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# 三阶求导的两种顺序

基线：HIPS/autograd 1.9.1，commit `f53a21734fdfae636f448744d9097d8d35a643a0`。对应 [Lesson 10](https://zdd14990.github.io/blog/applied-math/autograd/10-higher-order-ad/) 与既有实验 `experiments/higher_order.py`。这是对象/时序示意，R/P/B 都是教学标签。

## 调用栈先建立 0，再建立 1、2

```text
grad(grad(grad(f)))(2.0)
  outer grad begins trace0, B0(value=2,node=R0)
    middle grad begins trace1, B1(value=B0,node=R1)
      inner grad begins trace2, B2(value=B1,node=R2)
        original f receives B2
```

## 从原 f 看到的包裹层次是 2 -> 1 -> 0

```text
B2 [ArrayBox, _trace=2, _node=R2]
 |_value
 v
B1 [ArrayBox, _trace=1, _node=R1]
 |_value
 v
B0 [ArrayBox, _trace=0, _node=R0]
 |_value
 v
2.0 [float]
```

## primitive 递归进入与返回

```text
power(B2,3) -> select trace2 -> args=(B1,3), parents=(R2,)
  power(B1,3) -> select trace1 -> args=(B0,3), parents=(R1,)
    power(B0,3) -> select trace0 -> args=(2,3), parents=(R0,)
      raw power(2,3) -> 8
    create P0 -> return Box0(value=8,node=P0)
  create P1 -> return Box1(value=Box0,node=P1)
create P2 -> return Box2(value=Box1,node=P2)
```

## 反向计算仍在外层被观察

```text
inner forward trace2 finishes -> top=1
  inner VJP uses x=Box1 -> computes 3*x^2 -> derivative value12 in Box1
middle forward trace1 finishes -> top=0
  middle VJP differentiates the derivative program -> value12 in Box0
outer forward trace0 finishes -> top=-1
  outer VJP -> third derivative6 as an ordinary value
```

只解最高层让每个 trace 保留自己的依赖；VJP 使用可微 primitives 让 derivative program 本身可被外层观察。两者缺一不可。实验直接验证输入 Box 层次和12/12/6；P 节点的递归示意依据当前 primitive 源码推演。