# Partial derivative: deriving the partial-derivative formulas of a quadratic from the difference-quotient limit, and computing a gradient

[← Guide index](https://ocme.evemisslab.com/en/derivations/) · [Original MKO](https://ocme.evemisslab.com/en/mko/mko-partial-derivative/)

This is an owner-adopted natural-language teaching guide. Finite/arithmetic checks are not universal proofs; no Lean was run, and review scope is recorded per revision. Original source status labels are retained as historical data. Formulas retain exact TeX; no new semantic AST is invented.

derivation-mko-partial-derivative-01 · production-20261019-adopted-96



## Goal and scope

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.

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.





## Givens and symbols


- func_limit_def
- abs_rules
- order_basics
- distributive
- power_laws


- G1 — 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).
- G2 — Instance: c=(1,−2,3,1,2,−1), that is f(x,y)=1−2x+3y+x²+2xy−y², at the point (1,2).


- c=(c_0,c_1,c_2,c_3,c_4,c_5) — coefficients in the order [1,x,y,x²,xy,y²]
- f(x,y) — the quadratic polynomial
- h — a nonzero increment
- q_x(h) — the x difference quotient (f(x+h,y)−f(x,y))/h
- f_x,\ f_y — the x- and y-partial derivatives
- \nabla f=(f_x,f_y) — the gradient: the ordered collection of partials





## Rules and sources


- func_limit_def · prerequisite — 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).
- abs_rules · prerequisite — 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.
- order_basics · prerequisite — 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.
- distributive · prerequisite — Distributivity: u(v+w)=uv+uw; with commutativity also (v+w)u=vu+wu.
- power_laws · prerequisite — 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.
- def_unfold · prerequisite — Unfolding a definition: replacing a defined term by its definition (or folding it back) does not change whether a statement is true.
- add_inverse · prerequisite — Additive inverse: every u has −u with u+(−u)=0; u−v means u+(−v).
- mul_comm · prerequisite — Commutativity of multiplication: uv=vu (in commutative number systems such as ℤ, ℚ, ℝ, ℂ).
- add_comm · prerequisite — Commutativity of addition: u+v=v+u.
- mul_inverse_nonzero · prerequisite — 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.
- frac_rules · prerequisite — 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.
- exists_intro · prerequisite — ∃-introduction: if some a∈D satisfies P(a), then ∃x∈D, P(x).
- forall_intro · prerequisite — ∀-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.
- eq_replace · prerequisite — Replacement of equals: an expression inside a statement may be replaced by an equal expression without changing whether the statement is true.
- numeric_evaluation · prerequisite — Exact evaluation of an explicit integer or rational expression; independently recomputable.
- and_intro_elim · prerequisite — Conjunction: from P and Q infer P∧Q; from P∧Q infer P and infer Q.
- mko_partial_def · cited — 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. · mko-partial-derivative
- mko_mvf_def · cited — Explicit slices of bivariate real polynomials f:ℝ²→ℝ, with coefficients ordered as [1,x,y,x²,x*y,y²]. · mko-multivariable-real-function
- L_numerator_x · lemma — For all reals x, y, h: f(x+h,y)−f(x,y)=h(c_1+2c_3x+c_4y)+c_3h².
- L_affine_limit · lemma — For real constants A, B, the function h↦A+Bh on h≠0 satisfies lim_{h→0}(A+Bh)=A.





## Main derivation

S01 · definitionBefore

f_x(x,y)

After

\lim_{h\to0,\ h\neq0}\frac{f(x+h,y)-f(x,y)}{h}

Conditions: y\ \text{fixed}

Rule: mko_partial_def · Uses: mko_partial_def

Leaf basis: mko_partial_def + def_unfold

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.



S02 · definitionBefore

c=(c_0,c_1,c_2,c_3,c_4,c_5)

After

f(x,y)=c_0+c_1x+c_2y+c_3x^2+c_4xy+c_5y^2

Conditions: none

Rule: mko_mvf_def · Uses: G1, mko_mvf_def

Leaf basis: mko_mvf_def + def_unfold

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.



S03 · equivalentBefore

f(x+h,y)-f(x,y)

After

h(c_1+2c_3x+c_4y)+c_3h^2

Conditions: none

Rule: L_numerator_x · Uses: S02, L_numerator_x

Leaf basis: L_numerator_x + distributive

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).



S04 · equivalentBefore

\frac{h(c_1+2c_3x+c_4y)+c_3h^2}{h}

After

q_x(h)=(c_1+2c_3x+c_4y)+c_3h\quad(h\neq0)

Conditions: h\neq0

Rule: mul_inverse_nonzero · Uses: S03, mul_inverse_nonzero

Leaf basis: mul_inverse_nonzero + frac_rules + power_laws

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.



S05 · theorem_applicationBefore

\lim_{h\to0}\big(A+Bh\big)

After

A

Conditions: A=c_1+2c_3x+c_4y,\ B=c_3

Rule: L_affine_limit · Uses: S04, L_affine_limit

Leaf basis: L_affine_limit + func_limit_def

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.



S06 · impliesBefore

f_x(x,y)=\lim_{h\to0}q_x(h)

After

f_x(x,y)=c_1+2c_3x+c_4y

Conditions: none

Rule: L_affine_limit · Uses: S01, S04, S05

Leaf basis: L_affine_limit + def_unfold

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.



S07 · impliesBefore

\frac{f(x,y+h)-f(x,y)}{h}=(c_2+c_4x+2c_5y)+c_5h

After

f_y(x,y)=c_2+c_4x+2c_5y

Conditions: h\neq0; x\ \text{fixed}

Rule: L_affine_limit · Uses: S01, S02, S05

Leaf basis: L_affine_limit + distributive + mul_inverse_nonzero

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'.



S08 · equivalentBefore

c=(1,-2,3,1,2,-1)

After

f_x=-2+2x+2y,\quad f_y=3+2x-2y

Conditions: none

Rule: eq_replace · Uses: S06, S07, G2

Leaf basis: eq_replace + numeric_evaluation

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.



S09 · equivalentBefore

(x,y)=(1,2)

After

f_x(1,2)=4,\ \ f_y(1,2)=1,\ \ \nabla f(1,2)=(4,1)

Conditions: none

Rule: numeric_evaluation · Uses: S08

Leaf basis: numeric_evaluation

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).



S10 · impliesBefore

\frac{f(1+h,2)-f(1,2)}{h}

After

\frac{4h+h^2}{h}=4+h\ \xrightarrow{h\to0}\ 4

Conditions: h\neq0

Rule: L_affine_limit · Uses: G2, S04, S09

Leaf basis: L_affine_limit + numeric_evaluation + distributive

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.



S11 · impliesBefore

q_x(h)=4+h

After

q_x(1)=5,\ \ q_x(\tfrac1{10})=\tfrac{41}{10},\ \ q_x(h)\neq4\ \text{for every}\ h\neq0

Conditions: h\neq0

Rule: order_basics · Uses: S10

Leaf basis: order_basics + numeric_evaluation

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).



S12 · impliesBefore

\frac{f(1+h,2+h)-f(1,2)}{h}

After

\frac{5h+2h^2}{h}=5+2h\ \xrightarrow{h\to0}\ 5\neq4=f_x(1,2)

Conditions: h\neq0

Rule: L_affine_limit · Uses: G2, S09, L_affine_limit

Leaf basis: L_affine_limit + numeric_evaluation + distributive

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).



S13 · impliesBefore

\text{S06 to S12}

After

f_x=c_1+2c_3x+c_4y,\ f_y=c_2+c_4x+2c_5y;\ \nabla f(1,2)=(4,1)

Conditions: none

Rule: and_intro_elim · Uses: S06, S07, S09, S11, S12

Leaf basis: and_intro_elim

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.







## Substeps and lemmas

S03.1 · equivalentBefore

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

0,\ c_1h,\ 0,\ c_3(2xh+h^2),\ c_4hy,\ 0

Conditions: none

Rule: distributive · Uses: S02, power_laws, distributive

Leaf basis: distributive + power_laws + add_inverse

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.



S03.2 · equivalentBefore

c_1h+c_3(2xh+h^2)+c_4hy

After

h(c_1+2c_3x+c_4y)+c_3h^2

Conditions: none

Rule: distributive · Uses: S03.1, distributive

Leaf basis: distributive + mul_comm + add_comm

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.



S05.1 · impliesBefore

\varepsilon>0

After

\delta=\frac{\varepsilon}{|B|+1}>0

Conditions: \varepsilon>0

Rule: exists_intro · Uses: func_limit_def, abs_rules

Leaf basis: exists_intro + abs_rules + order_basics

Take any ε>0. |B|+1≥1>0, so δ=ε/(|B|+1) is positive. The +1 avoids division by 0 when B=0.



S05.2 · impliesBefore

0<|h|<\delta

After

|(A+Bh)-A|=|B||h|\le|B|\delta=\frac{|B|}{|B|+1}\varepsilon<\varepsilon

Conditions: 0<|h|<\delta

Rule: abs_rules · Uses: S05.1, abs_rules, order_basics

Leaf basis: abs_rules + order_basics + forall_intro

|(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 audits


- S13 · premise_audit · S06, S07, S09 — Only conjoins conclusions already proved; total differentiability or continuity of f is not claimed (true for polynomials, but not proved here).
- S12 · premise_audit · L_affine_limit, numeric_evaluation — 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.
- S11 · premise_audit · S10, order_basics — 4+h=4 ⇔ h=0, and h=0 is excluded; this conclusion concerns this instance only.
- S10 · premise_audit · L_affine_limit, numeric_evaluation — Cross-checks the formula of S06; for this instance only.
- S09 · inverse_equivalence · numeric_evaluation — Exact integer evaluation.
- S08 · inverse_equivalence · eq_replace — Substitution of equals, valid both ways.
- S07 · premise_audit · L_affine_limit, distributive, mul_inverse_nonzero — The same four steps as S03 to S06 with the roles of x and y exchanged; N2 checks the numerator identity independently.
- S06 · premise_audit · S04, S05, S01 — Two functions that agree for h≠0 have the same limit; the value at h=0 plays no role.
- S05.2 · premise_audit · abs_rules, order_basics, forall_intro — 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.
- S05.1 · premise_audit · abs_rules, order_basics — A positive number divided by a positive number is positive; δ depends only on ε and B.
- S05 · premise_audit · L_affine_limit, func_limit_def — 0 is a limit point of ℝ∖{0}, so the function limit makes sense.
- S04 · inverse_equivalence · mul_inverse_nonzero, frac_rules — Multiplying by 1/h and by h undo each other, given h≠0.
- S03.2 · inverse_equivalence · distributive, mul_comm — Factoring and expanding undo each other.
- S03.1 · inverse_equivalence · distributive, power_laws, add_inverse — Each term is an identity expansion; only the terms containing x survive, which is exactly the effect of fixing y.
- S03 · inverse_equivalence · L_numerator_x, distributive — This is an algebraic identity valid for all x, y, h, so it holds both ways.
- S02 · premise_audit · mko_mvf_def — 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.
- S01 · premise_audit · mko_partial_def — Only unfolds the definition; h=0 is excluded because the quotient is undefined there.





## Error controls and negative controls

Do not move both coordinatesOne 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.

One quotient is not the partialAnother 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).

Existence of partials does not imply continuityMKO 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.


- q_x(1)=5, so f_x(1,2)=5 → reject — The partial is the limit as h→0: q_x(h)=4+h→4 (S10, S11).
- (f(1+h,2+h)−f(1,2))/h→5, so the x-partial is 5 → reject — This moves y as well; with y=2 fixed the limit is 4 (S12).
- If both partials of a function exist at a point, it is continuous there → reject — 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).
- If the central difference (q(h)+q(−h))/2 gives a number for some h, the partial exists and equals it → reject — For u(x,y)=|x| at (0,0) the central difference is always 0, yet the x-partial does not exist (error control 2).





## Result and limitations

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.


- 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.
- The geometric meaning of the gradient, total differentiability and higher partials (the MKO's formal next) are out of scope.





## Verification and source binding

11 declared checks for this guide; only the listed expressions are covered.


- source commit: b28a5edbd46279b1db2a3f3f48bb3f447e18899f
- adopted source SHA-256: cec61d27ae749c4333888ce08539fdb4a14ce72bd62e3e9d85ee86ac92f9eab1
- original source SHA-256: 277ef1a5d5807fe90f5ab864bef246dbbc0d025aaa5183b9a9d7f44340c942e7
- Lean: NOT_RUN · independent AI: NOT_MEASURED

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