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](/uploads/image/z/7083767-47-7.jpg?t=%E8%AF%B7%E9%97%AE+%E8%AE%BEy%3Dy%28x%29%E6%9C%89%E6%96%B9%E7%A8%8B2x-tan%28x-y%29%3D%E2%88%AB%E4%B8%8A%E9%99%90x-y%E4%B8%8B%E9%99%900+%5Bsec%EF%BC%88t%EF%BC%89%5D%5E2d%E6%89%80%E7%A1%AE%E5%AE%9A%2C%E6%B1%82d%5E2y%2Fdx%5E2)
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)
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)
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