【正确答案】正确答案:(Ⅰ)先求y(0):由x=arctant知,x=0<=>t=0,x>0(<0)<=>t>0(<0).由y=ln(1一t
2
一siny知,x=0<=>y=一siny<=>

并判断它在x=0邻域的正负号.

其中δ>0是充分小的数.因此x=0是y=f(x)的极大值点. (Ⅱ)由隐函数求导法知y'(x)满足

令x=x
0
,相应地y=y
0
,由F
x
'(x
0
,y
0
)=0,F
y
'(x
0
,y
0
)≠0得y'(x
0
)=0.将上式再对x求导, 并注意y=y(x)即得

再令x=x
0
,相应地y=y
0
,y'(x
0
)=0,得
