{
  "schema_version": "ocme-derivation-reader-v1",
  "guide_id": "derivation-mko-partial-derivative-01",
  "mko_id": "mko-partial-derivative",
  "title": {
    "zh-Hant": "偏導數：由差商極限推出二次多項式的偏導公式，並計算一點的梯度",
    "en": "Partial derivative: deriving the partial-derivative formulas of a quadratic from the difference-quotient limit, and computing a gradient"
  },
  "contract": {
    "schema_version": "ocme-derivation-guide-v0.1",
    "guide_id": "derivation-mko-partial-derivative-01",
    "source": {
      "mko_id": "mko-partial-derivative",
      "commit": "b28a5edbd46279b1db2a3f3f48bb3f447e18899f",
      "sha256": "7159526c8e30cc8cb9086f9690d279fd3852328abe371206b3441ecf6ef69ce8",
      "english_sha256": "11ec37b647f687ce00aba1864579424c7da8dccdbac8051d553f0a9763c182fe"
    },
    "mode": "theorem_proof",
    "audience_prerequisites": [
      "func_limit_def",
      "abs_rules",
      "order_basics",
      "distributive",
      "power_laws"
    ],
    "goal": {
      "zh-Hant": "對係數依 [1,x,y,x²,xy,y²] 排列的二次多項式 f，從差商極限的定義證明 ∂f/∂x=c_1+2c_3x+c_4y、∂f/∂y=c_2+c_4x+2c_5y。再對 c=(1,−2,3,1,2,−1) 算出 (1,2) 的偏導與梯度 (4,1)，並說明單一非零 h 的差商（5 或 41/10）與同時移動兩個座標（極限 5）都不是 x 偏導。",
      "en": "For a quadratic f with coefficients in the order [1,x,y,x²,xy,y²], prove from the difference-quotient limit that ∂f/∂x=c_1+2c_3x+c_4y and ∂f/∂y=c_2+c_4x+2c_5y. Then for c=(1,−2,3,1,2,−1) compute the partials and the gradient (4,1) at (1,2), and show that a single nonzero-h quotient (5 or 41/10) and moving both coordinates at once (limit 5) are not the x-partial."
    },
    "givens": [
      {
        "id": "G1",
        "statement": {
          "zh-Hant": "f(x,y)=c_0+c_1x+c_2y+c_3x²+c_4xy+c_5y²，係數為整數，次序固定為 [1,x,y,x²,xy,y²]，f 在 ℝ² 上處處有定義（MKO 假設 a1、a2）。",
          "en": "f(x,y)=c_0+c_1x+c_2y+c_3x²+c_4xy+c_5y² with integer coefficients in the fixed order [1,x,y,x²,xy,y²]; f is defined on all of ℝ² (MKO assumptions a1, a2)."
        }
      },
      {
        "id": "G2",
        "statement": {
          "zh-Hant": "實例：c=(1,−2,3,1,2,−1)，即 f(x,y)=1−2x+3y+x²+2xy−y²，點 (1,2)。",
          "en": "Instance: c=(1,−2,3,1,2,−1), that is f(x,y)=1−2x+3y+x²+2xy−y², at the point (1,2)."
        }
      }
    ],
    "steps": [
      {
        "id": "S01",
        "uses": [
          "mko_partial_def"
        ],
        "before_tex": "f_x(x,y)",
        "after_tex": "\\lim_{h\\to0,\\ h\\neq0}\\frac{f(x+h,y)-f(x,y)}{h}",
        "relation": "definition",
        "rule": "mko_partial_def",
        "conditions": [
          "y\\ \\text{fixed}"
        ],
        "explanation": {
          "zh-Hant": "依 MKO 展開 x 偏導：y 固定不動，只讓第一個座標增加 h（h≠0），取差商後令 h→0。y 偏導則固定 x、只動第二個座標（MKO 段落 0）。偏導是一個極限，不是某一個 h 的差商值。",
          "en": "Unfold the x-partial by the MKO: y stays fixed, only the first coordinate changes by h (h≠0), and the difference quotient is taken as h→0. The y-partial fixes x and moves only the second coordinate (MKO paragraph 0). A partial derivative is a limit, not the quotient at one particular h."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "mko_partial_def"
          ],
          "explanation": {
            "zh-Hant": "只展開定義；h=0 被排除，因為差商在 h=0 沒有定義。",
            "en": "Only unfolds the definition; h=0 is excluded because the quotient is undefined there."
          }
        },
        "leaf_basis": "mko_partial_def + def_unfold"
      },
      {
        "id": "S02",
        "uses": [
          "G1",
          "mko_mvf_def"
        ],
        "before_tex": "c=(c_0,c_1,c_2,c_3,c_4,c_5)",
        "after_tex": "f(x,y)=c_0+c_1x+c_2y+c_3x^2+c_4xy+c_5y^2",
        "relation": "definition",
        "rule": "mko_mvf_def",
        "conditions": [],
        "explanation": {
          "zh-Hant": "依固定次序把係數向量讀成多項式。之後的推導對任意係數成立，最後再代入 G2 的實例。",
          "en": "Read the coefficient vector as a polynomial in the fixed order. The derivation below works for arbitrary coefficients, and the instance of G2 is substituted at the end."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "mko_mvf_def"
          ],
          "explanation": {
            "zh-Hant": "係數次序是輸入契約，錯置可能改變多項式與偏導；但不是每個置換都一定改變結果，例如交換相等係數不變。本題 c_1=−2、c_2=3 不相等，交換後確會改變所列偏導。",
            "en": "Coefficient order is part of the input contract. A permutation can change the polynomial and its partials, but not every permutation must do so: swapping equal coefficients changes nothing. The worked c_1=−2 and c_2=3 are unequal, and swapping them does change the displayed partials."
          }
        },
        "leaf_basis": "mko_mvf_def + def_unfold"
      },
      {
        "id": "S03",
        "uses": [
          "S02",
          "L_numerator_x"
        ],
        "before_tex": "f(x+h,y)-f(x,y)",
        "after_tex": "h(c_1+2c_3x+c_4y)+c_3h^2",
        "relation": "equivalent",
        "rule": "L_numerator_x",
        "conditions": [],
        "explanation": {
          "zh-Hant": "引理 L_numerator_x（S03.1、S03.2 證明）：分子是 h 的二次式，常數項為 0。N1 在 3^9 個格點上驗證這個多項式恆等式（每個變數次數至多 2，足以判定）。",
          "en": "Lemma L_numerator_x (proved in S03.1, S03.2): the numerator is quadratic in h with zero constant term. N1 checks this polynomial identity on 3^9 grid points (degree at most 2 in each variable, which decides it)."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "L_numerator_x",
            "distributive"
          ],
          "explanation": {
            "zh-Hant": "這是代數恆等式，對所有 x、y、h 成立，雙向皆可。",
            "en": "This is an algebraic identity valid for all x, y, h, so it holds both ways."
          }
        },
        "leaf_basis": "L_numerator_x + distributive"
      },
      {
        "id": "S03.1",
        "uses": [
          "S02",
          "power_laws",
          "distributive"
        ],
        "before_tex": "c_0,\\ c_1x,\\ c_2y,\\ c_3x^2,\\ c_4xy,\\ c_5y^2\\ \\text{at}\\ (x+h,y)\\ \\text{minus at}\\ (x,y)",
        "after_tex": "0,\\ c_1h,\\ 0,\\ c_3(2xh+h^2),\\ c_4hy,\\ 0",
        "relation": "equivalent",
        "rule": "distributive",
        "conditions": [],
        "explanation": {
          "zh-Hant": "逐項相減：常數項、c_2y、c_5y² 不含 x，因為 y 固定而相消為 0；c_1(x+h)−c_1x=c_1h；c_3((x+h)²−x²)=c_3(2xh+h²)，用到 (x+h)²=x²+2xh+h²；c_4((x+h)y−xy)=c_4hy。",
          "en": "Subtract term by term: the constant, c_2y and c_5y² do not involve x and cancel to 0 because y is fixed; c_1(x+h)−c_1x=c_1h; c_3((x+h)²−x²)=c_3(2xh+h²), using (x+h)²=x²+2xh+h²; c_4((x+h)y−xy)=c_4hy."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "distributive",
            "power_laws",
            "add_inverse"
          ],
          "explanation": {
            "zh-Hant": "每一項都是恆等展開；只有含 x 的項留下，這正是「固定 y」的效果。",
            "en": "Each term is an identity expansion; only the terms containing x survive, which is exactly the effect of fixing y."
          }
        },
        "leaf_basis": "distributive + power_laws + add_inverse"
      },
      {
        "id": "S03.2",
        "uses": [
          "S03.1",
          "distributive"
        ],
        "before_tex": "c_1h+c_3(2xh+h^2)+c_4hy",
        "after_tex": "h(c_1+2c_3x+c_4y)+c_3h^2",
        "relation": "equivalent",
        "rule": "distributive",
        "conditions": [],
        "explanation": {
          "zh-Hant": "把含 h 一次方的項提出 h：c_1h+2c_3xh+c_4yh=h(c_1+2c_3x+c_4y)，剩下 c_3h²。這證得引理 L_numerator_x。",
          "en": "Factor h out of the terms of first degree in h: c_1h+2c_3xh+c_4yh=h(c_1+2c_3x+c_4y), leaving c_3h². This proves Lemma L_numerator_x."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "distributive",
            "mul_comm"
          ],
          "explanation": {
            "zh-Hant": "提出公因子與展開互逆。",
            "en": "Factoring and expanding undo each other."
          }
        },
        "leaf_basis": "distributive + mul_comm + add_comm"
      },
      {
        "id": "S04",
        "uses": [
          "S03",
          "mul_inverse_nonzero"
        ],
        "before_tex": "\\frac{h(c_1+2c_3x+c_4y)+c_3h^2}{h}",
        "after_tex": "q_x(h)=(c_1+2c_3x+c_4y)+c_3h\\quad(h\\neq0)",
        "relation": "equivalent",
        "rule": "mul_inverse_nonzero",
        "conditions": [
          "h\\neq0"
        ],
        "explanation": {
          "zh-Hant": "因 h≠0，可以乘以 1/h：h·A/h=A、c_3h²/h=c_3h。所以對每個非零 h，差商是 h 的仿射式 A+Bh，其中 A=c_1+2c_3x+c_4y、B=c_3。h=0 時差商沒有定義，但極限不看 h=0。",
          "en": "Since h≠0, we may multiply by 1/h: h·A/h=A and c_3h²/h=c_3h. So for every nonzero h, the quotient is the affine expression A+Bh in h, with A=c_1+2c_3x+c_4y and B=c_3. At h=0 the quotient is undefined, but the limit does not look at h=0."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "mul_inverse_nonzero",
            "frac_rules"
          ],
          "explanation": {
            "zh-Hant": "乘 1/h 與乘 h 互逆，前提 h≠0。",
            "en": "Multiplying by 1/h and by h undo each other, given h≠0."
          }
        },
        "leaf_basis": "mul_inverse_nonzero + frac_rules + power_laws"
      },
      {
        "id": "S05",
        "uses": [
          "S04",
          "L_affine_limit"
        ],
        "before_tex": "\\lim_{h\\to0}\\big(A+Bh\\big)",
        "after_tex": "A",
        "relation": "theorem_application",
        "rule": "L_affine_limit",
        "conditions": [
          "A=c_1+2c_3x+c_4y,\\ B=c_3"
        ],
        "explanation": {
          "zh-Hant": "引理 L_affine_limit（S05.1、S05.2 以 ε-δ 證明）：仿射式 A+Bh 在 h→0 時的極限是 A。這把極限從「直覺代入 h=0」變成可核對的論證。",
          "en": "Lemma L_affine_limit (proved by ε-δ in S05.1, S05.2): the affine expression A+Bh has limit A as h→0. This turns 'just substitute h=0' into an argument that can be checked."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "L_affine_limit",
            "func_limit_def"
          ],
          "explanation": {
            "zh-Hant": "0 是 ℝ∖{0} 的極限點，所以函數極限有意義。",
            "en": "0 is a limit point of ℝ∖{0}, so the function limit makes sense."
          }
        },
        "leaf_basis": "L_affine_limit + func_limit_def"
      },
      {
        "id": "S05.1",
        "uses": [
          "func_limit_def",
          "abs_rules"
        ],
        "before_tex": "\\varepsilon>0",
        "after_tex": "\\delta=\\frac{\\varepsilon}{|B|+1}>0",
        "relation": "implies",
        "rule": "exists_intro",
        "conditions": [
          "\\varepsilon>0"
        ],
        "explanation": {
          "zh-Hant": "任取 ε>0。|B|+1≥1>0，所以 δ=ε/(|B|+1) 是正數。加 1 是為了在 B=0 時也不會除以 0。",
          "en": "Take any ε>0. |B|+1≥1>0, so δ=ε/(|B|+1) is positive. The +1 avoids division by 0 when B=0."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "abs_rules",
            "order_basics"
          ],
          "explanation": {
            "zh-Hant": "正數除以正數為正；δ 只依賴 ε 與 B。",
            "en": "A positive number divided by a positive number is positive; δ depends only on ε and B."
          }
        },
        "leaf_basis": "exists_intro + abs_rules + order_basics"
      },
      {
        "id": "S05.2",
        "uses": [
          "S05.1",
          "abs_rules",
          "order_basics"
        ],
        "before_tex": "0<|h|<\\delta",
        "after_tex": "|(A+Bh)-A|=|B||h|\\le|B|\\delta=\\frac{|B|}{|B|+1}\\varepsilon<\\varepsilon",
        "relation": "implies",
        "rule": "abs_rules",
        "conditions": [
          "0<|h|<\\delta"
        ],
        "explanation": {
          "zh-Hant": "|(A+Bh)−A|=|Bh|=|B||h|。|h|<δ 且 |B|≥0 給 |B||h|≤|B|δ=|B|ε/(|B|+1)。因 |B|<|B|+1，比值 |B|/(|B|+1)<1，再乘以正的 ε 得 <ε。ε 任意，所以極限是 A；S05.1 與 S05.2 證得引理 L_affine_limit。",
          "en": "|(A+Bh)−A|=|Bh|=|B||h|. |h|<δ and |B|≥0 give |B||h|≤|B|δ=|B|ε/(|B|+1). Since |B|<|B|+1, the ratio |B|/(|B|+1)<1, and multiplying by the positive ε gives <ε. ε was arbitrary, so the limit is A; S05.1 and S05.2 prove Lemma L_affine_limit."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "abs_rules",
            "order_basics",
            "forall_intro"
          ],
          "explanation": {
            "zh-Hant": "用到 |uv|=|u||v|、乘非負數保序、乘正數保持嚴格不等；只宣稱上界。",
            "en": "Uses |uv|=|u||v|, that multiplying by a nonnegative number preserves ≤, and that multiplying by a positive number preserves <; only an upper bound is claimed."
          }
        },
        "leaf_basis": "abs_rules + order_basics + forall_intro"
      },
      {
        "id": "S06",
        "uses": [
          "S01",
          "S04",
          "S05"
        ],
        "before_tex": "f_x(x,y)=\\lim_{h\\to0}q_x(h)",
        "after_tex": "f_x(x,y)=c_1+2c_3x+c_4y",
        "relation": "implies",
        "rule": "L_affine_limit",
        "conditions": [],
        "explanation": {
          "zh-Hant": "S04 說差商在 h≠0 時是 A+Bh，S05 說它的極限是 A。依 S01 的定義，x 偏導存在且等於 A=c_1+2c_3x+c_4y。這是對每一點 (x,y) 與任意係數成立的精確公式。",
          "en": "S04 says the quotient equals A+Bh for h≠0, and S05 says its limit is A. By the definition in S01, the x-partial exists and equals A=c_1+2c_3x+c_4y. This is an exact formula valid at every point (x,y) and for arbitrary coefficients."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S04",
            "S05",
            "S01"
          ],
          "explanation": {
            "zh-Hant": "兩個函數在 h≠0 上相同，極限就相同；h=0 的值不參與。",
            "en": "Two functions that agree for h≠0 have the same limit; the value at h=0 plays no role."
          }
        },
        "leaf_basis": "L_affine_limit + def_unfold"
      },
      {
        "id": "S07",
        "uses": [
          "S01",
          "S02",
          "S05"
        ],
        "before_tex": "\\frac{f(x,y+h)-f(x,y)}{h}=(c_2+c_4x+2c_5y)+c_5h",
        "after_tex": "f_y(x,y)=c_2+c_4x+2c_5y",
        "relation": "implies",
        "rule": "L_affine_limit",
        "conditions": [
          "h\\neq0",
          "x\\ \\text{fixed}"
        ],
        "explanation": {
          "zh-Hant": "y 偏導固定 x、只動 y。同樣逐項相減：f(x,y+h)−f(x,y)=h(c_2+c_4x+2c_5y)+c_5h²（N2 為恆等式），除以 h≠0 得仿射式 A'+B'h，A'=c_2+c_4x+2c_5y、B'=c_5。由 L_affine_limit，f_y=A'。",
          "en": "The y-partial fixes x and moves only y. Subtracting term by term in the same way, f(x,y+h)−f(x,y)=h(c_2+c_4x+2c_5y)+c_5h² (N2 is an identity); dividing by h≠0 gives the affine A'+B'h with A'=c_2+c_4x+2c_5y and B'=c_5. By L_affine_limit, f_y=A'."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "L_affine_limit",
            "distributive",
            "mul_inverse_nonzero"
          ],
          "explanation": {
            "zh-Hant": "與 S03 到 S06 相同的四步，只是 x、y 的角色互換；N2 獨立驗證分子恆等式。",
            "en": "The same four steps as S03 to S06 with the roles of x and y exchanged; N2 checks the numerator identity independently."
          }
        },
        "leaf_basis": "L_affine_limit + distributive + mul_inverse_nonzero"
      },
      {
        "id": "S08",
        "uses": [
          "S06",
          "S07",
          "G2"
        ],
        "before_tex": "c=(1,-2,3,1,2,-1)",
        "after_tex": "f_x=-2+2x+2y,\\quad f_y=3+2x-2y",
        "relation": "equivalent",
        "rule": "eq_replace",
        "conditions": [],
        "explanation": {
          "zh-Hant": "代入 c_1=−2、c_2=3、c_3=1、c_4=2、c_5=−1：f_x=−2+2·1·x+2y=−2+2x+2y；f_y=3+2x+2·(−1)·y=3+2x−2y。",
          "en": "Substitute c_1=−2, c_2=3, c_3=1, c_4=2, c_5=−1: f_x=−2+2·1·x+2y=−2+2x+2y; f_y=3+2x+2·(−1)·y=3+2x−2y."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "eq_replace"
          ],
          "explanation": {
            "zh-Hant": "等量代換，雙向成立。",
            "en": "Substitution of equals, valid both ways."
          }
        },
        "leaf_basis": "eq_replace + numeric_evaluation"
      },
      {
        "id": "S09",
        "uses": [
          "S08"
        ],
        "before_tex": "(x,y)=(1,2)",
        "after_tex": "f_x(1,2)=4,\\ \\ f_y(1,2)=1,\\ \\ \\nabla f(1,2)=(4,1)",
        "relation": "equivalent",
        "rule": "numeric_evaluation",
        "conditions": [],
        "explanation": {
          "zh-Hant": "f_x(1,2)=−2+2+4=4（N3）；f_y(1,2)=3+2−4=1（N4）。梯度只是依座標次序收集兩個偏導：(4,1)（MKO 段落 1）。",
          "en": "f_x(1,2)=−2+2+4=4 (N3); f_y(1,2)=3+2−4=1 (N4). The gradient just collects the two partials in coordinate order: (4,1) (MKO paragraph 1)."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "numeric_evaluation"
          ],
          "explanation": {
            "zh-Hant": "精確整數求值。",
            "en": "Exact integer evaluation."
          }
        },
        "leaf_basis": "numeric_evaluation"
      },
      {
        "id": "S10",
        "uses": [
          "G2",
          "S04",
          "S09"
        ],
        "before_tex": "\\frac{f(1+h,2)-f(1,2)}{h}",
        "after_tex": "\\frac{4h+h^2}{h}=4+h\\ \\xrightarrow{h\\to0}\\ 4",
        "relation": "implies",
        "rule": "L_affine_limit",
        "conditions": [
          "h\\neq0"
        ],
        "explanation": {
          "zh-Hant": "直接核對實例：f(1,2)=6（N5），f(1+h,2)−6=4h+h²（N6 為 h 的恆等式），差商為 4+h，極限 4，與 S09 一致。這是一般公式之外的獨立計算。",
          "en": "Check the instance directly: f(1,2)=6 (N5) and f(1+h,2)−6=4h+h² (N6 is an identity in h), so the quotient is 4+h with limit 4, agreeing with S09. This is a computation independent of the general formula."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "L_affine_limit",
            "numeric_evaluation"
          ],
          "explanation": {
            "zh-Hant": "與 S06 的公式互相核對；只對這個實例。",
            "en": "Cross-checks the formula of S06; for this instance only."
          }
        },
        "leaf_basis": "L_affine_limit + numeric_evaluation + distributive"
      },
      {
        "id": "S11",
        "uses": [
          "S10"
        ],
        "before_tex": "q_x(h)=4+h",
        "after_tex": "q_x(1)=5,\\ \\ q_x(\\tfrac1{10})=\\tfrac{41}{10},\\ \\ q_x(h)\\neq4\\ \\text{for every}\\ h\\neq0",
        "relation": "implies",
        "rule": "order_basics",
        "conditions": [
          "h\\neq0"
        ],
        "explanation": {
          "zh-Hant": "h=1 給差商 (f(2,2)−6)/1=5（N7）；h=1/10 給 41/10（N8）。事實上 q_x(h)=4+h 對每個非零 h 都不等於 4。所以任何單一非零 h 的差商都不是偏導；偏導是 h→0 的極限（MKO 段落 2：有限差分只是近似或差商樣本）。",
          "en": "h=1 gives the quotient (f(2,2)−6)/1=5 (N7); h=1/10 gives 41/10 (N8). In fact q_x(h)=4+h differs from 4 for every nonzero h. So no single nonzero-h quotient is the partial derivative; the partial is the limit as h→0 (MKO paragraph 2: a finite difference is an approximation or a sample of the quotient)."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S10",
            "order_basics"
          ],
          "explanation": {
            "zh-Hant": "4+h=4 ⇔ h=0，而 h=0 被排除；這個結論只針對本實例。",
            "en": "4+h=4 ⇔ h=0, and h=0 is excluded; this conclusion concerns this instance only."
          }
        },
        "leaf_basis": "order_basics + numeric_evaluation"
      },
      {
        "id": "S12",
        "uses": [
          "G2",
          "S09",
          "L_affine_limit"
        ],
        "before_tex": "\\frac{f(1+h,2+h)-f(1,2)}{h}",
        "after_tex": "\\frac{5h+2h^2}{h}=5+2h\\ \\xrightarrow{h\\to0}\\ 5\\neq4=f_x(1,2)",
        "relation": "implies",
        "rule": "L_affine_limit",
        "conditions": [
          "h\\neq0"
        ],
        "explanation": {
          "zh-Hant": "若同時移動兩個座標：f(1+h,2+h)−6=5h+2h²（N9 為恆等式），差商 5+2h，由 L_affine_limit 極限為 5。這是沿 (1,1) 方向的變化率，在本例等於 f_x+f_y=4+1，不是 x 偏導 4。偏導必須固定另一個座標（MKO 指出的誤解）。",
          "en": "Moving both coordinates at once: f(1+h,2+h)−6=5h+2h² (N9 is an identity), the quotient is 5+2h, and by L_affine_limit its limit is 5. This is the rate of change along the direction (1,1), which here equals f_x+f_y=4+1, not the x-partial 4. A partial derivative must keep the other coordinate fixed (the misconception named by the MKO)."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "L_affine_limit",
            "numeric_evaluation"
          ],
          "explanation": {
            "zh-Hant": "只說明這個極限不是 f_x；「方向導數等於 f_x+f_y」只是本例的數值觀察，本稿不證明一般公式。",
            "en": "Only shows that this limit is not f_x; 'directional derivative equals f_x+f_y' is merely a numerical observation for this example, and no general formula is proved here."
          }
        },
        "leaf_basis": "L_affine_limit + numeric_evaluation + distributive"
      },
      {
        "id": "S13",
        "uses": [
          "S06",
          "S07",
          "S09",
          "S11",
          "S12"
        ],
        "before_tex": "\\text{S06 to S12}",
        "after_tex": "f_x=c_1+2c_3x+c_4y,\\ f_y=c_2+c_4x+2c_5y;\\ \\nabla f(1,2)=(4,1)",
        "relation": "implies",
        "rule": "and_intro_elim",
        "conditions": [],
        "explanation": {
          "zh-Hant": "總結：由差商極限證得二次多項式偏導公式；實例在 (1,2) 的梯度為 (4,1)；單一 h 的差商（5、41/10）與同時移動兩座標的極限（5）都不是 x 偏導。",
          "en": "Summary: the partial-derivative formulas of a quadratic follow from the difference-quotient limit; the instance has gradient (4,1) at (1,2); single-h quotients (5, 41/10) and the limit with both coordinates moving (5) are not the x-partial."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S06",
            "S07",
            "S09"
          ],
          "explanation": {
            "zh-Hant": "只合取已證結論；不宣稱 f 全微分或連續（雖然多項式事實上如此，本稿不證）。",
            "en": "Only conjoins conclusions already proved; total differentiability or continuity of f is not claimed (true for polynomials, but not proved here)."
          }
        },
        "leaf_basis": "and_intro_elim"
      }
    ],
    "result": {
      "zh-Hant": "對 f=c_0+c_1x+c_2y+c_3x²+c_4xy+c_5y²，差商在 h≠0 時為仿射式，故 f_x=c_1+2c_3x+c_4y、f_y=c_2+c_4x+2c_5y。對 c=(1,−2,3,1,2,−1)：f_x=−2+2x+2y、f_y=3+2x−2y，在 (1,2) 得梯度 (4,1)。q_x(h)=4+h 對任何非零 h 都不等於 4；同時移動兩座標的極限為 5。",
      "en": "For f=c_0+c_1x+c_2y+c_3x²+c_4xy+c_5y², the quotient is affine in h for h≠0, so f_x=c_1+2c_3x+c_4y and f_y=c_2+c_4x+2c_5y. For c=(1,−2,3,1,2,−1): f_x=−2+2x+2y, f_y=3+2x−2y, giving the gradient (4,1) at (1,2). q_x(h)=4+h differs from 4 for every nonzero h; moving both coordinates gives the limit 5."
    },
    "negative_controls": [
      {
        "claim": {
          "zh-Hant": "q_x(1)=5，所以 f_x(1,2)=5",
          "en": "q_x(1)=5, so f_x(1,2)=5"
        },
        "expected": "reject",
        "reason": {
          "zh-Hant": "偏導是 h→0 的極限：q_x(h)=4+h→4（S10、S11）。",
          "en": "The partial is the limit as h→0: q_x(h)=4+h→4 (S10, S11)."
        }
      },
      {
        "claim": {
          "zh-Hant": "(f(1+h,2+h)−f(1,2))/h→5，所以 x 偏導是 5",
          "en": "(f(1+h,2+h)−f(1,2))/h→5, so the x-partial is 5"
        },
        "expected": "reject",
        "reason": {
          "zh-Hant": "這同時移動了 y；固定 y=2 的極限是 4（S12）。",
          "en": "This moves y as well; with y=2 fixed the limit is 4 (S12)."
        }
      },
      {
        "claim": {
          "zh-Hant": "函數在一點的兩個偏導都存在，所以它在該點連續",
          "en": "If both partials of a function exist at a point, it is continuous there"
        },
        "expected": "reject",
        "reason": {
          "zh-Hant": "g(x,y)=xy/(x²+y²)、g(0,0)=0 在原點兩偏導皆為 0，但 g(t,t)=1/2（錯誤控制 3）。",
          "en": "g(x,y)=xy/(x²+y²), g(0,0)=0 has both partials 0 at the origin, but g(t,t)=1/2 (error control 3)."
        }
      },
      {
        "claim": {
          "zh-Hant": "中心差分 (q(h)+q(−h))/2 對某個 h 給出一個數，就證明了偏導存在且等於它",
          "en": "If the central difference (q(h)+q(−h))/2 gives a number for some h, the partial exists and equals it"
        },
        "expected": "reject",
        "reason": {
          "zh-Hant": "u(x,y)=|x| 在 (0,0) 的中心差分恆為 0，但 x 偏導不存在（錯誤控制 2）。",
          "en": "For u(x,y)=|x| at (0,0) the central difference is always 0, yet the x-partial does not exist (error control 2)."
        }
      }
    ],
    "status": "candidate",
    "gaps": [],
    "verification": {
      "structure": "checked_local_script",
      "mathematics": "candidate_argument",
      "formalization": "not_formalized"
    },
    "rule_glossary": [
      {
        "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": "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": "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": "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_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": "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": "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": "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": "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": "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": "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": "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": "and_intro_elim",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "合取：由 P 與 Q 得 P∧Q；由 P∧Q 可分別得 P 與 Q。",
          "en": "Conjunction: from P and Q infer P∧Q; from P∧Q infer P and infer Q."
        }
      },
      {
        "id": "mko_partial_def",
        "kind": "cited",
        "source": {
          "mko_id": "mko-partial-derivative",
          "sha256": "7159526c8e30cc8cb9086f9690d279fd3852328abe371206b3441ecf6ef69ce8",
          "commit": "b28a5edbd46279b1db2a3f3f48bb3f447e18899f"
        },
        "statement": {
          "zh-Hant": "二變數函數 f 在點 (x,y) 的 x 偏導，以 y 固定並令 h≠0 趨近 0 的差商極限定義；y 偏導類似。本輪對二次多項式做精確符號微分，不用單一有限差分代替極限。",
          "en": "The x-partial derivative of a bivariate function f at (x,y) fixes y and is defined by the difference-quotient limit as nonzero h tends to 0; the y-partial is analogous. This slice differentiates quadratic polynomials symbolically and exactly, rather than replacing the limit with a single finite difference."
        }
      },
      {
        "id": "mko_mvf_def",
        "kind": "cited",
        "source": {
          "mko_id": "mko-multivariable-real-function",
          "sha256": "f96ebe8babf93fa8de6b32cb4156ad2b710b938550218a6aff16be79ba4ef928",
          "commit": "b28a5edbd46279b1db2a3f3f48bb3f447e18899f"
        },
        "statement": {
          "zh-Hant": "二變數實多項式 f:ℝ²→ℝ 的明示切片，係數依 [1,x,y,x²,xy,y²] 固定次序。",
          "en": "Explicit slices of bivariate real polynomials f:ℝ²→ℝ, with coefficients ordered as [1,x,y,x²,x*y,y²]."
        }
      },
      {
        "id": "L_numerator_x",
        "kind": "lemma",
        "proved_in": [
          "S03.1",
          "S03.2"
        ],
        "statement": {
          "zh-Hant": "對所有實數 x、y、h：f(x+h,y)−f(x,y)=h(c_1+2c_3x+c_4y)+c_3h²。",
          "en": "For all reals x, y, h: f(x+h,y)−f(x,y)=h(c_1+2c_3x+c_4y)+c_3h²."
        }
      },
      {
        "id": "L_affine_limit",
        "kind": "lemma",
        "proved_in": [
          "S05.1",
          "S05.2"
        ],
        "statement": {
          "zh-Hant": "對實常數 A、B，函數 h↦A+Bh 在 h≠0 上有 lim_{h→0}(A+Bh)=A。",
          "en": "For real constants A, B, the function h↦A+Bh on h≠0 satisfies lim_{h→0}(A+Bh)=A."
        }
      }
    ],
    "revision_log": [],
    "numeric_checks": [
      {
        "id": "N1",
        "step": "S03",
        "vars": [
          "c0",
          "c1",
          "c2",
          "c3",
          "c4",
          "c5",
          "x",
          "y",
          "h"
        ],
        "degree": 2,
        "lhs": "(c0+c1*(x+h)+c2*y+c3*(x+h)**2+c4*(x+h)*y+c5*y**2)-(c0+c1*x+c2*y+c3*x**2+c4*x*y+c5*y**2)",
        "rhs": "h*(c1+2*c3*x+c4*y)+c3*h**2"
      },
      {
        "id": "N2",
        "step": "S07",
        "vars": [
          "c0",
          "c1",
          "c2",
          "c3",
          "c4",
          "c5",
          "x",
          "y",
          "h"
        ],
        "degree": 2,
        "lhs": "(c0+c1*x+c2*(y+h)+c3*x**2+c4*x*(y+h)+c5*(y+h)**2)-(c0+c1*x+c2*y+c3*x**2+c4*x*y+c5*y**2)",
        "rhs": "h*(c2+c4*x+2*c5*y)+c5*h**2"
      },
      {
        "id": "N3",
        "step": "S09",
        "lhs": "-2+2*1+2*2",
        "rhs": "4"
      },
      {
        "id": "N4",
        "step": "S09",
        "lhs": "3+2*1-2*2",
        "rhs": "1"
      },
      {
        "id": "N5",
        "step": "S10",
        "lhs": "1-2*1+3*2+1**2+2*1*2-2**2",
        "rhs": "6"
      },
      {
        "id": "N6",
        "step": "S10",
        "vars": [
          "h"
        ],
        "degree": 2,
        "lhs": "(1-2*(1+h)+3*2+(1+h)**2+2*(1+h)*2-2**2)-6",
        "rhs": "4*h+h**2"
      },
      {
        "id": "N7",
        "step": "S11",
        "lhs": "((1-2*2+3*2+2**2+2*2*2-2**2)-6)/1",
        "rhs": "5"
      },
      {
        "id": "N8",
        "step": "S11",
        "lhs": "((1-2*(11/10)+3*2+(11/10)**2+2*(11/10)*2-2**2)-6)/(1/10)",
        "rhs": "41/10"
      },
      {
        "id": "N9",
        "step": "S12",
        "vars": [
          "h"
        ],
        "degree": 2,
        "lhs": "(1-2*(1+h)+3*(2+h)+(1+h)**2+2*(1+h)*(2+h)-(2+h)**2)-6",
        "rhs": "5*h+2*h**2"
      },
      {
        "id": "N10",
        "step": "S07",
        "vars": [
          "h"
        ],
        "degree": 2,
        "lhs": "(1-2*1+3*(2+h)+1**2+2*1*(2+h)-(2+h)**2)-6",
        "rhs": "h-h**2"
      },
      {
        "id": "N11",
        "lhs": "((1-2*0+3*2+0**2+2*0*2-2**2)-6)/(-1)",
        "rhs": "3"
      }
    ]
  },
  "presentation": {
    "md_scope": {
      "zh-Hant": "只處理 G1 的二變數二次多項式切片（MKO 假設 a1），偏導以差商極限精確推導。不以偏導存在推出一般函數的連續或全微分（a2）；錯誤控制中的非多項式例子只作對照。不改動 canonical MKO 或其英文 edition。",
      "en": "Treats only the bivariate quadratic slice of G1 (MKO assumption a1), deriving the partials exactly from difference-quotient limits. The existence of partials is not used to infer continuity or total differentiability of general functions (a2); the non-polynomial examples in the error controls are only contrasts. The canonical MKO and its English edition are not changed."
    },
    "md_symbols": [
      [
        "c=(c_0,c_1,c_2,c_3,c_4,c_5)",
        "依 [1,x,y,x²,xy,y²] 次序的係數",
        "coefficients in the order [1,x,y,x²,xy,y²]"
      ],
      [
        "f(x,y)",
        "二次多項式",
        "the quadratic polynomial"
      ],
      [
        "h",
        "非零增量",
        "a nonzero increment"
      ],
      [
        "q_x(h)",
        "x 方向差商 (f(x+h,y)−f(x,y))/h",
        "the x difference quotient (f(x+h,y)−f(x,y))/h"
      ],
      [
        "f_x,\\ f_y",
        "x 偏導與 y 偏導",
        "the x- and y-partial derivatives"
      ],
      [
        "\\nabla f=(f_x,f_y)",
        "梯度：偏導的有序收集",
        "the gradient: the ordered collection of partials"
      ]
    ],
    "md_error_controls": [
      {
        "title": "不要同時移動兩個座標 / Do not move both coordinates",
        "zh": "MKO 指出的誤解之一是同時改變兩個座標。S12 中 (f(1+h,2+h)−f(1,2))/h→5，而 x 偏導是 4。x 偏導的定義要求 y 固定，只有第一個座標加 h。",
        "en": "One misconception named by the MKO is to change both coordinates at once. In S12, (f(1+h,2+h)−f(1,2))/h→5, while the x-partial is 4. The definition of the x-partial requires y to stay fixed, with h added to the first coordinate only."
      },
      {
        "title": "一個 h 的差商不是偏導 / One quotient is not the partial",
        "zh": "MKO 的另一個誤解是以一個非零 h 的差商聲稱偏導存在。本例 q_x(h)=4+h，對每個非零 h 都不是 4（h=1 給 5，h=1/10 給 41/10）。即使某種有限差分碰巧給出正確數字（例如中心差分 (q_x(1)+q_x(−1))/2=(5+3)/2=4），那也只是二次多項式的特性，不是存在性的證明：對 u(x,y)=|x| 在 (0,0)，中心差分 (|h|−|−h|)/(2h)=0 對每個 h≠0 成立，但 x 偏導不存在（右差商為 1、左差商為 −1）。",
        "en": "Another misconception named by the MKO is to claim a partial exists from one nonzero-h quotient. Here q_x(h)=4+h is never 4 for nonzero h (h=1 gives 5, h=1/10 gives 41/10). Even when some finite difference happens to give the right number (for example the central difference (q_x(1)+q_x(−1))/2=(5+3)/2=4), that is a feature of quadratics, not a proof of existence: for u(x,y)=|x| at (0,0), the central difference (|h|−|−h|)/(2h)=0 for every h≠0, yet the x-partial does not exist (the right quotient is 1 and the left quotient is −1)."
      },
      {
        "title": "偏導存在不推出連續 / Existence of partials does not imply continuity",
        "zh": "MKO 假設 a2 提醒偏導存在不用來推出一般函數連續。例：g(x,y)=xy/(x²+y²)（(x,y)≠(0,0)）、g(0,0)=0。g(x,0)=0 與 g(0,y)=0 對所有 x、y 成立，所以兩個偏導在原點存在且為 0；但沿 y=x，g(t,t)=1/2 對所有 t≠0 成立，不趨近 g(0,0)=0，所以 g 在原點不連續。",
        "en": "MKO assumption a2 warns that existing partials are not used to infer continuity of general functions. Example: g(x,y)=xy/(x²+y²) for (x,y)≠(0,0) and g(0,0)=0. g(x,0)=0 and g(0,y)=0 for all x, y, so both partials exist at the origin and equal 0; but along y=x, g(t,t)=1/2 for all t≠0, which does not tend to g(0,0)=0, so g is not continuous at the origin."
      }
    ],
    "md_open_items": [
      {
        "zh-Hant": "錯誤控制 2、3 的非多項式例子以文字論證，沒有拆成編號步驟；本稿主結論不依賴它們。",
        "en": "The non-polynomial examples of error controls 2 and 3 are argued in prose, not in numbered steps; the main conclusions do not depend on them."
      },
      {
        "zh-Hant": "梯度的幾何意義、全微分與高階偏導（MKO 的 formal next）不在範圍內。",
        "en": "The geometric meaning of the gradient, total differentiability and higher partials (the MKO's formal next) are out of scope."
      }
    ]
  },
  "revision": {
    "id": "production-20261019-adopted-96",
    "source_commit": "b28a5edbd46279b1db2a3f3f48bb3f447e18899f",
    "adopted_source_sha256": "cec61d27ae749c4333888ce08539fdb4a14ce72bd62e3e9d85ee86ac92f9eab1",
    "original_source_sha256": "277ef1a5d5807fe90f5ab864bef246dbbc0d025aaa5183b9a9d7f44340c942e7",
    "owner_adopted": true,
    "bilingual_sync": "complete_source_bound",
    "formal_Lean": "NOT_RUN",
    "independent_AI_concurrence": "NOT_MEASURED",
    "issue_ids": [
      "BR-M064"
    ],
    "scope": "Owner-adopted natural-language teaching guide; not a formal proof, canonical MKO rewrite or evidence-status promotion."
  }
}