搜搜考试网设y=y(x) 由方程ysinx=cos(x-y) 所确定,则y'(0)=
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设y=y(x) 由方程ysinx=cos(x-y) 所确定,则y'(0)=
实现求出隐函数Y‘再将X=0带入吗？
实现求出隐函数Y‘再将X=0带入吗？
设y=y(x) 由方程ysinx=cos(x-y) 所确定,则y'(0)=
x=0时cos(-y)=cosy=0,故y=π/2+2kπ,k∈Z
F(x,y)=ysinx-cos(x-y)=0
dy/dx=-(∂F/∂x)/(∂F/∂y)=-[ycosx+sin(x-y)]/[sinx-sin(x-y)]
将x=0,y=π/2+2kπ,代入上式得：
y′(0)=-[π/2+2kπ+sin(-π/2-2kπ)]/[sin0-sin(-π/2-2kπ)]=-[π/2+2kπ-sin(π/2+2kπ)]/[sin(π/2+2kπ)]
=-[π/2+2kπ-sin(π/2)]/sin(π/2)=-(π/2+2kπ-1)=1-π/2-2kπ.k∈Z
x=0时cos(-y)=cosy=0,故y=π/2+2kπ,k∈Z
F(x,y)=ysinx-cos(x-y)=0
dy/dx=-(∂F/∂x)/(∂F/∂y)=-[ycosx+sin(x-y)]/[sinx-sin(x-y)]
将x=0,y=π/2+2kπ,代入上式得：
y′(0)=-[π/2+2kπ+sin(-π/2-2kπ)]/[sin0-sin(-π/2-2kπ)]=-[π/2+2kπ-sin(π/2+2kπ)]/[sin(π/2+2kπ)]
=-[π/2+2kπ-sin(π/2)]/sin(π/2)=-(π/2+2kπ-1)=1-π/2-2kπ.k∈Z
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