复合函数中的链式法则ƒ(g(x))对x求导得ƒ'(g(x)) • g'(x)或dy/dx = dy/du • du/dx在这里，e^(xlna)，令ƒ(u) = e^u，u = g(x) = xlnaƒ'(u) = e^u，g'(x) = lna则[ƒ(g(x))]' = ƒ'(u) • g'(x) = e^
u • lna = e^(xlna) • lna = a^
x • lna或令y = e^u，u = xlna则dy/du = e^u，du/dx = lna所以dy/dx = dy/du • du/dx = e^
u • lna = e^(xlna) • lna = a^
x • lna
