设二元函数f(x,y)满足丨f(x,y)丨≦x²+y².证明f(x,y)在(0,0)可微.

设二元函数f(x,y)满足丨f(x,y)丨≦x²+y².证明f(x,y)在(0,0)可微. 证明二元函数可微.设 lim [f(x,y)-f(0,0)+2x-y]/√x^2+y^2=0证明f(x,y)在点(0,0 设f(x)在x=0处可导,且对任意x.y满足f(x+y)=f(x)f(y),证明f(x)处处可导,且 设函数f(x)满足条件f(x+y)=f(x)+f(y),且f(x)在x=0处连续,证明f(x)在所有的点x0处连续 设f(x)是定义在0到正无穷大上的增函数,且对一切x.y>0满足f(x/y)f(x)-f(y),... 证明二元函数不可微设f(x,y)=xy/√x^2+y^2,(x,y)≠(0,0)0,(x,y)=(0,0)证明f(x,y 设函数y=f(x)是定义域在R,并且满足f(x+y)=f(x)+f(y) 二元函数f(x,y)=x+y/x-y,求f(y/x,x/y) 定义在R上的函数f(x)满足f (x + y) = f (x) + f ( y )(x,y∈R),当x>0时,f (x) 若定义域为R函数f(x)满足f(x+y)=2*f(x)*f(y),且f(0)不等于0,证明f（x）是偶函数 设函数f(x)=y在（0,+∞）上是增函数,并满足f(xy)=f(x)+f(y),f(4)=1 设u=f(x,y)可微,且满足方程x(σ f/σ x)+y(σ f/σ y)=0