求方程y=tan(x+y+1)确定的隐函数y=y(x)的二阶导数
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两边对x求导得：
y'=[sec(x+y+1)]^2*(1+y')
得：y'=[sec(x+y+1)^2/{1-[sec(x+y+1)]^2}=-[sec(x+y+1)]^2/[tan(x+y+1)]^2=-[csc(x+y+1)]^2
再求导,得：y"=2csc(x+y+1)*csc(x+y+1)*ctg(x+y+1)*(1+y')
代入y'到上式,得:y"=2[csc(x+y+1)]^2*ctg(x+y+1)*{1-[csc(x+y+1)]^2}
=-2[csc(x+y+1)]^2*ctg(x+y+1)*[ctg(x+y+1)]^2
=-2[csc(x+y+1)]^2*[ctg(x+y+1)]^3
y'=[sec(x+y+1)]^2*(1+y')
得：y'=[sec(x+y+1)^2/{1-[sec(x+y+1)]^2}=-[sec(x+y+1)]^2/[tan(x+y+1)]^2=-[csc(x+y+1)]^2
再求导,得：y"=2csc(x+y+1)*csc(x+y+1)*ctg(x+y+1)*(1+y')
代入y'到上式,得:y"=2[csc(x+y+1)]^2*ctg(x+y+1)*{1-[csc(x+y+1)]^2}
=-2[csc(x+y+1)]^2*ctg(x+y+1)*[ctg(x+y+1)]^2
=-2[csc(x+y+1)]^2*[ctg(x+y+1)]^3
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