设f(x+y,x-y)=x^2+y^2,df(x,y)/dx+df(x,y)/dy
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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,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
再问: 设f(x+y,x-y)=x^2-y^2,df(x,y)/dx+df(x,y)/dy 应该啊是减号
再答: 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=x
再问: 设f(x+y,x-y)=x^2-y^2,求аf(x,y)/аx+аf(x,y)/аy 这样会影响结果吗?
再答: 会的,方法完全不对,结果会不一样的
再问: f(x+y,x-y)=x^2-y^2,求аf(x,y)/аx+аf(x,y)/аy 这样要怎么求啊?
再答: 同上,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=x
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
再问: 设f(x+y,x-y)=x^2-y^2,df(x,y)/dx+df(x,y)/dy 应该啊是减号
再答: 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=x
再问: 设f(x+y,x-y)=x^2-y^2,求аf(x,y)/аx+аf(x,y)/аy 这样会影响结果吗?
再答: 会的,方法完全不对,结果会不一样的
再问: f(x+y,x-y)=x^2-y^2,求аf(x,y)/аx+аf(x,y)/аy 这样要怎么求啊?
再答: 同上,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=x
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