The multivariable chain rule and local estimates

Try

Let F(x,y)=(x+y,xy) and G(u,v)=u²+v. At (1,2), DF=[[1,1],[2,1]] and DG at F(1,2)=(3,2) is the row (6,1). Their product is (8,7). Direct substitution gives G(F(x,y))=(x+y)²+xy, whose derivative at (1,2) agrees. This numerical check establishes the order of the maps, while the general proof must control remainders.

Predict

Predict the dimensions before multiplying. DF maps input displacements into intermediate displacements; DG converts those into output displacements. Therefore the inner derivative acts first and appears on the right in the matrix product. Evaluating DG at the original input rather than at F(a) would compare different spaces and usually produce the wrong numbers.

Experiment

Trace one displacement.

Stage	Linear change	Remainder
Inner map F	Ah	r(h)=o(||h||)
Outer map G	Bk	s(k)=o(||k||)
Composition	BAh	Br(h)+s(k)

Here k=Ah+r(h). The missing estimate in many incomplete proofs is that ||k||=O(||h||). Without it, a quantity small relative to ||k|| has not yet been shown small relative to the original input scale.

Understand

Differentiability of F gives ||k||≤(||A||+1)||h|| for sufficiently small h. Substitute the inner expansion into the outer one: G(F(a+h))=G(F(a))+BAh+Br(h)+s(k). The first remainder is o(||h||) because B is bounded. For any η>0, sufficiently small k satisfies ||s(k)||≤η||k||, with s(0)=0. Combining with the bound on k makes the second remainder at most η(||A||+1)||h||. Thus the total remainder divided by ||h|| tends to zero, proving D(G∘F)(a)=DG(F(a))DF(a).

This proof also explains the zero-inner-increment case: no division by ||k|| is needed there, because the outer remainder at zero is zero. Treating that case explicitly avoids an undefined ratio hidden in a symbolic small-o manipulation.

Apply

A useful global estimate follows on a convex open region. If F is C¹ and ||DF||≤M throughout it, integrate the derivative along the segment from y to x: F(x)−F(y)=∫₀¹DF(y+t(x−y))(x−y)dt. The norm is at most M||x−y||. Convexity keeps the whole segment in the domain. One should not assert a vector-valued MVT saying a single intermediate derivative necessarily equals the entire vector difference; the integral argument is valid without that stronger and generally false statement.

Repair “a derivative bound zero makes a function globally constant on any open domain.” Different constants on disconnected components have zero derivative everywhere. The joining-segment condition or another connectedness argument is necessary for the global conclusion.

From a segment estimate to a connected domain

A hole can block the straight segment without blocking every path. In the square annulus, join (−2,0) to (2,0) through (−2,2) and (2,2). The segment lengths are 2, 4 and 2. A zero derivative on all three segments forces the same value at the endpoints even though the length-eight route is longer than the distance four. Predict what survives when the function is not constant or the domain is disconnected.

Can the derivative estimate follow this route?

Domain and function Square annulus; constant sevenSlit square annulus; principal angleTwo half-planes; constants −1 and +1 Candidate route Direct segmentOuter detourShortcut through the center Angle endpoint height ε 0.50.250.1250.0625

Predict which segment leaves the domain. Coordinates are dimensionless, angles in radians. Height affects only the slit example. The square annulus is 1 < max(|x|,|y|) < 3. The slit removes its negative x-axis; the half-plane example has |x| > 1.

Every segment stays inside the domain. Candidate length 8.0000; endpoint distance 4.0000; actual value difference 0.000000.

Derivative norm bound M=0. Valid path estimate M × length = 0.0000. The unsupported shortcut M × endpoint distance would give 0.0000.

Segment	From → to	Length	Verdict
1	(-2, 0) → (-2, 2)	2.0000	Valid: inside domain
2	(-2, 2) → (2, 2)	4.0000	Valid: inside domain
3	(2, 2) → (2, 0)	2.0000	Valid: inside domain

The domain is not convex, yet the valid detour transports the zero-derivative conclusion. A failed straight segment does not mean no path exists.

Reset

At most five segments are checked by exact interval inequalities on fixed binary coordinates. Angle evaluation uses floating point. These temporary controls illustrate proofs; they do not test arbitrary paths or award mastery.

The mean-value inequality needs only differentiability, not continuity of the derivative. Suppose F maps an open subset of Euclidean m-space to Euclidean n-space, the segment from a to b stays in the domain, and the operator norm of DF is bounded by M there. If d=F(b)−F(a) is zero the conclusion is immediate. Otherwise set v=d/||d|| and apply the scalar mean value theorem to g(t)=v·F(a+t(b−a)). The chain rule gives |g′(t)| ≤ M||b−a||. Since g(1)−g(0)=||d||, the scalar theorem proves ||F(b)−F(a)|| ≤ M||b−a||. The intermediate point depends on the chosen scalar projection; this does not assert a vector equality with one derivative matrix.

Now assume DF=0 everywhere on an open domain U. Each point has a ball contained in U. Any two points in that ball are joined by a segment in it, so the inequality with M=0 makes F constant on the ball. Thus F is locally constant. If γ is any continuous path in U, then F∘γ is locally constant on [0,1]: continuity of γ carries a sufficiently small parameter neighborhood into one such ball. The parameters where F∘γ equals its initial value form a nonempty relatively open set, and its complement is open by the same argument. Connectedness of [0,1] forces the complement to be empty. If U is path connected, every endpoint pair has such a path, so F is constant on U.

Notice what the proof did not do: γ need not be differentiable, so writing (F∘γ)′=DF(γ)γ′ would be unjustified. Openness supplies local balls; connectedness of the parameter interval transports their constant values.

For a finite geometric version, cover the path image by balls contained in U. Their inverse images form an open cover of the compact interval. Choose a positive Lebesgue number for a finite subcover and partition the interval into shorter pieces. Each piece of the path lies in one ball, so its endpoints can be joined by a straight segment in that ball. The finitely many segments produce a polygonal route in U. Applying the segment inequality and the triangle inequality gives an endpoint difference at most M times the sum of segment lengths. With M=0 that sum disappears. With positive M it generally cannot be replaced by the endpoint distance.

Bounded derivatives can still fail a global distance estimate

Remove the negative x-axis from the open square annulus. The principal angle θ is smooth there and its gradient is (−y,x)/(x²+y²), whose Euclidean norm is below one because max(|x|,|y|)>1. At points (−2,ε) and (−2,−ε), their distance is 2ε, while the angle difference is 2π − 2 arctan(ε/2). As ε tends to zero this difference tends to 2π, so no finite global Euclidean Lipschitz constant works. The domain is still path connected: travel outward or inward along each ray to the square of radius two, then along that square without crossing the removed point. The valid five-segment detour in the inspector has length 16−2ε. Its path estimate remains valid, though coarse.

By contrast, on the two open half-planes |x|>1, the function taking −1 on the left and +1 on the right has derivative zero. It is constant on each component, not on the whole domain. Any supposed joining path has an x-coordinate taking every intermediate value, hence crosses the excluded strip. This impossibility argument uses the intermediate value theorem; a list of failed candidate routes would not suffice.

Continue

Optional depth differentiates a local inverse identity. If F and G are differentiable local inverses, the chain rule gives DG(F(a))DF(a)=I. In equal finite dimensions, this identifies DG(F(a)) as the inverse matrix. This computation does not by itself prove an inverse exists or is differentiable; those are the substantive claims of the inverse theorem.

Save both the numerical composition check and the remainder proof. Label where finite-dimensional boundedness of linear maps enters. Next, second-order expansions will use the same segment method to turn a multivariable Taylor problem into a one-variable theorem with controlled Hessian errors. For an affine inner map, its remainder is exactly zero, so only the outer remainder needs scaling. For a constant inner map, its displacement is zero and the composition is constant near the point. Checking these limiting cases helps expose an unjustified division by the inner displacement norm.