设f(x+y,x-y)=x^2+y^2,df(x,y)/dx+df(x,y)/dy设f(x+y,x-y)=x^2-y^2,аf(x,y)/аx+аf(x,y)/аy 这样会影响结果吗?

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设f(x+y,x-y)=x^2+y^2,df(x,y)/dx+df(x,y)/dy设f(x+y,x-y)=x^2-y^2,аf(x,y)/аx+аf(x,y)/аy 这样会影响结果吗?
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设f(x+y,x-y)=x^2+y^2,df(x,y)/dx+df(x,y)/dy设f(x+y,x-y)=x^2-y^2,аf(x,y)/аx+аf(x,y)/аy 这样会影响结果吗?
设f(x+y,x-y)=x^2+y^2,df(x,y)/dx+df(x,y)/dy
设f(x+y,x-y)=x^2-y^2,аf(x,y)/аx+аf(x,y)/аy
这样会影响结果吗?

设f(x+y,x-y)=x^2+y^2,df(x,y)/dx+df(x,y)/dy设f(x+y,x-y)=x^2-y^2,аf(x,y)/аx+аf(x,y)/аy 这样会影响结果吗?
令x+y=u,x-y=v,则x=(u+v)/2,y=(u-v)/2
所以:f(u,v)=(u+v)^2/4+(u-v)^2/4,即:f(x,x)=(x+y)^2/4+(x-y)^2/4
所以:df(x,y)/dx+df(x,y)/dy =(x+y)/2-(x-y)/2=y

会诶,你把符号都变了嘛,正号变成负号了
解答过程:f(x+y,x-y)=x^2-y^2=(x+y)(x-y)
设u=x+y,v=x-y,则f(u,v)=uv,au/ax=1,av/ax=1,au/ay=1,av/ay=-1
由链式法则可知 af/ax=af/au*au/ax=v=x-y
同理 af/ay=af/av*av/ay=-u=-x-y
^2代表平方,*代表乘