f(x+y)=f(x)+f(y)+2xy,
移项,有f(x+y)-f(x) = f(y) + 2x
两边同除以y,有[f(x+y)- f(x)]/y = f(y)/y + 2x
也就是[f(x+y)- f(x)]/[(x+y)-x] = f(y)/y + 2x
对y→0取极限,有f'(x) = 0 + 2x
也就是f'(x) = 2x
于是f(x) = x²+C (其中C为常数）
代入f(x+y)=f(x)+f(y)+2xy验证,有C=0.
综上,f'(x) = 2x,f(x) = x².