f(u,x)具有二阶连续偏导数,
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f(u,x)具有二阶连续偏导数,
f(u,x)具有二阶连续偏导数,且满足δ^2f/δu^2+δ^2f/δv^2=1,
又g(x,y)=f[xy,(x^2-y^2)/2],
求δ^2g/δx^2+δ^2g/δy^2.
我做到δg/δx=y(δf/δu)+x(δf/δv),
符号是偏导数的符号啊,不是全微分,δ^2g/δx^2这个是2阶x偏导数
f(u,x)具有二阶连续偏导数,且满足δ^2f/δu^2+δ^2f/δv^2=1,
又g(x,y)=f[xy,(x^2-y^2)/2],
求δ^2g/δx^2+δ^2g/δy^2.
我做到δg/δx=y(δf/δu)+x(δf/δv),
符号是偏导数的符号啊,不是全微分,δ^2g/δx^2这个是2阶x偏导数
以下以df/dx表示一阶偏导数,d2f/dx2表示二阶偏导数
g(x,y)看作是由f(u,v),u=xy,v=1/2×(x^2-y^2)复合而成
则
dg/dx=df/du×y+df/dv×x
d2g/dx2
=y×d(df/du)/dx+df/dv+x×d(df/dv)/dx
=y×[d2f/du2×y+d2f/dudv×x]+df/dv+x×[d2f/dudv×y+d2f/dv2×x]
=y^2×d2f/du2+2xy×d2f/dudv+df/dv+x^2×d2f/dv2
dg/dy=df/du×x-df/dv×y
d2g/dx2
=x×d(df/du)/dy-df/dv-y×d(df/dv)/dy
=x×[d2f/du2×x-d2f/dudv×y]-df/dv-y×[d2f/dudv×x-d2f/dv2×y]
=x^2×d2f/du2-2xy×d2f/dudv-df/dv+y^2×d2f/dv2
所以,
d2g/dx2+d2g/dy2=(x^2+y^2)×[d2f/du2+d2f/dv2]=0
g(x,y)看作是由f(u,v),u=xy,v=1/2×(x^2-y^2)复合而成
则
dg/dx=df/du×y+df/dv×x
d2g/dx2
=y×d(df/du)/dx+df/dv+x×d(df/dv)/dx
=y×[d2f/du2×y+d2f/dudv×x]+df/dv+x×[d2f/dudv×y+d2f/dv2×x]
=y^2×d2f/du2+2xy×d2f/dudv+df/dv+x^2×d2f/dv2
dg/dy=df/du×x-df/dv×y
d2g/dx2
=x×d(df/du)/dy-df/dv-y×d(df/dv)/dy
=x×[d2f/du2×x-d2f/dudv×y]-df/dv-y×[d2f/dudv×x-d2f/dv2×y]
=x^2×d2f/du2-2xy×d2f/dudv-df/dv+y^2×d2f/dv2
所以,
d2g/dx2+d2g/dy2=(x^2+y^2)×[d2f/du2+d2f/dv2]=0
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