y = ln√(1-x³)
= (1/2)ln(1-x³)
dy/dx = (1/2) * 1/(1-x³) * (-3x²)
= -3x²/[2(1-x³)]
= 3x²/[2(x³-1)]
dy = {3x²/[2(x³-1)]}dx能帮我写下y'的过程吗。我y'=-3x^2/2(1-x^3)^2其實y' = dy/dx，這兩個符號一樣的，都是表示導數微分就是dy=f(x)dx形式