Question

设y=3次根号下1+x³/1-x³,求y'

有人解答的出来不

Answer 1

请用括号将题目表达清楚!

假定为y = ³√[(1 + x³)/(1 - x³)]
y' = (1/3)³√[(1 - x³)²/(1 + x³)²]*[(1 + x³)/(1 - x³)]'
= (1/3)³√[(1 - x³)²/(1 + x³)²]*[(1 + x³)'(1 - x³) - (1 + x³)(1 - x³)']/(1 - x³)²
= (1/3)³√[(1 - x³)²/(1 + x³)²]*(6x²)/(1 - x³)²
= 2x²/{³√[(1 - x³)⁴(1 + x³)²]}