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方程y=tAn(x%y)所确定的函数的二阶导数

y=tan(x+y)y'=sec(x+y)*(x+y)'=sec(x+y)*(1+y')=sec(x+y)+y'sec(x+y)y'-y'sec(x+y)=sec(x+y)y'=sec(x+y)/[1-sec(x+y)]=sec(x+y)/{-[sec(x+y)-1]}=sec(x+y)/[-tan(x+y)]=-1/cos(x+y)*cos(x+y)/sin(x+y)=-csc(x+y)y''=-2csc(x+y)*[-

由方程y=tan(x+y)两边直接对x求导,得 y'=(1+y')sec2(x+y) ∴两边继续对x求导,得 y″=y″sec2(x+y)+2(1+y′)2sec2(x+y)tan(x+y) 将y'=(1+y')sec2(x+y)代入,化简得 y''=-2csc2(x+y)cot3(x+y).

y=tan(x+y)y'=tan'(x+y)=sec^2(x+y)(x+y)'=sec^2(x+y)*(1+y')y'=sec^2(x+y)/[1-sec^2(x+y))=-sec^2(x+y)/tan^2(x+y)=-1/sin^2(x+y)=-csc^2(x+y)y''=-2csc(x+y)*[csc(x+y)]'=-2csc(x+y)*[-csc(x+y)cot(x+y)](x+y)'=2csc^2(x+y)cot(x+y)(1+y')=2csc^2(x+y)cot(x+y)[1-csc^2(x+y)].1-csc^2(x+y)=cot^2(x+y)

y=tan(x+y) =>arctany=x+y=>y=arctany-x (对方程两边 一阶求导)=>y'=y'/(1+y^2)-1 (对方程两边 二阶求导)=>y''=y''/(1+y^2)-2yy'/(1+y^2)^2

y=tan(x+y)y'=[sec(x+y)]^2*(1+y')则y'=[sec(x+y)]^2/{1-[sec(x+y)]^2}=-[sec(x+y)]^2/tan(x+y)]^2=-1/[sin(x+y)]^2则y''={-1/[sin(x+y)]^2}'={-2[sin(x+y)]cos(x+y)}*(1+y')1/[sin(x+y)]^4则y''={-2[sin(x+y)]cos(

本题运用隐函数求导法则和导数的四则运算,再进行代入即可求得答案:

y=tan(x+y)两边求导:y'=(sec(x+y))^2*(1+y'),所以y'=-(csc(x+y))^2=-1-1/y^2 所以,y''=2y'/y^3=-2(1+y^2)/y^5

如图

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