Question

求教一个高数二次求导问题

12题，前面一次求导化简都明白，就是二次求导是怎么求出来的，没看懂，求大神指点





Answer 1

将备用y’代入

供参考，请笑纳。

Answer 2

我说明白了吗？

Answer 3

Answer 4

可以令分母 = g(x) = x（1-f'(y), y' = 1/g(x)
用 chain rule,
y'' = -(1/g(x))^2 g'(x)
= -(1/g(x))^2 [1-f''(y) - xf''(y) y']
= -(1/g(x))^2 [1-f''(y) - xf''(y)/g(x)]
= -(1/g(x))^3 [(1-f''(y))g(x) - xf''(y)]
= -(1/（x(1-f'(y)))^3 [(1-f''(y)^2 - f''(y)]x, after substitute g(x) = x(1-f'(y))
= 答案