请问 设y=y(x)有方程2x-tan(x-y)=∫上限x-y下限0 [sec（t）]^2d所确定,求d^2y/dx^2

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请问 设y=y(x)有方程2x-tan(x-y)=∫上限x-y下限0 [sec（t）]^2d所确定,求d^2y/dx^2
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请问 设y=y(x)有方程2x-tan(x-y)=∫上限x-y下限0 [sec（t）]^2d所确定,求d^2y/dx^2
请问 设y=y(x)有方程2x-tan(x-y)=∫上限x-y下限0 [sec（t）]^2d所确定,求d^2y/dx^2

请问 设y=y(x)有方程2x-tan(x-y)=∫上限x-y下限0 [sec（t）]^2d所确定,求d^2y/dx^2
2x-tan(x-y)=∫(0,x-y) [sec（t）]^2dt
两边对x求导得：
2-sec²(x-y)(1-y')=sec²(x-y)(1-y')
sec²(x-y)(1-y')=1
y'=1-cos²(x-y)=sin²(x-y)
y''=2sin(x-y)cos(x-y)(1-y')
=sin(2x-2y)*cos²(x-y)

2x-tan(x-y)=∫(上限x-y下限0) [sec（t）]^2dt=tan(x-y)
=>x-tan(x-y)=0
=>dy/dx=1-sec²(x-y)
=>d^2y/dx^2=-2sec(x-y)tan(x-y)sec(x-y)=-2sec²(x-y)tan(x-y)

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