sin(2x+y)+y^2=1确定隐函数y=y(x),求y',y"
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sin(2x+y)+y^2=1确定隐函数y=y(x),求y',y"
请具体的写出来,
请具体的写出来,
sin(2x+y)+y^2=1
(2x+y)′*cos(2x+y)+2y*y′=0
(2+y′)*cos(2x+y)+2y*y′=0
2cos(2x+y)+ y′*cos(2x+y) )+2y*y′=0
y′*(cos(2x+y) +2y)=- 2cos(2x+y)
y′=- 2cos(2x+y)/ (cos(2x+y) +2y)
sin(2x+y)+y^2=1
(2x+y)′*cos(2x+y)+2y*y′=0
(2+y′)*cos(2x+y)+2y*y′=0
y〃*cos(2x+y)-(2+y′)*(2x+y) ′ *sin(2x+y)+2y′*y′+2y*y〃=0
y〃*cos(2x+y)-(2+y′)*(2+y′) *sin(2x+y)+2y′*y′+2y*y〃=0
y〃*(cos(2x+y) +2y) =(2+y′)*(2+y′) *sin(2x+y)-2y′*y′
y〃=((2+y′)*(2+y′) *sin(2x+y)-2y′*y′)/(cos(2x+y) +2y)
最后把算出的y′带入上面的式子,在化简一下就行
(2x+y)′*cos(2x+y)+2y*y′=0
(2+y′)*cos(2x+y)+2y*y′=0
2cos(2x+y)+ y′*cos(2x+y) )+2y*y′=0
y′*(cos(2x+y) +2y)=- 2cos(2x+y)
y′=- 2cos(2x+y)/ (cos(2x+y) +2y)
sin(2x+y)+y^2=1
(2x+y)′*cos(2x+y)+2y*y′=0
(2+y′)*cos(2x+y)+2y*y′=0
y〃*cos(2x+y)-(2+y′)*(2x+y) ′ *sin(2x+y)+2y′*y′+2y*y〃=0
y〃*cos(2x+y)-(2+y′)*(2+y′) *sin(2x+y)+2y′*y′+2y*y〃=0
y〃*(cos(2x+y) +2y) =(2+y′)*(2+y′) *sin(2x+y)-2y′*y′
y〃=((2+y′)*(2+y′) *sin(2x+y)-2y′*y′)/(cos(2x+y) +2y)
最后把算出的y′带入上面的式子,在化简一下就行
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