{
  "schema_version": "ocme-derivation-reader-v1",
  "guide_id": "derivation-mko-derivative-rules-01",
  "mko_id": "mko-derivative-rules",
  "title": {
    "zh-Hant": "微分法則：以展開求導核對乘積與鏈式規則",
    "en": "Derivative rules: checking the product and chain rules against differentiating the expansion"
  },
  "contract": {
    "schema_version": "ocme-derivation-guide-v0.1",
    "guide_id": "derivation-mko-derivative-rules-01",
    "source": {
      "mko_id": "mko-derivative-rules",
      "commit": "b28a5edbd46279b1db2a3f3f48bb3f447e18899f",
      "sha256": "4756c1f6226c8cecc330975377ab5d42ed0e461b0fe039c05959fc60ab7c628f",
      "english_sha256": "ed443b25aa7217b775b464078d3230bca85b98df615f3517b6284c5f828ded44"
    },
    "mode": "worked_example",
    "audience_prerequisites": [
      "distributive",
      "power_laws",
      "abs_rules",
      "order_basics",
      "frac_rules",
      "composition_def",
      "numeric_evaluation"
    ],
    "goal": {
      "zh-Hant": "以升冪係數列表 f=[2,0,1]、g=[1,3]（即 f(x)=2+x²、g(x)=1+3x）：先由導數定義證明三次以下多項式的求導公式；再以兩條獨立路徑求 (fg)' 與 (f∘g)'——展開後求導，以及使用乘積、鏈式規則——並確認兩者對每個 x 相同；同時核對和與純量倍的規則，指出 f'g'、漏乘內導數、外導數取錯點、反讀係數等錯誤，最後求 x=1 的值。",
      "en": "With the ascending coefficient lists f=[2,0,1] and g=[1,3] (that is f(x)=2+x² and g(x)=1+3x): first prove from the definition of the derivative the differentiation formula for polynomials of degree at most three; then find (fg)' and (f∘g)' along two independent routes, differentiating the expansion and applying the product and chain rules, and confirm they agree at every x; also check the sum and scalar rules, expose the errors f'g', a missing inner derivative, an outer derivative at the wrong point and reversed coefficient reading, and evaluate at x=1."
    },
    "givens": [
      {
        "id": "G1",
        "statement": {
          "zh-Hant": "多項式以升冪係數列表表示：[a0,a1,a2,a3] 表示 a0+a1x+a2x²+a3x³；末端零正規化；全實數論域。",
          "en": "Polynomials are ascending coefficient lists: [a0,a1,a2,a3] means a0+a1x+a2x²+a3x³; trailing zeros are normalized; the domain is all of ℝ."
        }
      },
      {
        "id": "G2",
        "statement": {
          "zh-Hant": "f=[2,0,1]，g=[1,3]。",
          "en": "f=[2,0,1], g=[1,3]."
        }
      },
      {
        "id": "G3",
        "statement": {
          "zh-Hant": "純量 c=−4；求值點 x=1。",
          "en": "Scalar c=−4; evaluation point x=1."
        }
      }
    ],
    "steps": [
      {
        "id": "S01",
        "uses": [
          "G1"
        ],
        "before_tex": "[a_0,a_1,a_2,a_3]",
        "after_tex": "a_0+a_1x+a_2x^2+a_3x^3",
        "relation": "definition",
        "rule": "mko_deriv_rules",
        "conditions": [],
        "explanation": {
          "zh-Hant": "依 MKO，列表由常數項起算、升冪排列：第 k 個位置（由 0 起算）是 x^k 的係數。末端補零是同一個多項式。",
          "en": "By the MKO, a list starts with the constant term in ascending order: position k (counting from 0) holds the coefficient of x^k. Appending trailing zeros gives the same polynomial."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "G1",
            "mko_deriv_rules"
          ],
          "explanation": {
            "zh-Hant": "此步只固定讀法。",
            "en": "This step only fixes the reading convention."
          }
        },
        "leaf_basis": "mko_deriv_rules + def_unfold"
      },
      {
        "id": "S02",
        "uses": [
          "G2",
          "S01"
        ],
        "before_tex": "f=[2,0,1],\\ g=[1,3]",
        "after_tex": "f(x)=2+x^2,\\ g(x)=1+3x",
        "relation": "equivalent",
        "rule": "def_unfold",
        "conditions": [],
        "explanation": {
          "zh-Hant": "依升冪讀法，[2,0,1] 是 2+0x+1x²=2+x²，[1,3] 是 1+3x。若誤以降冪讀，[2,0,1] 會變成 2x²+1，[1,3] 會變成 x+3，是不同的多項式。",
          "en": "By the ascending reading, [2,0,1] is 2+0x+1x²=2+x² and [1,3] is 1+3x. Misread in descending order, [2,0,1] would become 2x²+1 and [1,3] would become x+3, different polynomials."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "def_unfold"
          ],
          "explanation": {
            "zh-Hant": "由多項式讀回升冪列表得原列表，雙向一致。",
            "en": "Reading the polynomials back into ascending lists returns the original lists, so it holds both ways."
          }
        },
        "leaf_basis": "def_unfold"
      },
      {
        "id": "S03",
        "uses": [
          "mko_derivative_def"
        ],
        "before_tex": "p(x)=c_0+c_1x+c_2x^2+c_3x^3",
        "after_tex": "p'(x)=c_1+2c_2x+3c_3x^2\\ (\\forall x\\in\\mathbb{R})",
        "relation": "implies",
        "rule": "L_poly_deriv3",
        "conditions": [],
        "explanation": {
          "zh-Hant": "依引理 L_poly_deriv3，三次以下多項式處處可微，導數列表為 [c1,2c2,3c3]。這是「展開後求導」路徑的依據，由導數定義直接證明（S03.1 至 S03.4），不引用乘積或鏈式規則，所以可獨立核對那些規則。",
          "en": "By Lemma L_poly_deriv3, a polynomial of degree at most three is differentiable everywhere with derivative list [c1,2c2,3c3]. This underlies the 'differentiate the expansion' route and is proved directly from the definition (S03.1 to S03.4) without the product or chain rules, so it can check those rules independently."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "L_poly_deriv3",
            "mko_derivative_def"
          ],
          "explanation": {
            "zh-Hant": "引理只涵蓋三次以下；本稿所有求導的多項式都在此範圍內。",
            "en": "The lemma covers degree at most three only; every polynomial differentiated in this guide is in that range."
          }
        },
        "leaf_basis": "mko_derivative_def + distributive + abs_rules + order_basics"
      },
      {
        "id": "S03.1",
        "uses": [
          "distributive",
          "power_laws"
        ],
        "before_tex": "(x+h)^2-x^2,\\ (x+h)^3-x^3",
        "after_tex": "h(2x+h),\\ h(3x^2+3xh+h^2)",
        "relation": "equivalent",
        "rule": "distributive",
        "conditions": [],
        "explanation": {
          "zh-Hant": "展開 (x+h)²=x²+2xh+h²、(x+h)³=x³+3x²h+3xh²+h³，減去 x²、x³ 後提出 h（N1）。",
          "en": "Expand (x+h)²=x²+2xh+h² and (x+h)³=x³+3x²h+3xh²+h³, subtract x² and x³, and factor out h (N1)."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "distributive",
            "power_laws"
          ],
          "explanation": {
            "zh-Hant": "恆等式，雙向成立。",
            "en": "Identities, valid both ways."
          }
        },
        "leaf_basis": "distributive + power_laws + add_inverse"
      },
      {
        "id": "S03.2",
        "uses": [
          "S03.1",
          "frac_rules"
        ],
        "before_tex": "\\frac{p(x+h)-p(x)}{h}\\ (h\\neq 0)",
        "after_tex": "(c_1+2c_2x+3c_3x^2)+h\\,R,\\quad R=c_2+c_3(3x+h)",
        "relation": "equivalent",
        "rule": "frac_rules",
        "conditions": [
          "h\\neq 0"
        ],
        "explanation": {
          "zh-Hant": "由 S03.1，p(x+h)−p(x)=c1h+c2h(2x+h)+c3h(3x²+3xh+h²)=h[(c1+2c2x+3c3x²)+h(c2+3c3x+c3h)]（N2 以六個變數的多項式恆等式核對）。h≠0 時除以 h。令 D=c1+2c2x+3c3x²，差商為 D+hR。",
          "en": "By S03.1, p(x+h)−p(x)=c1h+c2h(2x+h)+c3h(3x²+3xh+h²)=h[(c1+2c2x+3c3x²)+h(c2+3c3x+c3h)] (N2 checks this as a polynomial identity in six variables). For h≠0 divide by h. With D=c1+2c2x+3c3x², the quotient is D+hR."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "frac_rules",
            "distributive"
          ],
          "explanation": {
            "zh-Hant": "h≠0 時乘除 h 互逆。",
            "en": "For h≠0, multiplying and dividing by h undo each other."
          }
        },
        "leaf_basis": "frac_rules + distributive + mul_inverse_nonzero"
      },
      {
        "id": "S03.3",
        "uses": [
          "S03.2",
          "abs_rules"
        ],
        "before_tex": "0<|h|\\le 1",
        "after_tex": "|R|\\le M=|c_2|+|c_3|(3|x|+1)",
        "relation": "implies",
        "rule": "abs_rules",
        "conditions": [
          "|h|\\le 1"
        ],
        "explanation": {
          "zh-Hant": "三角不等式與 |uv|=|u||v|：|R|≤|c2|+|c3||3x+h|≤|c2|+|c3|(3|x|+|h|)≤|c2|+|c3|(3|x|+1)，最後一步用 |h|≤1。M 只依 x 與係數，不依 h。",
          "en": "Triangle inequality and |uv|=|u||v|: |R|≤|c2|+|c3||3x+h|≤|c2|+|c3|(3|x|+|h|)≤|c2|+|c3|(3|x|+1), the last step using |h|≤1. M depends only on x and the coefficients, not on h."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "abs_rules",
            "order_basics"
          ],
          "explanation": {
            "zh-Hant": "只給上界；需要 |h|≤1，這在 S03.4 由 δ≤1 保證。",
            "en": "Only an upper bound; it needs |h|≤1, which S03.4 ensures through δ≤1."
          }
        },
        "leaf_basis": "abs_rules + order_basics"
      },
      {
        "id": "S03.4",
        "uses": [
          "S03.2",
          "S03.3",
          "mko_derivative_def"
        ],
        "before_tex": "\\varepsilon>0,\\ \\delta=\\min\\left(1,\\tfrac{\\varepsilon}{M+1}\\right)",
        "after_tex": "0<|h|<\\delta\\Rightarrow|h||R|\\le|h|M<\\delta(M+1)\\le\\varepsilon;\\ p'(x)=D",
        "relation": "implies",
        "rule": "mko_derivative_def",
        "conditions": [
          "M+1>0"
        ],
        "explanation": {
          "zh-Hant": "M+1>0，δ>0。若 0<|h|<δ，則 |h|<1，S03.3 適用：|(D+hR)−D|=|h||R|≤|h|M≤|h|(M+1)<δ(M+1)≤ε。ε 任意，依導數定義 p'(x)=D=c1+2c2x+3c3x²。S03.1 至 S03.4 合起來證得引理 L_poly_deriv3。",
          "en": "M+1>0 and δ>0. If 0<|h|<δ then |h|<1 and S03.3 applies: |(D+hR)−D|=|h||R|≤|h|M≤|h|(M+1)<δ(M+1)≤ε. As ε is arbitrary, the definition gives p'(x)=D=c1+2c2x+3c3x². S03.1 to S03.4 together prove Lemma L_poly_deriv3."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S03.2",
            "S03.3",
            "mko_derivative_def",
            "forall_intro"
          ],
          "explanation": {
            "zh-Hant": "用 M+1 而非 M，避免 M=0 時除以 0；嚴格不等來自 |h|<δ。",
            "en": "M+1 is used instead of M to avoid dividing by 0 when M=0; strictness comes from |h|<δ."
          }
        },
        "leaf_basis": "mko_derivative_def + forall_intro + order_basics + abs_rules"
      },
      {
        "id": "S04",
        "uses": [
          "S02",
          "S03"
        ],
        "before_tex": "f=[2,0,1],\\ g=[1,3]",
        "after_tex": "f'(x)=2x\\ ([0,2]),\\quad g'(x)=3\\ ([3])",
        "relation": "implies",
        "rule": "L_poly_deriv3",
        "conditions": [],
        "explanation": {
          "zh-Hant": "依引理：f 的 c1=0、c2=1，f'=[0,2·1]=[0,2]，即 2x；g 的 c1=3，g'=[3]。兩者處處可微，滿足乘積與鏈式規則的可微前提。",
          "en": "By the lemma: f has c1=0, c2=1, so f'=[0,2·1]=[0,2], that is 2x; g has c1=3, so g'=[3]. Both are differentiable everywhere, meeting the differentiability hypotheses of the product and chain rules."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S02",
            "S03"
          ],
          "explanation": {
            "zh-Hant": "直接代入引理。",
            "en": "A direct substitution into the lemma."
          }
        },
        "leaf_basis": "L_poly_deriv3 + numeric_evaluation"
      },
      {
        "id": "S05",
        "uses": [
          "S02",
          "distributive"
        ],
        "before_tex": "f(x)g(x)=(2+x^2)(1+3x)",
        "after_tex": "2+6x+x^2+3x^3\\ \\to\\ [2,6,1,3]",
        "relation": "equivalent",
        "rule": "distributive",
        "conditions": [],
        "explanation": {
          "zh-Hant": "展開：2·1+2·3x+x²·1+x²·3x=2+6x+x²+3x³（N3）。乘積升到三次，新的 x³ 項不能截掉。",
          "en": "Expand: 2·1+2·3x+x²·1+x²·3x=2+6x+x²+3x³ (N3). The product rises to degree three, and the new x³ term may not be truncated."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "distributive"
          ],
          "explanation": {
            "zh-Hant": "恆等式，雙向成立。",
            "en": "An identity, valid both ways."
          }
        },
        "leaf_basis": "distributive + power_laws + mul_comm"
      },
      {
        "id": "S06",
        "uses": [
          "S05",
          "S03"
        ],
        "before_tex": "[2,6,1,3]",
        "after_tex": "(fg)'(x)=6+2x+9x^2\\ \\to\\ [6,2,9]",
        "relation": "implies",
        "rule": "L_poly_deriv3",
        "conditions": [],
        "explanation": {
          "zh-Hant": "路徑一（展開後求導）：依引理，[c1,2c2,3c3]=[6,2·1,3·3]=[6,2,9]。",
          "en": "Route one (differentiate the expansion): by the lemma, [c1,2c2,3c3]=[6,2·1,3·3]=[6,2,9]."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S05",
            "S03"
          ],
          "explanation": {
            "zh-Hant": "不用乘積規則，只用 S03 的引理。",
            "en": "No product rule is used, only the lemma of S03."
          }
        },
        "leaf_basis": "L_poly_deriv3 + numeric_evaluation"
      },
      {
        "id": "S07",
        "uses": [
          "S02",
          "S04",
          "S06",
          "mko_deriv_rules"
        ],
        "before_tex": "f'(x)g(x)+f(x)g'(x)",
        "after_tex": "2x(1+3x)+(2+x^2)\\cdot 3=6+2x+9x^2=(fg)'(x)",
        "relation": "theorem_application",
        "rule": "mko_deriv_rules",
        "conditions": [
          "f,g\\ \\text{differentiable at}\\ x"
        ],
        "explanation": {
          "zh-Hant": "路徑二（乘積規則）：f'g+fg'=2x+6x²+6+3x²=6+2x+9x²（N4）。與 S06 是同一個多項式，所以兩條路徑對每個實數 x 都一致；這是多項式恆等式，不是有限點抽樣。",
          "en": "Route two (product rule): f'g+fg'=2x+6x²+6+3x²=6+2x+9x² (N4). It is the same polynomial as in S06, so the two routes agree at every real x; this is a polynomial identity, not finite sampling."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S04",
            "S06",
            "mko_deriv_rules"
          ],
          "explanation": {
            "zh-Hant": "規則的前提（f、g 在 x 可微）由 S04 提供；一致性只對這組 f、g 驗證，不證明一般的乘積規則。",
            "en": "The rule's hypothesis (f and g differentiable at x) comes from S04; agreement is verified for this f and g only and does not prove the product rule in general."
          }
        },
        "leaf_basis": "mko_deriv_rules + distributive + eq_refl_symm_trans"
      },
      {
        "id": "S08",
        "uses": [
          "S04",
          "S06"
        ],
        "before_tex": "f'(x)g'(x)",
        "after_tex": "2x\\cdot 3=6x;\\ x=0:\\ 0\\neq 6=(fg)'(0)",
        "relation": "implies",
        "rule": "quantifier_negation",
        "conditions": [],
        "explanation": {
          "zh-Hant": "常見錯誤是把乘積的導數寫成 f'g'=6x。在 x=0 它是 0，而 (fg)'(0)=6（N7），一個點不同就說明 (fg)'≠f'g'。",
          "en": "A common error writes the derivative of a product as f'g'=6x. At x=0 it is 0 while (fg)'(0)=6 (N7); one differing point shows (fg)'≠f'g'."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S04",
            "S06",
            "quantifier_negation"
          ],
          "explanation": {
            "zh-Hant": "一個反例否定「兩函數相等」。",
            "en": "One counterexample refutes equality of the two functions."
          }
        },
        "leaf_basis": "quantifier_negation + numeric_evaluation"
      },
      {
        "id": "S09",
        "uses": [
          "S02",
          "composition_def"
        ],
        "before_tex": "(f\\circ g)(x)=f(1+3x)",
        "after_tex": "2+(1+3x)^2=3+6x+9x^2\\ \\to\\ [3,6,9]",
        "relation": "equivalent",
        "rule": "composition_def",
        "conditions": [],
        "explanation": {
          "zh-Hant": "依合成定義 (f∘g)(x)=f(g(x))，把 f 的變數換成 1+3x：2+(1+3x)²=2+1+6x+9x²=3+6x+9x²（N5）。",
          "en": "By the definition of composition (f∘g)(x)=f(g(x)); replace the variable of f by 1+3x: 2+(1+3x)²=2+1+6x+9x²=3+6x+9x² (N5)."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "composition_def",
            "distributive"
          ],
          "explanation": {
            "zh-Hant": "定義代入與展開都是等式，雙向成立。",
            "en": "Substitution by definition and expansion are equalities, valid both ways."
          }
        },
        "leaf_basis": "composition_def + distributive + power_laws"
      },
      {
        "id": "S10",
        "uses": [
          "S09",
          "S03"
        ],
        "before_tex": "[3,6,9]",
        "after_tex": "(f\\circ g)'(x)=6+18x\\ \\to\\ [6,18]",
        "relation": "implies",
        "rule": "L_poly_deriv3",
        "conditions": [],
        "explanation": {
          "zh-Hant": "路徑一：依引理，[c1,2c2]=[6,2·9]=[6,18]。",
          "en": "Route one: by the lemma, [c1,2c2]=[6,2·9]=[6,18]."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S09",
            "S03"
          ],
          "explanation": {
            "zh-Hant": "不用鏈式規則。",
            "en": "No chain rule is used."
          }
        },
        "leaf_basis": "L_poly_deriv3 + numeric_evaluation"
      },
      {
        "id": "S11",
        "uses": [
          "S02",
          "S04",
          "S10",
          "mko_deriv_rules"
        ],
        "before_tex": "f'(g(x))\\,g'(x)",
        "after_tex": "2(1+3x)\\cdot 3=6+18x=(f\\circ g)'(x)",
        "relation": "theorem_application",
        "rule": "mko_deriv_rules",
        "conditions": [
          "g\\ \\text{differentiable at}\\ x",
          "f\\ \\text{differentiable at}\\ g(x)"
        ],
        "explanation": {
          "zh-Hant": "路徑二（鏈式規則）：外導數 f'(u)=2u 在 u=g(x)=1+3x 取值，得 2(1+3x)，再乘內導數 g'(x)=3：6+18x（N6），與 S10 相同，對每個 x 都一致。",
          "en": "Route two (chain rule): the outer derivative f'(u)=2u is evaluated at u=g(x)=1+3x, giving 2(1+3x), then multiplied by the inner derivative g'(x)=3: 6+18x (N6), the same as S10, agreeing at every x."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S04",
            "S10",
            "mko_deriv_rules"
          ],
          "explanation": {
            "zh-Hant": "前提：g 在 x 可微、f 在 g(x) 可微，S04 給出兩者處處可微。",
            "en": "Hypotheses: g differentiable at x and f at g(x); S04 gives differentiability everywhere for both."
          }
        },
        "leaf_basis": "mko_deriv_rules + distributive + eq_refl_symm_trans"
      },
      {
        "id": "S12",
        "uses": [
          "S04",
          "S10"
        ],
        "before_tex": "f'(x)g'(x),\\ f'(g(x))",
        "after_tex": "6x,\\ 2+6x;\\ x=0:\\ 0\\neq 6,\\ 2\\neq 6",
        "relation": "implies",
        "rule": "quantifier_negation",
        "conditions": [],
        "explanation": {
          "zh-Hant": "兩種錯誤：外導數在 x 而非 g(x) 取值，得 f'(x)g'(x)=6x；漏乘內導數，得 f'(g(x))=2+6x。在 x=0，正確值是 6，兩個錯誤分別給 0 與 2（N8、N9）。",
          "en": "Two errors: evaluating the outer derivative at x instead of g(x) gives f'(x)g'(x)=6x; omitting the inner derivative gives f'(g(x))=2+6x. At x=0 the correct value is 6, while the errors give 0 and 2 (N8, N9)."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S04",
            "S10",
            "quantifier_negation"
          ],
          "explanation": {
            "zh-Hant": "每個錯誤公式各以一點的反例否定。",
            "en": "Each wrong formula is refuted by a one-point counterexample."
          }
        },
        "leaf_basis": "quantifier_negation + numeric_evaluation"
      },
      {
        "id": "S13",
        "uses": [
          "S02",
          "S03",
          "S04",
          "G3",
          "mko_deriv_rules"
        ],
        "before_tex": "f+g=[3,3,1],\\ -4f=[-8,0,-4]",
        "after_tex": "(f+g)'=[3,2]=[0,2]+[3],\\ (-4f)'=[0,-8]=-4\\cdot[0,2]",
        "relation": "implies",
        "rule": "L_poly_deriv3",
        "conditions": [],
        "explanation": {
          "zh-Hant": "f+g=(2+1)+3x+x²=[3,3,1]（N12），依引理導數 [3,2]，即 3+2x，等於 f'+g'=2x+3。−4f=[−8,0,−4]，導數 [0,−8]，即 −8x，等於 −4·f'=−4·2x。和規則與純量規則在此成立；純量 0 會給零多項式 [0]，也要保留，不可刪除。",
          "en": "f+g=(2+1)+3x+x²=[3,3,1] (N12), with derivative [3,2] by the lemma, that is 3+2x, equal to f'+g'=2x+3. −4f=[−8,0,−4] has derivative [0,−8], that is −8x, equal to −4·f'=−4·2x. The sum and scalar rules hold here; a scalar 0 would give the zero polynomial [0], which must be kept, not deleted."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S03",
            "S04",
            "mko_deriv_rules"
          ],
          "explanation": {
            "zh-Hant": "兩條路徑對照；不證明一般的和、純量規則。",
            "en": "The two routes are compared; the general sum and scalar rules are not proved."
          }
        },
        "leaf_basis": "L_poly_deriv3 + mko_deriv_rules + numeric_evaluation"
      },
      {
        "id": "S14",
        "uses": [
          "S06",
          "S10",
          "G3"
        ],
        "before_tex": "(fg)'(1),\\ (f\\circ g)'(1)",
        "after_tex": "6+2+9=17,\\ 6+18=24",
        "relation": "equivalent",
        "rule": "numeric_evaluation",
        "conditions": [],
        "explanation": {
          "zh-Hant": "在 x=1 求值：(fg)'(1)=6+2·1+9·1²=17，(f∘g)'(1)=6+18·1=24（N10、N11）。",
          "en": "Evaluate at x=1: (fg)'(1)=6+2·1+9·1²=17 and (f∘g)'(1)=6+18·1=24 (N10, N11)."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "numeric_evaluation"
          ],
          "explanation": {
            "zh-Hant": "精確求值。",
            "en": "Exact evaluation."
          }
        },
        "leaf_basis": "numeric_evaluation"
      }
    ],
    "result": {
      "zh-Hant": "f=[2,0,1]=2+x²、g=[1,3]=1+3x，f'=2x、g'=3。fg=[2,6,1,3]，(fg)'=[6,2,9]=6+2x+9x²，與 f'g+fg' 對每個 x 相同；f'g'=6x 錯誤。f∘g=[3,6,9]，(f∘g)'=[6,18]=6+18x，與 f'(g(x))g'(x)=2(1+3x)·3 相同；f'(x)g'(x)=6x 與 f'(g(x))=2+6x 都錯誤。(f+g)'=[3,2]、(−4f)'=[0,−8]。在 x=1：(fg)'(1)=17，(f∘g)'(1)=24。",
      "en": "f=[2,0,1]=2+x² and g=[1,3]=1+3x, with f'=2x and g'=3. fg=[2,6,1,3] and (fg)'=[6,2,9]=6+2x+9x², the same as f'g+fg' at every x; f'g'=6x is wrong. f∘g=[3,6,9] and (f∘g)'=[6,18]=6+18x, the same as f'(g(x))g'(x)=2(1+3x)·3; both f'(x)g'(x)=6x and f'(g(x))=2+6x are wrong. (f+g)'=[3,2] and (−4f)'=[0,−8]. At x=1: (fg)'(1)=17 and (f∘g)'(1)=24."
    },
    "negative_controls": [
      {
        "claim": {
          "zh-Hant": "(fg)'=f'g'=6x",
          "en": "(fg)'=f'g'=6x"
        },
        "expected": "reject",
        "reason": {
          "zh-Hant": "(fg)'=6+2x+9x²；x=0 時 6≠0。",
          "en": "(fg)'=6+2x+9x²; at x=0, 6≠0."
        }
      },
      {
        "claim": {
          "zh-Hant": "(f∘g)'(x)=f'(x)g'(x)=6x",
          "en": "(f∘g)'(x)=f'(x)g'(x)=6x"
        },
        "expected": "reject",
        "reason": {
          "zh-Hant": "外導數要在 g(x) 取值；正確為 6+18x。",
          "en": "The outer derivative must be evaluated at g(x); the correct result is 6+18x."
        }
      },
      {
        "claim": {
          "zh-Hant": "(f∘g)'(x)=f'(g(x))=2+6x",
          "en": "(f∘g)'(x)=f'(g(x))=2+6x"
        },
        "expected": "reject",
        "reason": {
          "zh-Hant": "漏乘內導數 g'(x)=3。",
          "en": "The inner derivative g'(x)=3 is missing."
        }
      },
      {
        "claim": {
          "zh-Hant": "[2,0,1] 表示 2x²+0x+1",
          "en": "[2,0,1] means 2x²+0x+1"
        },
        "expected": "reject",
        "reason": {
          "zh-Hant": "列表是升冪，[2,0,1]=2+x²。",
          "en": "Lists are ascending: [2,0,1]=2+x²."
        }
      },
      {
        "claim": {
          "zh-Hant": "f 是二次、g 是一次，所以 (fg)' 至多一次，可以略去 x² 項",
          "en": "f is quadratic and g has degree one, so (fg)' has degree at most one and the x² term can be dropped"
        },
        "expected": "reject",
        "reason": {
          "zh-Hant": "fg 是三次，(fg)'=6+2x+9x² 是二次；不能截掉新項。",
          "en": "fg has degree three and (fg)'=6+2x+9x² has degree two; new terms may not be truncated."
        }
      }
    ],
    "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": "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": "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": "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": "composition_def",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "合成：對 f:A→B 與 g:B→C，g∘f:A→C 定義為 (g∘f)(a)=g(f(a))。",
          "en": "Composition: for f:A→B and g:B→C, g∘f:A→C is defined by (g∘f)(a)=g(f(a))."
        }
      },
      {
        "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": "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": "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": "mul_comm",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "乘法交換律：uv=vu（在交換的數系中，例如 ℤ、ℚ、ℝ、ℂ）。",
          "en": "Commutativity of multiplication: uv=vu (in commutative number systems such as ℤ, ℚ, ℝ, ℂ)."
        }
      },
      {
        "id": "eq_refl_symm_trans",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "等號的性質：u=u；若 u=v 則 v=u；若 u=v 且 v=w 則 u=w。",
          "en": "Properties of equality: u=u; if u=v then v=u; if u=v and v=w then u=w."
        }
      },
      {
        "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_deriv_rules",
        "kind": "cited",
        "source": {
          "mko_id": "mko-derivative-rules",
          "sha256": "4756c1f6226c8cecc330975377ab5d42ed0e461b0fe039c05959fc60ab7c628f",
          "commit": "b28a5edbd46279b1db2a3f3f48bb3f447e18899f"
        },
        "statement": {
          "zh-Hant": "在相應點可微的實函數有 (f+g)'=f'+g'、(cf)'=cf'、(fg)'=f'g+fg'；若 g 在 x 可微且 f 在 g(x) 可微，(f∘g)'(x)=f'(g(x))g'(x)。本輪 f、g 是全 ℝ 多項式，以常數項起算的升冪係數列表表示。",
          "en": "For real functions differentiable at the relevant points, (f+g)'=f'+g', (cf)'=cf' and (fg)'=f'g+fg'; if g is differentiable at x and f at g(x), then (f∘g)'(x)=f'(g(x))g'(x). Here f and g are polynomials defined on all of ℝ given by ascending coefficient lists starting with the constant term."
        }
      },
      {
        "id": "mko_derivative_def",
        "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。",
          "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."
        }
      },
      {
        "id": "L_poly_deriv3",
        "kind": "lemma",
        "proved_in": [
          "S03.1",
          "S03.2",
          "S03.3",
          "S03.4"
        ],
        "statement": {
          "zh-Hant": "對 p(x)=c0+c1x+c2x²+c3x³（實係數）與每個實數 x：p 在 x 可微且 p'(x)=c1+2c2x+3c3x²；即列表 [c0,c1,c2,c3] 的導數列表為 [c1,2c2,3c3]。",
          "en": "For p(x)=c0+c1x+c2x²+c3x³ with real coefficients and every real x: p is differentiable at x and p'(x)=c1+2c2x+3c3x²; that is, the list [c0,c1,c2,c3] has derivative list [c1,2c2,3c3]."
        }
      }
    ],
    "revision_log": [],
    "numeric_checks": [
      {
        "id": "N1",
        "step": "S03.1",
        "vars": [
          "x",
          "h"
        ],
        "degree": 3,
        "lhs": "(x+h)**3-x**3",
        "rhs": "h*(3*x**2+3*x*h+h**2)"
      },
      {
        "id": "N2",
        "step": "S03.2",
        "vars": [
          "c0",
          "c1",
          "c2",
          "c3",
          "x",
          "h"
        ],
        "degree": 3,
        "lhs": "(c0+c1*(x+h)+c2*(x+h)**2+c3*(x+h)**3)-(c0+c1*x+c2*x**2+c3*x**3)",
        "rhs": "h*((c1+2*c2*x+3*c3*x**2)+h*(c2+c3*(3*x+h)))"
      },
      {
        "id": "N3",
        "step": "S05",
        "vars": [
          "x"
        ],
        "degree": 3,
        "lhs": "(2+x**2)*(1+3*x)",
        "rhs": "2+6*x+x**2+3*x**3"
      },
      {
        "id": "N4",
        "step": "S07",
        "vars": [
          "x"
        ],
        "degree": 2,
        "lhs": "2*x*(1+3*x)+(2+x**2)*3",
        "rhs": "6+2*x+9*x**2"
      },
      {
        "id": "N5",
        "step": "S09",
        "vars": [
          "x"
        ],
        "degree": 2,
        "lhs": "2+(1+3*x)**2",
        "rhs": "3+6*x+9*x**2"
      },
      {
        "id": "N6",
        "step": "S11",
        "vars": [
          "x"
        ],
        "degree": 1,
        "lhs": "2*(1+3*x)*3",
        "rhs": "6+18*x"
      },
      {
        "id": "N7",
        "step": "S08",
        "lhs": "2*0*3",
        "relation": "!=",
        "rhs": "6+2*0+9*0**2"
      },
      {
        "id": "N8",
        "step": "S12",
        "lhs": "2*0*3",
        "relation": "!=",
        "rhs": "6+18*0"
      },
      {
        "id": "N9",
        "step": "S12",
        "lhs": "2*(1+3*0)",
        "relation": "!=",
        "rhs": "6+18*0"
      },
      {
        "id": "N10",
        "step": "S14",
        "lhs": "6+2*1+9*1**2",
        "rhs": "17"
      },
      {
        "id": "N11",
        "step": "S14",
        "lhs": "6+18*1",
        "rhs": "24"
      },
      {
        "id": "N12",
        "step": "S13",
        "vars": [
          "x"
        ],
        "degree": 2,
        "lhs": "(2+x**2)+(1+3*x)",
        "rhs": "3+3*x+x**2"
      },
      {
        "id": "N13",
        "lhs": "2+5**2",
        "rhs": "27"
      }
    ]
  },
  "presentation": {
    "md_scope": {
      "zh-Hant": "只處理 MKO 的多項式切片（輸入至多三次、升冪列表）。和、純量倍、乘積、鏈式規則本身引自 MKO，本稿只對這組 f、g 以多項式恆等式核對它們，並由定義證明三次以下多項式的求導公式；不證明一般可微函數的規則。不改動 canonical MKO 或英文 edition。",
      "en": "Stays within the MKO's polynomial slice (inputs of degree at most three, ascending lists). The sum, scalar, product and chain rules themselves are cited from the MKO; this guide only checks them for this f and g as polynomial identities and proves from the definition the derivative formula for polynomials of degree at most three; it does not prove the rules for general differentiable functions. It does not change the canonical MKO or its English edition."
    },
    "md_symbols": [
      [
        "[a_0,a_1,a_2,a_3]",
        "升冪係數列表 a0+a1x+a2x²+a3x³",
        "the ascending coefficient list for a0+a1x+a2x²+a3x³"
      ],
      [
        "f',\\ g'",
        "f、g 的導函數",
        "the derivatives of f and g"
      ],
      [
        "fg,\\ f\\circ g",
        "乘積與合成 f(g(x))",
        "the product and the composition f(g(x))"
      ],
      [
        "R,\\ M",
        "差商的餘項與其界",
        "the remainder of the quotient and its bound"
      ]
    ],
    "md_error_controls": [
      {
        "title": "乘積的導數不是導數的乘積 / The derivative of a product is not the product of derivatives",
        "zh": "MKO 的誤解之一是乘積導數直接乘兩導數。f'g'=6x，而 (fg)'=6+2x+9x²；x=0 時 0≠6（S08）。正確的是 f'g+fg'（S07），它與展開後求導（S06）一致。",
        "en": "One misconception the MKO lists is multiplying the two derivatives. f'g'=6x while (fg)'=6+2x+9x²; at x=0, 0≠6 (S08). The correct form is f'g+fg' (S07), which agrees with differentiating the expansion (S06)."
      },
      {
        "title": "鏈式規則：外導數在 g(x) 取值，再乘內導數 / Chain rule: outer derivative at g(x), times the inner derivative",
        "zh": "另兩個誤解：外導數在 x 取值得 6x，漏乘內導數得 2+6x（S12），都與正確的 6+18x 不同。若內函數是常數，例如 g=[5]，則 f∘g 是常數 f(5)=27，導數 0，鏈式規則也給 f'(5)·0=0，這個零結果要保留。",
        "en": "Two more misconceptions: evaluating the outer derivative at x gives 6x, and omitting the inner derivative gives 2+6x (S12); both differ from the correct 6+18x. If the inner function is constant, for example g=[5], then f∘g is the constant f(5)=27 with derivative 0, and the chain rule also gives f'(5)·0=0; this zero result must be kept."
      },
      {
        "title": "係數列表是升冪 / Coefficient lists are ascending",
        "zh": "把 [2,0,1] 反讀成 2x²+1 會得到導數 4x 而不是 2x，輸入多項式因而改變，後續公式或數值可能錯；某些測點即使偶然相同，也不能使反讀約定正確。末端補零（[2,0,1,0]）是同一個多項式，不是新題；乘積與合成可能升次，新項不能截掉（S05）。",
        "en": "Reading [2,0,1] backwards as 2x²+1 gives the derivative 4x instead of 2x, and this changes the input polynomial and can invalidate later formulas or values; accidental agreement at a test point does not validate the reversed convention. Appending a trailing zero ([2,0,1,0]) gives the same polynomial, not a new exercise; products and compositions may raise the degree, and new terms may not be truncated (S05)."
      }
    ],
    "md_open_items": [
      {
        "zh-Hant": "和、純量、乘積、鏈式規則引自 MKO，本稿只對這組 f、g 以多項式恆等式核對，不證明一般可微函數的規則；引理 L_poly_deriv3 只涵蓋三次以下多項式，三次輸入的乘積（至多六次）不在本稿求導範圍內。",
        "en": "The sum, scalar, product and chain rules are cited from the MKO; this guide checks them only for this f and g as polynomial identities and does not prove them for general differentiable functions. Lemma L_poly_deriv3 covers degree at most three, so products of cubic inputs (up to degree six) are outside what this guide differentiates."
      }
    ]
  },
  "revision": {
    "id": "production-20261019-adopted-96",
    "source_commit": "b28a5edbd46279b1db2a3f3f48bb3f447e18899f",
    "adopted_source_sha256": "d018b0ec61f9ad721df0ccc4872b72a742b428f70ec5a6bc3709008d1983cb9f",
    "original_source_sha256": "28b99c11ef6db059516f55a9a5cd9265cb5ca4d59d4d271549486abdf6c12833",
    "owner_adopted": true,
    "bilingual_sync": "complete_source_bound",
    "formal_Lean": "NOT_RUN",
    "independent_AI_concurrence": "NOT_MEASURED",
    "issue_ids": [
      "BR-M026"
    ],
    "scope": "Owner-adopted natural-language teaching guide; not a formal proof, canonical MKO rewrite or evidence-status promotion."
  }
}