然为0了。