{
  "schema_version": "ocme-derivation-reader-v1",
  "guide_id": "derivation-mko-derivative-01",
  "mko_id": "mko-derivative",
  "title": {
    "zh-Hant": "導數：由差商極限求二次式導數並判定尖點無導數",
    "en": "The derivative: a quadratic's derivative from the difference-quotient limit, and no derivative at a corner"
  },
  "contract": {
    "schema_version": "ocme-derivation-guide-v0.1",
    "guide_id": "derivation-mko-derivative-01",
    "source": {
      "mko_id": "mko-derivative",
      "commit": "b28a5edbd46279b1db2a3f3f48bb3f447e18899f",
      "sha256": "b523a35b9e5c20d66a7346f84ac5bece2c4167b17d46208fdb9b40115deeaf64",
      "english_sha256": "5b384deb770b949487d4788df980511f8844bd2e59d136696e0681c2ee60db1c"
    },
    "mode": "worked_example",
    "audience_prerequisites": [
      "distributive",
      "frac_rules",
      "abs_rules",
      "order_basics",
      "func_limit_def",
      "numeric_evaluation"
    ],
    "goal": {
      "zh-Hant": "對 f(t)=3t²−2t+5：先對 h≠0 化簡差商 (f(1+h)−f(1))/h=4+3h，再以 ε-δ 證明 f'(1)=4；推導一般公式：差商為 2a2x+a1+a2h、導數為 2a2x+a1；說明有限步 h=1/10 的差商 4.3 不是導數、a2=0 的仿射邊界；最後證明 g(t)=|t−2| 在 t=2 沒有導數（不是 0），而在 t>2 處導數為 1。",
      "en": "For f(t)=3t²−2t+5: first simplify the difference quotient for h≠0, (f(1+h)−f(1))/h=4+3h, then prove f'(1)=4 by ε-δ; derive the general formulas: the quotient is 2a2x+a1+a2h and the derivative 2a2x+a1; explain why the finite-step quotient 4.3 at h=1/10 is not the derivative, and the affine boundary a2=0; finally prove that g(t)=|t−2| has no derivative at t=2 (not 0), while its derivative is 1 for t>2."
    },
    "givens": [
      {
        "id": "G1",
        "statement": {
          "zh-Hant": "函數 ℝ→ℝ，全實數論域；有限步 h 必須非零。",
          "en": "Functions ℝ→ℝ on the full real domain; a finite step h must be nonzero."
        }
      },
      {
        "id": "G2",
        "statement": {
          "zh-Hant": "f(t)=3t²−2t+5（a2=3、a1=−2、a0=5），求 x=1 的導數。",
          "en": "f(t)=3t²−2t+5 (a2=3, a1=−2, a0=5); find the derivative at x=1."
        }
      },
      {
        "id": "G3",
        "statement": {
          "zh-Hant": "比較用：有限步 h=1/10；仿射 e(t)=−2t+5（a2=0）；尖點函數 g(t)=|t−2|。",
          "en": "For comparison: the finite step h=1/10; the affine e(t)=−2t+5 (a2=0); the corner function g(t)=|t−2|."
        }
      }
    ],
    "steps": [
      {
        "id": "S01",
        "uses": [
          "G1",
          "func_limit_def"
        ],
        "before_tex": "f'(x)",
        "after_tex": "f'(x)=D\\iff\\forall\\varepsilon>0\\ \\exists\\delta>0\\ \\forall h:\\ 0<|h|<\\delta\\Rightarrow\\left|\\frac{f(x+h)-f(x)}{h}-D\\right|<\\varepsilon",
        "relation": "definition",
        "rule": "mko_derivative",
        "conditions": [
          "h\\neq 0"
        ],
        "explanation": {
          "zh-Hant": "依 MKO，導數是差商在 h→0（h≠0）的兩側有限極限。差商本身只對 h≠0 有定義；導數是極限，不是某一個有限步的差商。若極限不存在，導數就不存在。",
          "en": "By the MKO, the derivative is the finite two-sided limit of the difference quotient as h→0 with h≠0. The quotient itself is defined only for h≠0; the derivative is a limit, not the quotient at some finite step. If the limit does not exist, neither does the derivative."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "G1",
            "mko_derivative",
            "func_limit_def"
          ],
          "explanation": {
            "zh-Hant": "此步只展開定義。",
            "en": "This step only unfolds the definition."
          }
        },
        "leaf_basis": "mko_derivative + func_limit_def + def_unfold"
      },
      {
        "id": "S02",
        "uses": [
          "G2"
        ],
        "before_tex": "f(1)",
        "after_tex": "3\\cdot 1^2-2\\cdot 1+5=6",
        "relation": "equivalent",
        "rule": "numeric_evaluation",
        "conditions": [],
        "explanation": {
          "zh-Hant": "中心值 f(1)=3−2+5=6（N3）。這是差商分子中被減去的量。",
          "en": "The centre value f(1)=3−2+5=6 (N3). It is the amount subtracted in the numerator of the quotient."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "numeric_evaluation"
          ],
          "explanation": {
            "zh-Hant": "精確求值，兩邊同一數。",
            "en": "Exact evaluation; both sides name the same number."
          }
        },
        "leaf_basis": "numeric_evaluation"
      },
      {
        "id": "S03",
        "uses": [
          "G2",
          "distributive"
        ],
        "before_tex": "f(1+h)",
        "after_tex": "3(1+h)^2-2(1+h)+5=6+4h+3h^2",
        "relation": "equivalent",
        "rule": "distributive",
        "conditions": [],
        "explanation": {
          "zh-Hant": "(1+h)²=1+2h+h²，所以 3(1+h)²=3+6h+3h²；−2(1+h)=−2−2h；相加 3+6h+3h²−2−2h+5=6+4h+3h²（N1 以多項式恆等式核對）。",
          "en": "(1+h)²=1+2h+h², so 3(1+h)²=3+6h+3h²; −2(1+h)=−2−2h; adding, 3+6h+3h²−2−2h+5=6+4h+3h² (N1 checks this as a polynomial identity)."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "distributive",
            "power_laws"
          ],
          "explanation": {
            "zh-Hant": "對所有 h 成立的恆等式，雙向成立。",
            "en": "An identity for every h, valid both ways."
          }
        },
        "leaf_basis": "distributive + power_laws + add_comm + add_assoc"
      },
      {
        "id": "S04",
        "uses": [
          "S02",
          "S03"
        ],
        "before_tex": "f(1+h)-f(1)",
        "after_tex": "4h+3h^2",
        "relation": "equivalent",
        "rule": "eq_replace",
        "conditions": [],
        "explanation": {
          "zh-Hant": "代入 S03 與 S02：(6+4h+3h²)−6=4h+3h²。",
          "en": "Substitute S03 and S02: (6+4h+3h²)−6=4h+3h²."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "eq_replace",
            "add_inverse"
          ],
          "explanation": {
            "zh-Hant": "等值代換，雙向成立。",
            "en": "Replacement of equals, valid both ways."
          }
        },
        "leaf_basis": "eq_replace + add_inverse + add_zero"
      },
      {
        "id": "S05",
        "uses": [
          "S04",
          "frac_rules"
        ],
        "before_tex": "\\frac{4h+3h^2}{h}\\ (h\\neq 0)",
        "after_tex": "4+3h",
        "relation": "equivalent",
        "rule": "frac_rules",
        "conditions": [
          "h\\neq 0"
        ],
        "explanation": {
          "zh-Hant": "分子 4h+3h²=h(4+3h)，對 h≠0 約去 h 得 4+3h。h=0 時差商是 0/0，沒有定義；所以不能把 h=0 直接代入原差商，只能先在 h≠0 處化簡，再看極限。",
          "en": "The numerator is 4h+3h²=h(4+3h); for h≠0 cancel h to get 4+3h. At h=0 the quotient is 0/0, undefined; so h=0 may not be substituted into the original quotient; one simplifies for h≠0 first and then examines the limit."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "frac_rules",
            "distributive"
          ],
          "explanation": {
            "zh-Hant": "反向是乘回 h/h，需要 h≠0；條件由 G1 保證。",
            "en": "Backward, multiply by h/h, which needs h≠0; G1 guarantees this."
          }
        },
        "leaf_basis": "frac_rules + distributive + mul_inverse_nonzero"
      },
      {
        "id": "S06",
        "uses": [
          "S01",
          "S05"
        ],
        "before_tex": "\\varepsilon>0\\ \\text{arbitrary},\\ \\delta=\\tfrac{\\varepsilon}{3}",
        "after_tex": "0<|h|<\\delta\\Rightarrow|(4+3h)-4|=3|h|<\\varepsilon;\\ f'(1)=4",
        "relation": "implies",
        "rule": "forall_intro",
        "conditions": [
          "\\varepsilon>0"
        ],
        "explanation": {
          "zh-Hant": "取任意 ε>0，令 δ=ε/3。若 0<|h|<δ，由 S05，|Q(h)−4|=|3h|=3|h|<3δ=ε。ε 任意，依 S01，f'(1)=4。",
          "en": "Take an arbitrary ε>0 and let δ=ε/3. If 0<|h|<δ, then by S05 |Q(h)−4|=|3h|=3|h|<3δ=ε. As ε is arbitrary, S01 gives f'(1)=4."
        },
        "reverse_audit": {
          "kind": "sufficient_goal_reduction",
          "required_ids": [
            "S01",
            "S05",
            "forall_intro"
          ],
          "explanation": {
            "zh-Hant": "目標 G：f'(1)=4。足夠條件 H：對任意 ε，δ=ε/3 有效；H⇒G 由定義。H 由 S05 正向證成，未把 G 當作前提，也未代入 h=0。",
            "en": "Goal G: f'(1)=4. Sufficient condition H: for arbitrary ε, δ=ε/3 works; H⇒G by the definition. H is proved forwards from S05, without assuming G and without substituting h=0."
          }
        },
        "leaf_basis": "forall_intro + exists_intro + abs_rules + order_basics"
      },
      {
        "id": "S07",
        "uses": [
          "G2"
        ],
        "before_tex": "\\frac{f(x+h)-f(x)}{h},\\ f(t)=a_2t^2+a_1t+a_0,\\ h\\neq 0",
        "after_tex": "2a_2x+a_1+a_2h",
        "relation": "equivalent",
        "rule": "L_diff_quot",
        "conditions": [
          "h\\neq 0"
        ],
        "explanation": {
          "zh-Hant": "一般化：依引理 L_diff_quot，任何二次以下多項式在任意 x 的差商都是 2a2x+a1+a2h。代入 a2=3、a1=−2、x=1 得 6−2+3h=4+3h，與 S05 一致（N6）。引理在 S07.1、S07.2 證明。",
          "en": "Generalize: by Lemma L_diff_quot, the difference quotient of any polynomial of degree at most two at any x is 2a2x+a1+a2h. With a2=3, a1=−2, x=1 this is 6−2+3h=4+3h, agreeing with S05 (N6). The lemma is proved in S07.1 and S07.2."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "L_diff_quot"
          ],
          "explanation": {
            "zh-Hant": "引理給出 h≠0 時兩式相等，雙向成立。",
            "en": "The lemma gives equality of the two expressions for h≠0, valid both ways."
          }
        },
        "leaf_basis": "distributive + frac_rules + mul_inverse_nonzero"
      },
      {
        "id": "S07.1",
        "uses": [
          "distributive",
          "power_laws"
        ],
        "before_tex": "f(x+h)-f(x)",
        "after_tex": "a_2(2xh+h^2)+a_1h=h(2a_2x+a_1+a_2h)",
        "relation": "equivalent",
        "rule": "distributive",
        "conditions": [],
        "explanation": {
          "zh-Hant": "(x+h)²−x²=2xh+h²，a1(x+h)−a1x=a1h，a0 相消；提出公因子 h 得 h(2a2x+a1+a2h)（N2 以五個變數的多項式恆等式核對）。",
          "en": "(x+h)²−x²=2xh+h², a1(x+h)−a1x=a1h, and a0 cancels; factoring out h gives h(2a2x+a1+a2h) (N2 checks this as a polynomial identity in five variables)."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "distributive",
            "power_laws"
          ],
          "explanation": {
            "zh-Hant": "恆等式，雙向成立。",
            "en": "An identity, valid both ways."
          }
        },
        "leaf_basis": "distributive + power_laws + add_inverse"
      },
      {
        "id": "S07.2",
        "uses": [
          "S07.1",
          "frac_rules"
        ],
        "before_tex": "\\frac{h(2a_2x+a_1+a_2h)}{h}",
        "after_tex": "2a_2x+a_1+a_2h",
        "relation": "equivalent",
        "rule": "frac_rules",
        "conditions": [
          "h\\neq 0"
        ],
        "explanation": {
          "zh-Hant": "h≠0，約去 h。S07.1、S07.2 合起來證得引理 L_diff_quot。",
          "en": "h≠0, so cancel h. S07.1 and S07.2 together prove Lemma L_diff_quot."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "frac_rules",
            "mul_inverse_nonzero"
          ],
          "explanation": {
            "zh-Hant": "需要 h≠0；h=0 時兩邊不能比較。",
            "en": "It needs h≠0; at h=0 the two sides cannot be compared."
          }
        },
        "leaf_basis": "frac_rules + mul_inverse_nonzero"
      },
      {
        "id": "S08",
        "uses": [
          "S01",
          "S07"
        ],
        "before_tex": "\\lim_{h\\to 0,\\,h\\neq 0}(2a_2x+a_1+a_2h)",
        "after_tex": "f'(x)=2a_2x+a_1\\ \\text{for every real}\\ x",
        "relation": "implies",
        "rule": "L_lin_limit",
        "conditions": [],
        "explanation": {
          "zh-Hant": "取 A=2a2x+a1、B=a2，依引理 L_lin_limit，差商在 h→0 時趨向 2a2x+a1；依 S01，這就是 f'(x)。x 任意，所以公式對每個實數 x 成立。引理在 S08.1、S08.2 分 B=0 與 B≠0 證明。",
          "en": "With A=2a2x+a1 and B=a2, Lemma L_lin_limit gives that the quotient tends to 2a2x+a1 as h→0; by S01 this is f'(x). As x is arbitrary, the formula holds for every real x. The lemma is proved in S08.1 and S08.2 for B=0 and B≠0."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S01",
            "S07",
            "L_lin_limit"
          ],
          "explanation": {
            "zh-Hant": "兩個情況涵蓋所有 a2，包括 a2=0。",
            "en": "The two cases cover every a2, including a2=0."
          }
        },
        "leaf_basis": "forall_intro + order_basics + abs_rules"
      },
      {
        "id": "S08.1",
        "uses": [
          "mul_zero"
        ],
        "before_tex": "B=0",
        "after_tex": "|(A+0\\cdot h)-A|=0<\\varepsilon\\ \\text{for every}\\ h",
        "relation": "implies",
        "rule": "mul_zero",
        "conditions": [
          "B=0"
        ],
        "explanation": {
          "zh-Hant": "B=0 時 A+Bh=A 對每個 h 成立，誤差為 0<ε，任何 δ>0 都可以。",
          "en": "If B=0 then A+Bh=A for every h; the error is 0<ε, so any δ>0 works."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "mul_zero"
          ],
          "explanation": {
            "zh-Hant": "只需 ε>0。",
            "en": "Only ε>0 is needed."
          }
        },
        "leaf_basis": "mul_zero + add_zero + order_basics"
      },
      {
        "id": "S08.2",
        "uses": [
          "abs_rules",
          "order_basics"
        ],
        "before_tex": "B\\neq 0,\\ \\delta=\\tfrac{\\varepsilon}{|B|}",
        "after_tex": "0<|h|<\\delta\\Rightarrow|Bh|=|B||h|<|B|\\delta=\\varepsilon",
        "relation": "implies",
        "rule": "order_basics",
        "conditions": [
          "|B|>0"
        ],
        "explanation": {
          "zh-Hant": "B≠0 時 |B|>0，令 δ=ε/|B|>0。若 0<|h|<δ，兩邊乘正數 |B| 得 |B||h|<ε。S08.1、S08.2 合起來證得引理 L_lin_limit。",
          "en": "If B≠0 then |B|>0; let δ=ε/|B|>0. If 0<|h|<δ, multiplying by the positive |B| gives |B||h|<ε. S08.1 and S08.2 together prove Lemma L_lin_limit."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "abs_rules",
            "order_basics"
          ],
          "explanation": {
            "zh-Hant": "除以 |B| 需要 B≠0，故分情況。",
            "en": "Dividing by |B| needs B≠0, hence the case split."
          }
        },
        "leaf_basis": "abs_rules + order_basics + mul_inverse_nonzero"
      },
      {
        "id": "S09",
        "uses": [
          "S05",
          "S06",
          "G3"
        ],
        "before_tex": "h=\\tfrac{1}{10}",
        "after_tex": "Q(\\tfrac{1}{10})=4+\\tfrac{3}{10}=\\tfrac{43}{10}\\neq 4=f'(1)",
        "relation": "implies",
        "rule": "numeric_evaluation",
        "conditions": [],
        "explanation": {
          "zh-Hant": "有限步 h=1/10 的差商是 4+3/10=43/10（N4，直接由 f(1.1)、f(1) 計算），與導數 4 相差 a2h=3/10。有限差商只是平均變化率；導數是 h→0 的極限。只有當 a2h=0 時兩者才相等（見 S10）。",
          "en": "The quotient at the finite step h=1/10 is 4+3/10=43/10 (N4, computed directly from f(1.1) and f(1)), differing from the derivative 4 by a2h=3/10. A finite quotient is only an average rate; the derivative is the limit as h→0. The two agree only when a2h=0 (see S10)."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S05",
            "S06",
            "numeric_evaluation"
          ],
          "explanation": {
            "zh-Hant": "N4、N5 可重算；本步不宣稱任何有限差商等於導數。",
            "en": "N4 and N5 can be recomputed; the step does not claim that any finite quotient equals the derivative."
          }
        },
        "leaf_basis": "numeric_evaluation + frac_rules"
      },
      {
        "id": "S10",
        "uses": [
          "S07",
          "S08",
          "G3"
        ],
        "before_tex": "e(t)=-2t+5\\ (a_2=0)",
        "after_tex": "\\frac{e(x+h)-e(x)}{h}=-2\\ (h\\neq 0),\\ e'(x)=-2",
        "relation": "implies",
        "rule": "L_diff_quot",
        "conditions": [
          "h\\neq 0"
        ],
        "explanation": {
          "zh-Hant": "MKO 假設 a2 允許 a2=0（次數至多二，不必是真正的二次式）。由 S07，a2=0、a1=−2 的差商是 −2+0·h=−2，對每個 h≠0 都相同（N11）；由 S08，e'(x)=−2。這裡每個有限差商都等於導數，只因 a2h=0；這不能推廣到 a2≠0 的情形（S09）。常數 a2=a1=0 時導數為 0，也是合法的低階邊界。",
          "en": "MKO assumption a2 allows a2=0 (degree at most two, not necessarily a genuine quadratic). By S07 the quotient with a2=0, a1=−2 is −2+0·h=−2 for every h≠0 (N11); by S08, e'(x)=−2. Here every finite quotient equals the derivative only because a2h=0; this does not extend to a2≠0 (S09). A constant, a2=a1=0, has derivative 0, also a legitimate low-degree boundary."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S07",
            "S08",
            "L_diff_quot"
          ],
          "explanation": {
            "zh-Hant": "直接代入一般公式；不宣稱有限差商一般等於導數。",
            "en": "A direct substitution into the general formulas; no claim that finite quotients equal derivatives in general."
          }
        },
        "leaf_basis": "L_diff_quot + L_lin_limit + mul_zero"
      },
      {
        "id": "S11",
        "uses": [
          "G3",
          "abs_rules"
        ],
        "before_tex": "g(t)=|t-2|,\\ t=2",
        "after_tex": "\\frac{g(2+h)-g(2)}{h}=\\frac{|h|-0}{h}=\\frac{|h|}{h}\\ (h\\neq 0)",
        "relation": "equivalent",
        "rule": "eq_replace",
        "conditions": [
          "h\\neq 0"
        ],
        "explanation": {
          "zh-Hant": "g(2)=|0|=0，g(2+h)=|h|，所以差商為 |h|/h，對 h≠0 有定義。",
          "en": "g(2)=|0|=0 and g(2+h)=|h|, so the quotient is |h|/h, defined for h≠0."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "eq_replace",
            "abs_rules"
          ],
          "explanation": {
            "zh-Hant": "等值代換，雙向成立。",
            "en": "Replacement of equals, valid both ways."
          }
        },
        "leaf_basis": "eq_replace + abs_rules"
      },
      {
        "id": "S12",
        "uses": [
          "S11",
          "abs_rules"
        ],
        "before_tex": "\\frac{|h|}{h}",
        "after_tex": "1\\ (h>0),\\quad -1\\ (h<0)",
        "relation": "case_split",
        "rule": "abs_rules",
        "conditions": [
          "h\\neq 0"
        ],
        "explanation": {
          "zh-Hant": "h>0 時 |h|=h，差商 h/h=1；h<0 時 |h|=−h，差商 −h/h=−1（N9、N10）。兩個情況涵蓋所有 h≠0。右導數 1、左導數 −1。",
          "en": "For h>0, |h|=h and the quotient is h/h=1; for h<0, |h|=−h and the quotient is −h/h=−1 (N9, N10). The two cases cover every h≠0. The right derivative is 1 and the left derivative −1."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "abs_rules",
            "case_analysis"
          ],
          "explanation": {
            "zh-Hant": "分情況窮盡；h=0 不在考慮之內。",
            "en": "The cases are exhaustive; h=0 is not considered."
          }
        },
        "leaf_basis": "abs_rules + case_analysis + frac_rules"
      },
      {
        "id": "S13",
        "uses": [
          "S01",
          "S12"
        ],
        "before_tex": "g'(2)=D\\ \\text{(assumed)},\\ \\varepsilon=\\tfrac12,\\ h=\\pm\\tfrac{\\delta}{2}",
        "after_tex": "|1-D|<\\tfrac12,\\ |-1-D|<\\tfrac12\\Rightarrow 2\\le|1-D|+|D+1|<1:\\ \\text{contradiction}",
        "relation": "implies",
        "rule": "contradiction",
        "conditions": [],
        "explanation": {
          "zh-Hant": "反證：假設 g'(2)=D 存在。對 ε=1/2 取得 δ>0，令 h=δ/2 與 h=−δ/2，都滿足 0<|h|<δ。由 S12，|1−D|<1/2 且 |−1−D|<1/2。三角不等式：2=|1−(−1)|≤|1−D|+|D−(−1)|<1/2+1/2=1，矛盾。所以 g 在 2 沒有導數。",
          "en": "By contradiction: suppose g'(2)=D exists. For ε=1/2 obtain δ>0 and take h=δ/2 and h=−δ/2, both with 0<|h|<δ. By S12, |1−D|<1/2 and |−1−D|<1/2. Triangle inequality: 2=|1−(−1)|≤|1−D|+|D−(−1)|<1/2+1/2=1, a contradiction. So g has no derivative at 2."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S01",
            "S12",
            "abs_rules",
            "contradiction"
          ],
          "explanation": {
            "zh-Hant": "矛盾只否定「兩側極限存在」；兩個單側導數 1 與 −1 仍然存在。",
            "en": "The contradiction refutes only the existence of the two-sided limit; the one-sided derivatives 1 and −1 still exist."
          }
        },
        "leaf_basis": "contradiction + exists_elim + abs_rules + order_basics"
      },
      {
        "id": "S14",
        "uses": [
          "G3",
          "S01"
        ],
        "before_tex": "t_0>2,\\ 0<|h|<t_0-2",
        "after_tex": "\\frac{g(t_0+h)-g(t_0)}{h}=\\frac{(t_0+h-2)-(t_0-2)}{h}=1;\\ g'(t_0)=1",
        "relation": "implies",
        "rule": "abs_rules",
        "conditions": [
          "t_0>2"
        ],
        "explanation": {
          "zh-Hant": "t0>2 時，若 |h|<t0−2，則 t0+h>2，兩個絕對值都等於裡面的式子，分子為 h（N12），差商恰為 1。所以取 δ=t0−2，對任何 ε 誤差都是 0，g'(t0)=1。同理 t0<2 時 g(t)=2−t 在 t0 附近成立，g'(t0)=−1。",
          "en": "For t0>2, if |h|<t0−2 then t0+h>2, both absolute values equal their arguments, the numerator is h (N12), and the quotient is exactly 1. So with δ=t0−2 the error is 0 for every ε, and g'(t0)=1. Likewise for t0<2, g(t)=2−t near t0, so g'(t0)=−1."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "abs_rules",
            "S01"
          ],
          "explanation": {
            "zh-Hant": "需要 h 小到不跨過 2；這說明導數不存在只發生在尖點本身。",
            "en": "h must be small enough not to cross 2; this shows non-existence happens only at the corner itself."
          }
        },
        "leaf_basis": "abs_rules + order_basics + frac_rules"
      },
      {
        "id": "S15",
        "uses": [
          "S12",
          "S13"
        ],
        "before_tex": "\\text{a totalized API returning}\\ 0\\ \\text{at}\\ t=2",
        "after_tex": "g'(2)\\ \\text{does not exist};\\ 0\\ \\text{is not a limit of}\\ Q",
        "relation": "implies",
        "rule": "quantifier_negation",
        "conditions": [],
        "explanation": {
          "zh-Hant": "某些形式庫把導數做成全定義函數，在不可導點填預設值 0。但 0 不是差商的極限：對每個 h≠0，|Q(h)−0|=1，取 ε=1/2 就失敗。所以本條目的答案是「g'(2) 不存在」，不是 0；預設值是函數的記帳慣例，不代表導數存在。",
          "en": "Some formal libraries make the derivative a total function and fill the default 0 where none exists. But 0 is not a limit of the quotient: |Q(h)−0|=1 for every h≠0, so ε=1/2 already fails. The answer of this entry is therefore 'g'(2) does not exist', not 0; a default value is a bookkeeping convention, not the existence of a derivative."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S12",
            "S13",
            "quantifier_negation"
          ],
          "explanation": {
            "zh-Hant": "直接由 S12 的值與 S13 的不存在得出。",
            "en": "It follows directly from the values in S12 and the non-existence in S13."
          }
        },
        "leaf_basis": "quantifier_negation + abs_rules"
      }
    ],
    "result": {
      "zh-Hant": "對 h≠0，f(t)=3t²−2t+5 在 1 的差商為 4+3h，f'(1)=4（δ=ε/3）。一般地，二次以下多項式的差商為 2a2x+a1+a2h，導數為 2a2x+a1；h=1/10 的有限差商 4.3 不是導數，仿射情形 a2=0 才使有限差商等於導數。g(t)=|t−2| 在 2 的右差商為 1、左差商為 −1，所以 g'(2) 不存在（不是 0）；t>2 時 g'=1，t<2 時 g'=−1。",
      "en": "For h≠0 the quotient of f(t)=3t²−2t+5 at 1 is 4+3h, and f'(1)=4 (with δ=ε/3). In general a polynomial of degree at most two has quotient 2a2x+a1+a2h and derivative 2a2x+a1; the finite quotient 4.3 at h=1/10 is not the derivative, and only the affine case a2=0 makes finite quotients equal the derivative. g(t)=|t−2| has right quotient 1 and left quotient −1 at 2, so g'(2) does not exist (it is not 0); g'=1 for t>2 and g'=−1 for t<2."
    },
    "negative_controls": [
      {
        "claim": {
          "zh-Hant": "(f(1.1)−f(1))/0.1=4.3，所以 f'(1)=4.3",
          "en": "(f(1.1)−f(1))/0.1=4.3, so f'(1)=4.3"
        },
        "expected": "reject",
        "reason": {
          "zh-Hant": "這是有限差商；導數是極限 4（S06、S09）。",
          "en": "That is a finite quotient; the derivative is the limit 4 (S06, S09)."
        }
      },
      {
        "claim": {
          "zh-Hant": "在 (f(1+h)−f(1))/h 中直接令 h=0 求導數",
          "en": "Find the derivative by setting h=0 in (f(1+h)−f(1))/h"
        },
        "expected": "reject",
        "reason": {
          "zh-Hant": "h=0 時是 0/0，沒有定義；有限步必須非零。",
          "en": "At h=0 it is 0/0, undefined; a finite step must be nonzero."
        }
      },
      {
        "claim": {
          "zh-Hant": "|t−2| 在 2 的左右導數 −1 與 1 平均為 0，所以 g'(2)=0",
          "en": "The one-sided derivatives −1 and 1 of |t−2| at 2 average to 0, so g'(2)=0"
        },
        "expected": "reject",
        "reason": {
          "zh-Hant": "兩側極限不存在（S13）；|Q(h)−0|=1 對所有 h≠0。",
          "en": "The two-sided limit does not exist (S13); |Q(h)−0|=1 for every h≠0."
        }
      },
      {
        "claim": {
          "zh-Hant": "a2=0 時 f 不是本模型的函數",
          "en": "With a2=0, f is not a function of this model"
        },
        "expected": "reject",
        "reason": {
          "zh-Hant": "MKO 允許 a2=0（次數至多二）；e(t)=−2t+5 的導數為 −2（S10）。",
          "en": "The MKO allows a2=0 (degree at most two); e(t)=−2t+5 has derivative −2 (S10)."
        }
      },
      {
        "claim": {
          "zh-Hant": "|t−2| 在任何點都沒有導數",
          "en": "|t−2| has no derivative at any point"
        },
        "expected": "reject",
        "reason": {
          "zh-Hant": "t>2 時導數為 1，t<2 時為 −1（S14）；只有 t=2 不存在。",
          "en": "Its derivative is 1 for t>2 and −1 for t<2 (S14); only at t=2 does it fail to exist."
        }
      }
    ],
    "status": "candidate",
    "gaps": [],
    "verification": {
      "structure": "checked_local_script",
      "mathematics": "candidate_argument",
      "formalization": "not_formalized"
    },
    "rule_glossary": [
      {
        "id": "distributive",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "分配律：u(v+w)=uv+uw；配合乘法交換律也有 (v+w)u=vu+wu。",
          "en": "Distributivity: u(v+w)=uv+uw; with commutativity also (v+w)u=vu+wu."
        }
      },
      {
        "id": "frac_rules",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "分數運算（分母皆非零）：a/c+b/c=(a+b)/c；(a/b)(c/d)=(ac)/(bd)；(a/b)/(c/d)=(ad)/(bc)（另需 c≠0）；a/b=(ak)/(bk)（k≠0）。",
          "en": "Fraction rules (all denominators nonzero): a/c+b/c=(a+b)/c; (a/b)(c/d)=(ac)/(bd); (a/b)/(c/d)=(ad)/(bc) (also needing c≠0); a/b=(ak)/(bk) for k≠0."
        }
      },
      {
        "id": "abs_rules",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "絕對值（實數）：|u|=u（u≥0）、|u|=−u（u<0）；|u|≥0；|uv|=|u||v|；三角不等式 |u+v|≤|u|+|v|；|u|<r ⇔ −r<u<r。",
          "en": "Absolute value (reals): |u|=u if u≥0 and |u|=−u if u<0; |u|≥0; |uv|=|u||v|; triangle inequality |u+v|≤|u|+|v|; |u|<r ⇔ −r<u<r."
        }
      },
      {
        "id": "order_basics",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "實數的序：任兩數恰有 u<v、u=v、u>v 之一；u<v 且 v<w 推出 u<w；u<v 推出 u+w<v+w；u<v 且 w>0 推出 uw<vw；u<v 且 w<0 推出 uw>vw。",
          "en": "Order of the reals: exactly one of u<v, u=v, u>v holds; u<v and v<w give u<w; u<v gives u+w<v+w; u<v and w>0 give uw<vw; u<v and w<0 give uw>vw."
        }
      },
      {
        "id": "func_limit_def",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "函數極限：lim_{x→a} f(x)=L 表示對每個 ε>0，存在 δ>0，使所有滿足 0<|x−a|<δ 的定義域點 x 都有 |f(x)−L|<ε（a 須為定義域的極限點）。",
          "en": "Limit of a function: lim_{x→a} f(x)=L means that for every ε>0 there is δ>0 such that |f(x)−L|<ε for every domain point x with 0<|x−a|<δ (a must be a limit point of the domain)."
        }
      },
      {
        "id": "numeric_evaluation",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "明確整數或有理數算式的精確求值；可以獨立重算。",
          "en": "Exact evaluation of an explicit integer or rational expression; independently recomputable."
        }
      },
      {
        "id": "def_unfold",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "展開定義：把一個已定義的詞換成它的定義內容（或反向收回），命題的真假不變。",
          "en": "Unfolding a definition: replacing a defined term by its definition (or folding it back) does not change whether a statement is true."
        }
      },
      {
        "id": "power_laws",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "整數指數律（底數非零或指數為正整數）：a^m·a^n=a^{m+n}，(a^m)^n=a^{mn}，(ab)^n=a^n b^n，a^0=1。",
          "en": "Integer exponent laws (base nonzero, or exponents positive integers): a^m·a^n=a^{m+n}, (a^m)^n=a^{mn}, (ab)^n=a^n b^n, a^0=1."
        }
      },
      {
        "id": "add_comm",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "加法交換律：u+v=v+u。",
          "en": "Commutativity of addition: u+v=v+u."
        }
      },
      {
        "id": "add_assoc",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "加法結合律：(u+v)+w=u+(v+w)。",
          "en": "Associativity of addition: (u+v)+w=u+(v+w)."
        }
      },
      {
        "id": "eq_replace",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "等值代換：命題中的一個式子可換成與它相等的式子，命題的真假不變。",
          "en": "Replacement of equals: an expression inside a statement may be replaced by an equal expression without changing whether the statement is true."
        }
      },
      {
        "id": "add_inverse",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "加法反元素：每個 u 都有 −u，使 u+(−u)=0；u−v 定義為 u+(−v)。",
          "en": "Additive inverse: every u has −u with u+(−u)=0; u−v means u+(−v)."
        }
      },
      {
        "id": "add_zero",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "加法單位元：u+0=u。",
          "en": "Additive identity: u+0=u."
        }
      },
      {
        "id": "mul_inverse_nonzero",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "非零數的倒數：在 ℚ、ℝ、ℂ 中，若 u≠0，存在 1/u 使 u·(1/u)=1；對 u≠0 定義 v/u 為 v·(1/u)。u=0 時沒有倒數，v/0 不定義。",
          "en": "Inverse of a nonzero number: in ℚ, ℝ or ℂ, if u≠0 there is 1/u with u·(1/u)=1, and v/u means v·(1/u). 0 has no inverse, so v/0 is undefined."
        }
      },
      {
        "id": "forall_intro",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "∀引入（全稱推廣）：若某命題對一個「任意的」元素 x 證成，而證明沒有對 x 加任何額外假設，則它對論域中所有 x 成立。",
          "en": "∀-introduction (generalization): if a statement is proved for an arbitrary element x, with no extra assumption on x, it holds for every x in the domain."
        }
      },
      {
        "id": "exists_intro",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "∃引入：若找到 a∈D 使 P(a) 成立，則 ∃x∈D, P(x)。",
          "en": "∃-introduction: if some a∈D satisfies P(a), then ∃x∈D, P(x)."
        }
      },
      {
        "id": "mul_zero",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "乘以零：u·0=0。理由：u·0=u·(0+0)=u·0+u·0，兩邊同減 u·0 得 0=u·0。",
          "en": "Multiplying by zero: u·0=0. Reason: u·0=u·(0+0)=u·0+u·0; subtracting u·0 from both sides gives 0=u·0."
        }
      },
      {
        "id": "case_analysis",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "分情況論證：若 P∨Q 成立，且 P⇒R、Q⇒R 都已證，則 R 成立。所有情況必須涵蓋全部可能。",
          "en": "Proof by cases: if P∨Q holds and both P⇒R and Q⇒R are proved, then R holds. The cases must cover every possibility."
        }
      },
      {
        "id": "contradiction",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "反證法：若假設 ¬P 推出矛盾（某命題與其否定同時成立），則 P 成立。",
          "en": "Proof by contradiction: if assuming ¬P leads to a contradiction (a statement and its negation both hold), then P holds."
        }
      },
      {
        "id": "exists_elim",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "∃消去：由 ∃x∈D, P(x)，可取一個新名字 a 表示某個滿足 P(a) 的元素；只要最後結論不含 a，就可用 a 推理。",
          "en": "∃-elimination: from ∃x∈D, P(x) we may name some element a with P(a) by a fresh name; reasoning with a is allowed as long as the final conclusion does not mention a."
        }
      },
      {
        "id": "quantifier_negation",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "量詞否定：¬(∀x, P(x)) 等價於 ∃x, ¬P(x)；¬(∃x, P(x)) 等價於 ∀x, ¬P(x)。",
          "en": "Negating quantifiers: ¬(∀x, P(x)) is equivalent to ∃x, ¬P(x), and ¬(∃x, P(x)) is equivalent to ∀x, ¬P(x)."
        }
      },
      {
        "id": "mko_derivative",
        "kind": "cited",
        "source": {
          "mko_id": "mko-derivative",
          "sha256": "b523a35b9e5c20d66a7346f84ac5bece2c4167b17d46208fdb9b40115deeaf64",
          "commit": "b28a5edbd46279b1db2a3f3f48bb3f447e18899f"
        },
        "statement": {
          "zh-Hant": "對 f:ℝ→ℝ，f 在 x 的導數是 lim_(h→0,h≠0)(f(x+h)−f(x))/h，若此兩側有限極限存在。f(t)=a2t²+a1t+a0（允許 a2=0）的導數為 2a2x+a1，有限差商為該值加 a2h；|t−c| 在 t>c 為 1、t<c 為 −1，t=c 不存在，不填 0。",
          "en": "For f:ℝ→ℝ, the derivative at x is lim_(h→0,h≠0)(f(x+h)−f(x))/h when this finite two-sided limit exists. For f(t)=a2t²+a1t+a0 (a2=0 allowed) the derivative is 2a2x+a1 and the finite quotient is that value plus a2h; for |t−c| the derivative is 1 above c and −1 below c, and at c it does not exist rather than being 0."
        }
      },
      {
        "id": "L_diff_quot",
        "kind": "lemma",
        "proved_in": [
          "S07.1",
          "S07.2"
        ],
        "statement": {
          "zh-Hant": "對 f(t)=a2t²+a1t+a0、實數 x 與 h≠0：(f(x+h)−f(x))/h=2a2x+a1+a2h。",
          "en": "For f(t)=a2t²+a1t+a0, a real x and h≠0: (f(x+h)−f(x))/h=2a2x+a1+a2h."
        }
      },
      {
        "id": "L_lin_limit",
        "kind": "lemma",
        "proved_in": [
          "S08.1",
          "S08.2"
        ],
        "statement": {
          "zh-Hant": "對實數 A、B：每個 ε>0 都有 δ>0，使 0<|h|<δ 時 |(A+Bh)−A|<ε；即 h→0 時 A+Bh→A。",
          "en": "For reals A and B: every ε>0 admits δ>0 with |(A+Bh)−A|<ε whenever 0<|h|<δ; that is, A+Bh→A as h→0."
        }
      }
    ],
    "revision_log": [],
    "numeric_checks": [
      {
        "id": "N1",
        "step": "S03",
        "vars": [
          "h"
        ],
        "degree": 2,
        "lhs": "3*(1+h)**2-2*(1+h)+5",
        "rhs": "6+4*h+3*h**2"
      },
      {
        "id": "N2",
        "step": "S07.1",
        "vars": [
          "a2",
          "a1",
          "a0",
          "x",
          "h"
        ],
        "degree": 2,
        "lhs": "(a2*(x+h)**2+a1*(x+h)+a0)-(a2*x**2+a1*x+a0)",
        "rhs": "h*(2*a2*x+a1+a2*h)"
      },
      {
        "id": "N3",
        "step": "S02",
        "lhs": "3*1**2-2*1+5",
        "rhs": "6"
      },
      {
        "id": "N4",
        "step": "S09",
        "lhs": "((3*(1+1/10)**2-2*(1+1/10)+5)-6)/(1/10)",
        "rhs": "43/10"
      },
      {
        "id": "N5",
        "step": "S09",
        "lhs": "43/10",
        "relation": "!=",
        "rhs": "4"
      },
      {
        "id": "N6",
        "step": "S07",
        "lhs": "2*3*1+(-2)",
        "rhs": "4"
      },
      {
        "id": "N7",
        "step": "S13",
        "lhs": "abs(1-(-1))",
        "rhs": "2"
      },
      {
        "id": "N8",
        "step": "S13",
        "lhs": "1/2+1/2",
        "relation": "<",
        "rhs": "2"
      },
      {
        "id": "N9",
        "step": "S12",
        "lhs": "abs(1/10)/(1/10)",
        "rhs": "1"
      },
      {
        "id": "N10",
        "step": "S12",
        "lhs": "abs(-1/10)/(-1/10)",
        "rhs": "-1"
      },
      {
        "id": "N11",
        "step": "S10",
        "vars": [
          "x",
          "h"
        ],
        "degree": 1,
        "lhs": "(-2*(x+h)+5)-(-2*x+5)",
        "rhs": "-2*h"
      },
      {
        "id": "N12",
        "step": "S14",
        "vars": [
          "t",
          "h"
        ],
        "degree": 1,
        "lhs": "((t+h)-2)-(t-2)",
        "rhs": "h"
      }
    ]
  },
  "presentation": {
    "md_scope": {
      "zh-Hant": "只處理 MKO 的範圍：全實數論域、次數至多二的多項式（允許 a2=0）與 |t−c| 的尖點。不證明一般微分法則，不處理高次多項式或其他函數。不改動 canonical MKO 或英文 edition。",
      "en": "Stays inside the MKO: full real domain, polynomials of degree at most two (a2=0 allowed), and the corner of |t−c|. General differentiation rules, higher-degree polynomials and other functions are not treated. It does not change the canonical MKO or its English edition."
    },
    "md_symbols": [
      [
        "Q(h)=\\frac{f(x+h)-f(x)}{h}",
        "步長 h≠0 的差商",
        "the difference quotient at step h≠0"
      ],
      [
        "f'(x)",
        "Q(h) 在 h→0（h≠0）的兩側有限極限",
        "the finite two-sided limit of Q(h) as h→0 with h≠0"
      ],
      [
        "a_2,a_1,a_0",
        "f(t)=a2t²+a1t+a0 的係數",
        "the coefficients of f(t)=a2t²+a1t+a0"
      ],
      [
        "g(t)=|t-2|",
        "在 t=2 有尖點的函數",
        "the function with a corner at t=2"
      ]
    ],
    "md_error_controls": [
      {
        "title": "區分有限差商與導數 / Distinguish a finite difference quotient from the derivative",
        "zh": "MKO 的誤解之一是把有限差商直接叫導數。h=1/10 的差商是 4.3，導數是 4，相差 a2h=0.3（S09）。仿射 e 的有限差商恰等於導數，只因 a2=0（S10）；這不能升格為一般規則。",
        "en": "One misconception the MKO lists is calling a finite difference quotient the derivative. The quotient at h=1/10 is 4.3 while the derivative is 4, a difference of a2h=0.3 (S09). For the affine e, finite quotients equal the derivative only because a2=0 (S10); this cannot be promoted to a general rule."
      },
      {
        "title": "h=0 不能代入差商 / Do not substitute h=0 into the quotient",
        "zh": "另一個誤解是令 h=0 照算。原差商在 h=0 是 0/0，沒有定義。正確順序是：先在 h≠0 處化簡（S05、S07.2），再求 h→0 的極限（S06、S08）。化簡後的 4+3h 在 h=0 的值 4 恰是極限，這是由 S06 的 ε-δ 論證保證的，不是代入原式得到的。",
        "en": "Another misconception is setting h=0 and computing anyway. The original quotient at h=0 is 0/0, undefined. The correct order is: simplify for h≠0 first (S05, S07.2), then take the limit as h→0 (S06, S08). That the simplified 4+3h has value 4 at h=0, equal to the limit, is guaranteed by the ε-δ argument of S06, not by substituting into the original quotient."
      },
      {
        "title": "尖點不填 0 / Do not assign zero at the corner",
        "zh": "|t−2| 在 2 的左右差商分別為 −1 與 1，兩側極限不存在（S13）。不能以兩者平均 0 或以全定義 API 的預設 0 冒充導數（S15）。",
        "en": "The left and right quotients of |t−2| at 2 are −1 and 1, so the two-sided limit does not exist (S13). Neither their average 0 nor the default 0 of a totalized API may pose as the derivative (S15)."
      }
    ],
    "md_open_items": [
      {
        "zh-Hant": "本稿只處理次數至多二的多項式與 |t−c|；一般微分法則見 mko-derivative-rules。",
        "en": "This guide treats only polynomials of degree at most two and |t−c|; general differentiation rules belong to mko-derivative-rules."
      }
    ]
  },
  "revision": {
    "id": "production-20261019-adopted-96",
    "source_commit": "b28a5edbd46279b1db2a3f3f48bb3f447e18899f",
    "adopted_source_sha256": "a75f90a9f71b90af25dc527ee007abb8d54e43fd249e71f3a43474a0803b1085",
    "original_source_sha256": "a75f90a9f71b90af25dc527ee007abb8d54e43fd249e71f3a43474a0803b1085",
    "owner_adopted": true,
    "bilingual_sync": "complete_source_bound",
    "formal_Lean": "NOT_RUN",
    "independent_AI_concurrence": "NOT_MEASURED",
    "issue_ids": [],
    "scope": "Owner-adopted natural-language teaching guide; not a formal proof, canonical MKO rewrite or evidence-status promotion."
  }
}