可以啊。不过所代公式不同。
例如 函数 y = f(x) 是由隐函数 e^y-xy = 0 确定的隐函数。
记 F = e^y-xy, 则 F'x = y, F'y = e^y-x
y' = dy/dx = -Fx/Fy = y/(x-e^y) (1)
d^2y/dx^2 = [y'(x-e^y)-y(1-y'e^y)]/(x-e^y)^2
式(1)代入上式,得
d^2y/dx^2 = {y-y[1-ye^y/(x-e^y)]}/(x-e^y)^2
= y^2e^y/(x-e^y)^3
追问二阶导数也可以嘛
追答见补充解答。