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The change of variables formula expresses an integral in coordinates that make its integrand or its region simpler. Three ingredients change together: the region, the function being integrated, and the element of area or volume. The last change is measured by the absolute Jacobian determinant; forgetting it amounts to counting transformed cells as though their sizes had stayed the same.

In one dimension, substitution is the integrated chain rule. If $g$ is continuously differentiable on $[a,b]$ and $f$ is continuous on its image, thenThese are directed definite integrals: if $g$ decreases, the transformed endpoints reverse. This identity does not require $g$ to be injective, because an antiderivative composed with $g$ already accounts for any backtracking.^[ref]

For a concrete computation, set $u=1+x^2$, so $du=2xdx$. The endpoints $x=0,1$ become $u=1,2$:The numerator is precisely the derivative of the expression in the denominator. Substitution reveals the familiar natural logarithm hidden inside this rational function. Keeping the original endpoints after changing the integrand would describe a different integral.

For the elementary two-dimensional theorem, let $T:U\to V$ be a $C^1$ diffeomorphism between open sets in the plane, let $S\subset U$ be a bounded region with piecewise smooth boundary, and let $f$ be continuous on $T(S)$, with the displayed integrals finite. ThenHere $DT$ is the Jacobian matrix. The diffeomorphism assumption means a bijection with a continuously differentiable inverse, and ensures a nonzero Jacobian determinant. This excludes overlapping sheets; the more general area formula handles multiplicity.^[ref]

Take $S=[0,1{]}^2$ and the linear mapIts inverse is $(u,v)=((x-y)/2,y)$. The square becomes a parallelogram with vertices $(0,0),(2,0),(3,1),(1,1)$: base two, height one, area two. Consequently ${\iint }_{T(S)}1dxdy={\iint }_{S}2dudv=2$. A nonconstant integrand changes too: ${\iint }_{T(S)}xdxdy={\int }_0^1{\int }_0^1(2u+v)2dudv=3$.

The direction of the map matters when selecting the factor. Here $T$ goes from the new coordinates to the original coordinates, so the factor is two. Its inverse has Jacobian determinant $1/2$, appropriate for converting area in the opposite direction. Writing down the map before computing its Jacobian matrix prevents a reciprocal-factor error.

Now reflect the image vertically: $\tilde{T}(u,v)=(2u+v,-v)$ has Jacobian determinant $-2$, but still produces area two. The sign records reversal of the orientation of a vector space; ordinary area uses its absolute value. For oriented differential forms, the pullback of a differential form instead retains the signed determinant.^[ref]

The geometric beauty is that a nonlinear differentiable map has a local linear approximation: tiny rectangles behave approximately like these parallelograms. Summing their scaled contributions gives the change of variables formula. In polar coordinates this factor is $r$, explaining $dxdy=rdrd\theta$; equal angular steps cover more area farther from the origin. Choosing suitable coordinates can simplify both the boundary and the integrand.^[ref]

If a differentiable map $F$ takes values in a level set of a differentiable scalar function $H$, the chain rule gives $D{H}_{F(x)}D{F}_x=0$. At a regular point, the image of the derivative lies in the kernel of the nonzero covector $DH$: the tangent space of the level set. For a map into the unit circle this forces rank at most one.

On the open set of invertible square matrices, the inversion map is differentiable and

The Jacobian matrix collects all first partial derivatives of a differentiable map. When it is square, its Jacobian determinant gives the signed local volume factor.

A local diffeomorphism is a differentiable map that restricts near every point to a diffeomorphism onto an open subset of its target. The local inverse functions need not combine into a single global inverse function: different parts of the domain of a function can map onto the same output region.

For continuously differentiable maps between open subsets of ${\mathbb{R}}^n$, being a local diffeomorphism is equivalent to having an invertible derivative at every point. The inverse function theorem proves one direction. For the other, the chain rule for each local inverse function gives two inverse linear maps. The periodic angle example under the inverse function theorem has this property everywhere, even though it has infinitely many inputs for every point of the punctured plane. The noninjective map with constant Jacobian determinant gives another useful distinction: even a constant nonzero Jacobian determinant need not give global injectivity.

Every local diffeomorphism is an open map. To see this, take an open subset of the source and intersect it with the neighbourhoods where the map is a diffeomorphism. Each local image is open, and their union is the full image. Thus nearby outputs remain available locally, despite the possible global ambiguity.

For differentiable maps $f:{\mathbb{R}}^n\to {\mathbb{R}}^m$ and $g:{\mathbb{R}}^m\to {\mathbb{R}}^r$, the multivariable chain rule multiplies their Jacobian matrices:The outer $r\times m$ matrix multiplies the inner $m\times n$ matrix, producing an $r\times n$ matrix. This order records that the inner linear map acts first. Componentwise, the matrix multiplication sums the contributions of all intermediate coordinates. Existence of separate partial derivatives alone is not the required differentiability hypothesis.

For differentiable maps $F$ at $x$ and $G$ at $F(x)$, the chain rule assertsWrite $F(x+h)=F(x)+Ah+r(h)$ with $\Vert r(h)\Vert =o(\Vert h\Vert )$, and $G(F(x)+u)=G(F(x))+Bu+s(u)$ with $\Vert s(u)\Vert =o(\Vert u\Vert )$. Since $u=Ah+r(h)=O(\Vert h\Vert )$, substitution givesBoth remaining terms are $o(\Vert h\Vert )$, proving differentiability and the stated chain rule. The estimate for $s$ also holds when its argument is zero, by setting $s(0)=0$.

For the circle constraint take $H(u)=u_1^2+u_2^2$. The composite $H\circ F$ is constant, so the chain rule givesSince $\Vert F(x)\Vert =1$, this is a nonzero row vector annihilating $D{F}_x$. Thus $D{F}_x$ cannot be invertible: $\det D{F}_x=0$ at every $x$. Geometrically, the derivative of a map into a level set takes values in the tangent space to the circle.

The implicit function theorem states that if $F(x_0,y_0)=0$ and the partial derivative $D_{y}F(x_0,y_0)$ is an invertible linear map, then near $(x_0,y_0)$ the zero set of $F$ is uniquely the graph $y=g(x)$ of a continuously differentiable function.

If $f$ is a differentiable bijection with differentiable inverse $g$, the chain rule applied to $g\circ f$ and $f\circ g$ givesThus $D{f}_x$ is an isomorphism with inverse $D{g}_{f(x)}$.

If a continuously differentiable map $F:{\mathbb{R}}^n\to {\mathbb{R}}^n$ has invertible derivative everywhere, the inverse function theorem makes it a local diffeomorphism. In particular it is an open map, so its image is open. Its image need not be closed: $x\mapsto \arctan x$ has nonzero derivative everywhere and image $(-\pi /2,\pi /2)$.

For the given map of elementary symmetric polynomials,and direct evaluation of the determinant givesHence the critical set isIts complement is the Zariski-open set on which the three coordinates are pairwise distinct. Each point has one of the six possible strict coordinate orderings, and each ordering defines a nonempty convex open region. A continuous path cannot change an ordering without crossing $C$. Therefore ${\mathbb{R}}^3\setminus C$ has exactlyconnected components.