设f (x)在(0,+∞)内有定义,f′(1)=2,又对于任意的x,y∈(0,+∞)恒有f(xy)=yf(x)+xf(y

令x=y = 1得f(1) = 0
令 y = 1/x得 0 = f(x) / x + x f(1/x) 所以 f(1/x) = -f(x) / x^2
对x求导得
yf'(xy) = yf'(x) + f(y)
令y = 1/x得f'(1)/x = f'(x)/x + f(1/x) = f'(x)/x - f(x) / x^2
代入f'(1) = 2得
f'(x) - f(x) / x = 2
解这个微分方程得
f(x) = 2xlnx + Cx
f'(x) = 2lnx + 2 + C令x=1得C = 0
所以f(x) = 2 x lnx