5.7 Inverse and implicit functions

A field whose Jacobian matrix is invertible at a point can be undone near that point by a differentiable inverse, because the best linear approximation can be undone and a contraction argument carries the undoing over to the field itself. Applied to the map (x, y) to (x, F(x, y)) it says that an equation F(x, y) = 0 defines y as a differentiable function of x wherever the partial derivative F_y is not zero, and it gives the formula dy/dx = -F_x/F_y.

Most equations cannot be solved for one variable in a formula. The circle $x^2 + y^2 = 1$ can, in two pieces; the equation $y^2 + xz + z^2 - e^z = 4$ cannot at all. Yet we still talk about the slope of the surface, the way $z$ changes when $x$ changes, and every JEE student has differentiated both sides of such an equation and solved for $\frac{dy}{dx}$ . This chapter asks the question under that habit. When does an equation really define one variable as a function of the others, so that the habit is legal? And when is a map $F:\mathbb R^n \to \mathbb R^n$ undoable, so that a function $G$ with $G(F(x)) = x$ exists and is differentiable? The two questions turn out to be one theorem seen from two sides. The answer ties the Jacobian determinant of 5.3 to the contraction principle of 4.2, and it is what makes level sets into smooth curves and surfaces, which 5.8 needs.

Before you start

5.4: the chain rule, and the derivative of an inverse field, which assumed that an inverse exists. 4.2: contractions and the Banach fixed point theorem. 5.3: the Jacobian matrix and its determinant, and the gradient being perpendicular to level sets.

Quick check

F:\mathbb R^2 \to \mathbb R^2

is differentiable with a differentiable inverse

G

, and

DF(a) = \begin{bmatrix}2 & 0\\ 0 & 3\end{bmatrix}

. What is

DG(F(a))

Quick checkA contraction

T

of a non-empty complete metric space has how many fixed points?

Quick checkAt a point of the level curve

f(x, y) = c

where

\nabla f \ne O

, which way does the curve run?

The picture: an equation is a curve, a function is a graph

Drag the ringed point round the circle. The blue curve is the set of solutions of $F(x, y) = 0$ . The green line is the tangent, the red arrow is $\nabla F$ . As long as the red arrow is not horizontal (that is, $F_y \ne 0$ ), the curve near the point is the graph of a function $y = g(x)$ and its slope is $-F_x/F_y$ . When the arrow becomes horizontal the tangent is vertical and the graph over the $x$ -axis breaks, though a graph over the $y$ -axis is still fine. Switch to the folium and drag to the origin, where the arrow shrinks to nothing and the curve crosses itself: the theorem has nothing to say. Switch to the cusp: the curve is still a graph over the $y$ -axis, but not a differentiable one.

For the second question, look at the field of 5.3 again:

A differentiable field looks like its linearization when you zoom in. If the linearization is a one-to-one map (an invertible matrix), then the field itself, seen through a microscope, is one-to-one. The theorem says that this survives outside the microscope, on a small but fixed neighbourhood.

The idea, in words

Inverting a field. The linear map $v \mapsto Av$ with $A$ square is undoable exactly when $A$ is invertible, that is $\det A \ne 0$ (2.7). A differentiable field $F$ near $a$ is $F(a) + A(x - a)$ plus a small error, where $A = DF(a)$ . If $A$ is invertible, the linear part can be undone. The error is small, so undoing the whole thing should be possible with a small correction, and the correction is again an undoing problem of the same kind but smaller. That is exactly the shape of a contraction: we guess $x$ from the linear part, check how wrong we were, and correct, and each correction is smaller than the last by a fixed factor.

Why· Why the iteration is a contraction

We want to solve $F(x) = y$ for $x$ . Rewrite it as a fixed point problem $x = \varphi_y(x)$ where $\varphi_y(x) = x + A^{-1}\big(y - F(x)\big)$ . If $x$ solves $F(x) = y$ then the correction term is $0$ and $\varphi_y(x) = x$ ; if $\varphi_y(x) = x$ then $A^{-1}(y - F(x)) = O$ , so $F(x) = y$ . The derivative of $\varphi_y$ is $I - A^{-1}DF(x) = A^{-1}\big(A - DF(x)\big)$ , and this is the zero matrix at $x = a$ . By continuity of $DF$ it is small near $a$ : the map $\varphi_y$ changes distances by a factor below $\frac12$ there. That is the whole proof, plus bookkeeping to check that $\varphi_y$ stays inside the ball where it is a contraction.

Implicit functions. The equation $F(x, y) = 0$ , with $x \in \mathbb R^n$ unknown input and $y \in \mathbb R^m$ unknown output, has $m$ equations. Think of the map $\Phi(x, y) = (x, F(x, y))$ from $\mathbb R^{n+m}$ to itself. It keeps $x$ and replaces $y$ by $F(x, y)$ . Its inverse (if it exists) takes $(x, 0)$ back to $(x, g(x))$ where $F(x, g(x)) = 0$ : the inverse of $\Phi$ solves the equation. So the implicit function theorem is the inverse function theorem applied to $\Phi$ , and the condition ” $\Phi$ has an invertible Jacobian” turns out to be “the $m \times m$ block $D_yF$ is invertible”.

Where this came from· From power series to a rigorous real theorem

Solving an equation for one variable in terms of another is as old as algebra, and Lagrange (1770) gave a formula for the power series of an inverse function. Cauchy is credited with the first rigorous form of the implicit function theorem, for functions given by power series in complex variables. The Italian analyst Ulisse Dini then extended it to real functions of any number of variables, and historians note that he published his proof only in his teaching notes (his Lezioni), which is one reason the theorem was long credited to others. The Pisa school still calls it Dini’s theorem. The inverse function theorem was proved by an iteration scheme in the style of Picard and Goursat. Today’s textbook proof is the same shrinking-map idea packaged by the fixed point theorem Banach stated in 1922 (4.2); it is also how Picard’s iteration proves existence of solutions of ODEs (7.1).

Definitions and theorems

Definition B

Local inverse

A neighbourhood of a point is any open set that contains it, so “near $a$ ” means “inside some neighbourhood of $a$ , however small”. A field is continuously differentiable ( $C^1$ ) if all its partial derivatives exist and are continuous. Now let $F: U \to \mathbb R^n$ with $U \subseteq \mathbb R^n$ open and $a \in U$ . A local inverse of $F$ at $a$ is a pair of open sets $U' \ni a$ and $V \ni F(a)$ such that $F$ maps $U'$ one-to-one onto $V$ , together with the inverse map $G: V \to U'$ . If $F$ and $G$ are both continuously differentiable, $F|_{U'}$ is a diffeomorphism (5.4) of $U'$ onto $V$ .

In one lineFor a field $F$ defined near $a$ , a differentiable field $G$ defined on a neighbourhood of $F(a)$ with $G(F(x)) = x$ for all $x$ near $a$ and $F(G(y)) = y$ for all $y$ near $F(a)$ .

Why it existsIt is the word for 'undoing $F$ near this point only', which is all that a derivative can ever promise; the exponential map shows that undoing it everywhere is a different and often false statement.

Before the main theorem we need one tool from one-variable calculus that was not stated for vector fields. In 1.4 the mean value theorem says $f(b) - f(a) = f'(c)(b - a)$ . For a map with vector values there is no single point $c$ , but an inequality survives.

Lemma B Rudin Thm 9.19

Mean value inequality for fields

Let $K \subseteq \mathbb R^n$ be convex (it contains the segment between any two of its points, for example a ball), let $H$ be defined and differentiable on an open set containing $K$ , and suppose the operator norm (3.5) of the Jacobian satisfies $\lVert DH(z)\rVert \le M$ for every $z \in K$ . Then $\lVert H(x_1) - H(x_2)\rVert \le M\,\lVert x_1 - x_2\rVert$ for all $x_1, x_2 \in K$ .

In one lineIf $H$ is differentiable on a convex set $K$ and $\lVert DH(z)\rVert \le M$ for all $z \in K$ , then $\lVert H(x_1) - H(x_2)\rVert \le M\lVert x_1 - x_2\rVert$ for $x_1, x_2 \in K$ .

Why it is trueIt converts a bound on the derivative into a bound on how much distances can change, which is the contraction constant of the next proof.

ProofRudin Thm 9.19

The idea: Choose a unit vector w pointing along H(x1) minus H(x2). The scalar function t to w dot H(x2 + t(x1 - x2)) is an ordinary function of one variable, so the one-variable mean value theorem applies to it. Its derivative is w dot DH times the step, which is at most M times the length of the step.

Assume

H(x_1) \ne H(x_2)

and put

w = \dfrac{H(x_1) - H(x_2)}{\lVert H(x_1) - H(x_2)\rVert}

, a unit vector.

H(x_1) = H(x_2)

the inequality is

0 \le M\lVert x_1 - x_2\rVert

, true. Otherwise divide by the length.Reducing a vector inequality to a scalar one by dotting with a well-chosen unit vector is the standard trick when the mean value theorem is needed for vector values.

Define

\psi(t) = w\cdot H\big(x_2 + t(x_1 - x_2)\big)

for

t \in [0, 1]

K

is convex, so the segment from

x_2

x_1

lies in

K

, where

H

is defined.

\psi'(t) = w\cdot DH(z_t)(x_1 - x_2)

with

z_t = x_2 + t(x_1 - x_2)

Chain rule along a segment (the path theorem of 5.3 applied to each component of

H

, then dotted with

w

): the derivative of

t \mapsto H(x_2 + tv)

DH(x_2 + tv)\,v

\psi(1) - \psi(0) = \psi'(\tau)

for some

\tau \in (0, 1)

Mean value theorem of 1.4 for the one-variable function

\psi

[0, 1]

The left side is

\psi(1) - \psi(0) = w\cdot\big(H(x_1) - H(x_2)\big) = \lVert H(x_1) - H(x_2)\rVert

w\cdot\big(H(x_1) - H(x_2)\big) = \lVert H(x_1) - H(x_2)\rVert\,(w\cdot w)

and

w\cdot w = 1

The right side satisfies

\lvert\psi'(\tau)\rvert \le \lVert DH(z_\tau)(x_1 - x_2)\rVert \le M\lVert x_1 - x_2\rVert

. Put the two sides together.

\blacksquare

Cauchy–Schwarz (

\lvert w\cdot u\rvert \le \lVert w\rVert\lVert u\rVert

with

\lVert w\rVert = 1

), then the definition of the operator norm.

Theorem B Rudin Thm 9.24; Spivak Thm 2-11

Inverse function theorem

Let $U \subseteq \mathbb R^n$ be open, $F: U \to \mathbb R^n$ continuously differentiable, $a \in U$ , and $\det DF(a) \ne 0$ . Then there are open sets $U' \ni a$ and $V \ni F(a)$ such that $F: U' \to V$ is a bijection, its inverse $G: V \to U'$ is continuously differentiable, and

DG(y) = \big(DF(G(y))\big)^{-1} \quad (y \in V).

In one lineIf $F: U \to \mathbb R^n$ is continuously differentiable on an open set $U$ and $\det DF(a) \ne 0$ , then $F$ has a continuously differentiable local inverse $G$ near $a$ , with $DG(F(x)) = DF(x)^{-1}$ .

Why it is trueIt says the linear approximation controls the field: if the best linear approximation is undoable, so is the field, near the point. It is the converse of the chain rule fact in 5.4 that a differentiable inverse forces an invertible Jacobian.

ProofRudin Thm 9.24

The idea: Solving F(x) = y is the same as finding a fixed point of the map x to x + A^{-1}(y - F(x)), where A is the Jacobian at a. Its derivative is small near a, so by the mean value inequality it is a contraction of a small closed ball. The Banach fixed point theorem gives a unique solution for each y close to F(a). The same inequality shows the solution depends continuously on y, and the differentiability of F then makes it differentiable.

Put

A = DF(a)

b = F(a)

and

\alpha = \lVert A^{-1}\rVert

. Choose

r \gt 0

so that the closed ball

K = \bar B(a, r)

lies in

U

and

\lVert DF(x) - A\rVert \le \dfrac1{2\alpha}

for all

x \in K

A

is invertible since

\det A \ne 0

. The operator norm

\alpha

A^{-1}

is a finite positive number (3.5). The entries of

DF

are continuous, so

\lVert DF(x) - A\rVert \to 0

x \to a

For each

y \in \mathbb R^n

define

\varphi_y(x) = x + A^{-1}\big(y - F(x)\big)

Fixed points of

\varphi_y

are exactly the solutions of

F(x) = y

: the correction

A^{-1}(y - F(x))

is zero if and only if

y = F(x)

, because

A^{-1}

is invertible.This is the 'where does the idea come from' step: Newton's method would use

DF(x)^{-1}

at each step, and here we use the fixed matrix

A^{-1}

so that the iteration map has a derivative we can bound.

D\varphi_y(x) = I - A^{-1}DF(x) = A^{-1}\big(A - DF(x)\big)

, so

\lVert D\varphi_y(x)\rVert \le \alpha\cdot\dfrac1{2\alpha} = \dfrac12

for

x \in K

Differentiate:

D(x) = I

D(F) = DF(x)

and

A^{-1}

is constant. Factor out

A^{-1}

on the left. Operator norm of a product is at most the product of the norms.

\lVert\varphi_y(x_1) - \varphi_y(x_2)\rVert \le \dfrac12\lVert x_1 - x_2\rVert

for all

x_1, x_2 \in K

Mean value inequality for fields, applied to

H = \varphi_y

on the convex set

K

with

M = \frac12

For

x_1, x_2 \in K

\lVert x_1 - x_2\rVert \le \dfrac12\lVert x_1 - x_2\rVert + \alpha\lVert F(x_1) - F(x_2)\rVert

, hence

\lVert x_1 - x_2\rVert \le 2\alpha\,\lVert F(x_1) - F(x_2)\rVert

(\star)

The difference

\varphi_y(x_1) - \varphi_y(x_2) = (x_1 - x_2) - A^{-1}\big(F(x_1) - F(x_2)\big)

does not contain

y

. Rearrange:

x_1 - x_2

equals that difference plus

A^{-1}(F(x_1) - F(x_2))

, then use the triangle inequality, the previous step, and the operator norm.

In particular

F

is one-to-one on

K

F(x_1) = F(x_2)

then

(\star)

forces

\lVert x_1 - x_2\rVert \le 0

Fix

y \in V

. Then

\varphi_y

maps

K

into

B(a, r) \subseteq K

Let

\rho = \dfrac{r}{2\alpha}

and

V = B(b, \rho)

, an open ball around

b = F(a)

. For

y \in V

and

x \in K

\varphi_y(a) - a = A^{-1}(y - b)

has length less than

\alpha\rho = \frac r2

, and

\lVert\varphi_y(x) - \varphi_y(a)\rVert \le \frac12\lVert x - a\rVert \le \frac r2

. Add them with the triangle inequality.Strict inequality: the sum is below

r

, so

\varphi_y(x)

is in the open ball

B(a, r)

and in particular in

K

\varphi_y

has exactly one fixed point

x

K

, that is, exactly one

x \in K

with

F(x) = y

, and

x \in B(a, r)

K

is a closed subset of the complete space

\mathbb R^n

, hence complete (4.2);

\varphi_y: K \to K

is a contraction with constant

\frac12

. Banach fixed point theorem.

Put

U' = B(a, r) \cap F^{-1}(V)

. Then

U'

is open, contains

a

, and

F: U' \to V

is a bijection. Let

G: V \to U'

be its inverse.

F

is differentiable, hence continuous (5.3), so

F^{-1}(V)

is open (open sets from continuous fields, 5.1). The intersection of two open sets is open.

\lVert G(y_1) - G(y_2)\rVert \le 2\alpha\lVert y_1 - y_2\rVert

, so

G

is continuous (Lipschitz).

Apply

(\star)

x_1 = G(y_1)

and

x_2 = G(y_2)

The Jacobian

DF(x)

is invertible for every

x \in K

: if

DF(x)v = O

then

\lVert v\rVert \le \frac12\lVert v\rVert

, so

v = O

Suppose

DF(x)v = O

for some

x \in K

. Then

v = v - A^{-1}DF(x)v = A^{-1}(A - DF(x))v

, whose length is at most

\frac12\lVert v\rVert

. A square matrix with trivial null space is invertible (rank–nullity, 2.3).

Write

y - y_0 = B(x - x_0) + \lVert x - x_0\rVert E(x)

Fix

y_0 \in V

x_0 = G(y_0)

B = DF(x_0)

, and let

y \to y_0

with

x = G(y)

. Definition of differentiability of

F

x_0

F(x) - F(x_0) = B(x - x_0) + \lVert x - x_0\rVert E(x)

with

E(x) \to O

x \to x_0

\lVert G(y) - G(y_0) - B^{-1}(y - y_0)\rVert \le \lVert B^{-1}\rVert\,\lVert x - x_0\rVert\,\lVert E(x)\rVert

Multiply by

B^{-1}

and rearrange:

x - x_0 - B^{-1}(y - y_0) = -\lVert x - x_0\rVert B^{-1}E(x)

.This is the shape of the definition of differentiability for

G

y_0

with the candidate derivative

B^{-1}

: the error is the right side.

So the error is at most

2\alpha\lVert B^{-1}\rVert\,\lVert y - y_0\rVert\,\lVert E(G(y))\rVert

, which is

\lVert y - y_0\rVert

times something tending to

0

. Hence

G

is differentiable at

y_0

with

DG(y_0) = B^{-1} = DF(G(y_0))^{-1}

(\star)

\lVert x - x_0\rVert \le 2\alpha\lVert y - y_0\rVert

. And

G

is continuous, so

x = G(y) \to x_0

, hence

E(G(y)) \to O

y \to y_0

DG = (DF\circ G)^{-1}

is continuous. So

G

is continuously differentiable.

\blacksquare

The entries of the inverse of a

2 \times 2

n \times n

matrix are polynomials in its entries divided by the determinant (Cramer's rule, 2.7), so inversion is continuous on invertible matrices. Compose continuous maps:

G

, then

DF

, then inversion.

Why· The picture behind the proof

In the visualiser of 5.3 you zoom in on $F$ at $a$ and see a linear map. The proof is a quantitative version of that zoom. The radius $r$ is how far you may zoom out before $DF$ stops being within $\frac1{2\alpha}$ of $A$ . Inside that ball, $F$ never moves two points closer together than half the linear map would (that is $(\star)$ ), so $F$ cannot fold the ball onto itself: it is one-to-one. And $F$ cannot leave holes: the ball $V$ of radius $\rho = r/(2\alpha)$ around $F(a)$ is entirely covered, since the contraction $\varphi_y$ finds a pre-image for each of its points.

Proof workshopRebuild the contraction proof of the inverse function theoremRudin Thm 9.24

Tap the lines in the order a proof would write them. Each line can only use what is already above it.

Quick checkWhich step of the inverse function theorem's proof uses completeness of

\mathbb R^n

Quick check

F(x, y) = (e^x\cos y,\ e^x\sin y)

has

\det DF = e^{2x} \ne 0

at every point. What follows?

What the hypotheses are for

Watch out· Continuity of the derivative cannot be dropped

The function $f(x) = \dfrac x2 + x^2\sin\dfrac1x$ with $f(0) = 0$ is differentiable everywhere, with $f'(0) = \frac12 \ne 0$ . For $x \ne 0$ , $f'(x) = \frac12 + 2x\sin\frac1x - \cos\frac1x$ . At the points $x_k = \frac1{2\pi k}$ this is $\frac12 + 0 - 1 = -\frac12$ , which is negative, and these points get arbitrarily close to $0$ . So $f'$ takes both signs in every interval around $0$ ; $f$ goes up and down there and is not one-to-one on any interval around $0$ . The derivative exists and is invertible at $0$ , but it is not continuous at $0$ (it oscillates between about $-\frac12$ and $\frac32$ near $0$ ), and the proof’s step ” $\lVert DF(x) - A\rVert$ is small near $a$ ” fails.

Watch out· The converse is false: det DF(a) = 0 does not forbid an inverse

$F(x) = x^3$ has $F'(0) = 0$ , yet $F$ is one-to-one with the inverse $y^{1/3}$ . What fails is only the differentiability of the inverse, and that is forced to fail by the chain rule of 5.4 ( $G'(0)F'(0) = 1$ is impossible). So $\det DF(a) \ne 0$ is exactly the condition for a differentiable local inverse; a field with $\det DF(a) = 0$ may or may not be locally one-to-one. The map $(x, y) \mapsto (x^2 - y^2,\ 2xy)$ (that is $z \mapsto z^2$ on the complex numbers of 0.6) has determinant $4(x^2 + y^2)$ , which is $0$ only at the origin, and there it is two-to-one: $z$ and $-z$ have the same square, so no local inverse exists.

Definition T

Implicit function

Let $F$ be a field of $n + m$ variables, written $F(x, y)$ with $x \in \mathbb R^n$ , $y \in \mathbb R^m$ , with $m$ components (so there are as many equations as unknowns $y$ ). A function $g: W \to \mathbb R^m$ on an open set $W \subseteq \mathbb R^n$ is an implicit function defined by $F(x, y) = O$ if $F(x, g(x)) = O$ for every $x \in W$ . The equation does not tell us a formula for $g$ . Example: the upper half of the unit circle, $g(x) = \sqrt{1 - x^2}$ , is an implicit function defined by $x^2 + y^2 - 1 = 0$ , and so is $-\sqrt{1 - x^2}$ ; each is one of the possible choices of $g$ .

In one lineA function $g$ with $F(x, g(x)) = 0$ : it is 'defined implicitly' by the equation $F(x, y) = 0$ , with no formula given.

Why it existsIt is the exact meaning of the phrase 'assume the equation defines $y$ as a function of $x$ ' that every implicit differentiation starts with.

Theorem T MATH 376 Worksheet 1 (statement); Rudin Thm 9.28; Apostol 9.6 (formulas)

Implicit function theorem

Let $F: \Omega \to \mathbb R^m$ be continuously differentiable on an open set $\Omega \subseteq \mathbb R^{n+m}$ , with points written $(x, y)$ , $x \in \mathbb R^n$ , $y \in \mathbb R^m$ . Let $(a, b) \in \Omega$ with $F(a, b) = O$ , and suppose that the $m \times m$ matrix $D_yF(a, b)$ of partial derivatives with respect to the $y$ -variables is invertible. Then there are an open set $W \ni a$ , an open set $N \ni (a, b)$ and a unique function $g: W \to \mathbb R^m$ such that

$g(a) = b$ ,
for $(x, y) \in N$ : $F(x, y) = O$ if and only if $x \in W$ and $y = g(x)$ ,
$g$ is continuously differentiable, and

Dg(x) = -\big(D_yF(x, g(x))\big)^{-1}\,D_xF(x, g(x)).

In one lineIf $F(x, y) = O$ at $(a, b)$ , $F$ is continuously differentiable and the $m \times m$ matrix $D_yF(a, b)$ is invertible, then near $(a, b)$ the equation defines exactly one $y = g(x)$ , it is continuously differentiable, and $Dg(x) = -D_yF^{-1}D_xF$ .

Why it is trueIt is the licence for implicit differentiation: it says when the function being differentiated exists and has a derivative. The single condition is a Jacobian determinant that is not zero, the same condition as the inverse function theorem.

Here $D_xF$ is the $m \times n$ matrix of partials with respect to the $x$ -variables and $D_yF$ the $m \times m$ matrix of partials with respect to the $y$ -variables, so that $J_F = [\,D_xF\ \ D_yF\,]$ is the Jacobian of $F$ as a field on $\mathbb R^{n+m}$ . The statement is the one used as a black box in MATH 376 Worksheet 1; the proof below is not in the course (label B).

ProofRudin Thm 9.28; Spivak Thm 2-12

The idea: Build the map Phi(x, y) = (x, F(x, y)) from R^(n+m) to itself. Its Jacobian is block triangular with determinant det D_yF, so the inverse function theorem inverts it near (a, b). The inverse sends (x, 0) to (x, g(x)); that is the implicit function. The derivative formula is the chain rule applied to the identity F(x, g(x)) = 0.

Define

\Phi: \Omega \to \mathbb R^{n+m}

\Phi(x, y) = \big(x, F(x, y)\big)

Phi keeps

x

and replaces

y

by the value

F(x, y)

. It is continuously differentiable because

F

is.

D\Phi = \begin{bmatrix} I & 0\\ D_xF & D_yF\end{bmatrix}

, so

\det D\Phi(a, b) = \det D_yF(a, b) \ne 0

Jacobian of

\Phi

in block form: the first

n

rows are

[\,I\ \ 0\,]

(the derivative of

x

) and the last

m

rows are

[\,D_xF\ \ D_yF\,]

. A block lower-triangular matrix has determinant equal to the product of the determinants of its diagonal blocks.

There are open sets

N \ni (a, b)

and

V \ni (a, O)

such that

\Phi: N \to V

is a bijection with continuously differentiable inverse

\Psi: V \to N

Inverse function theorem (this chapter) for the field

\Phi

at the point

(a, b)

, where

\Phi(a, b) = (a, O)

Write

\Psi(x, z) = \big(x, h(x, z)\big)

\Phi

does not change the first

n

coordinates, so neither does its inverse:

\Psi(x, z) = (x, h(x, z))

for some continuously differentiable

h

Let

W = \{x \in \mathbb R^n : (x, O) \in V\}

and

g(x) = h(x, O)

for

x \in W

V

is open and contains

(a, O)

, so

W = \{x : (x, O) \in V\}

is an open set containing

a

(it is the preimage of

V

under the continuous map

x \mapsto (x, O)

F(x, g(x)) = O

for every

x \in W

, and

g(a) = b

because

\Psi(a, O) = (a, b)

\Psi(x, O) = (x, g(x))

lies in

N

. Apply

\Phi

to both sides:

\Phi(\Psi(x, O)) = (x, O)

, that is

(x, F(x, g(x))) = (x, O)

.This is the step that produces the implicit function: solving

F(x, y) = O

for

y

is the same as inverting

\Phi

along the slice where the second coordinate is

O

Uniqueness: for

(x, y) \in N

F(x, y) = O

if and only if

x \in W

and

y = g(x)

(x, y) \in N

and

F(x, y) = O

then

\Phi(x, y) = (x, O)

, so

(x, O) \in V

(so

x \in W

) and, since

\Phi

is one-to-one on

N

and

\Phi(x, g(x)) = (x, O)

too,

y = g(x)

. Conversely

x \in W

gives a solution

y = g(x)

g

is continuously differentiable.

g

is built from

h

and the inclusion

x \mapsto (x, O)

, both continuously differentiable.

D_xF + D_yF\;Dg = 0

Chain rule applied to the identity

F(x, g(x)) = O

: the field

x \mapsto (x, g(x))

has Jacobian

\begin{bmatrix} I\\ Dg\end{bmatrix}

, and the outer Jacobian is

[\,D_xF\ \ D_yF\,]

. The derivative of the constant

O

is the zero matrix.

Dg(x) = -D_yF^{-1}D_xF

\blacksquare

Multiply on the left by

D_yF^{-1}

, which exists at

(x, g(x))

for

x

near

a

because

D\Phi

is invertible at every point of

N

(it has the inverse

D\Psi

there), and

\det D\Phi = \det D_yF

Proof workshopGive the reason for each line: the implicit function theorem from the inverse function theoremRudin Thm 9.28

Each line of the proof is shown in order. Choose the reason that justifies it, then press Check.

Let $\Phi(x, y) = (x, F(x, y))$ , a continuously differentiable map of $\mathbb R^{n+m}$ to itself.
$D\Phi = \begin{bmatrix} I & 0\\ D_xF & D_yF\end{bmatrix}$ and $\det D\Phi(a, b) = \det D_yF(a, b) \ne 0$ .
By the inverse function theorem, $\Phi$ has a continuously differentiable inverse $\Psi$ from a neighbourhood $V$ of $(a, O)$ onto a neighbourhood $N$ of $(a, b)$ .
$\Psi(x, z) = (x, h(x, z))$ ; put $g(x) = h(x, O)$ .
$\Phi(\Psi(x, O)) = (x, O)$ says $F(x, g(x)) = O$ .
Differentiating $F(x, g(x)) = O$ gives $D_xF + D_yF\,Dg = 0$ , so $Dg = -D_yF^{-1}D_xF$ .

Implicit differentiation, with permission

Now the habit of every JEE student is a theorem. If the equation defines $y = g(x)$ , differentiate the identity $F(x, g(x)) = 0$ with the chain rule and solve for $g'$ . The division by $F_y$ is exactly the hypothesis that makes the implicit function theorem work.

Corollary T MATH 376 Worksheet 1; Apostol Thm 9.3 and (9.24)

Implicit differentiation formulas

(a) If $F(x, y) = 0$ with $F$ continuously differentiable and $F_y(a, b) \ne 0$ , then near $a$ the implicit function $y = g(x)$ satisfies $g'(x) = -\dfrac{F_x}{F_y}$ and, if $F$ is $C^2$ ,

g''(x) = -\frac{F_{xx}F_y^2 - 2F_{xy}F_xF_y + F_{yy}F_x^2}{F_y^3}.

(b) If $F(x, y, z) = 0$ and $F_z \ne 0$ , then $z = z(x, y)$ has $z_x = -F_x/F_z$ and $z_y = -F_y/F_z$ .

(c) If $F(x, y, u, v) = 0$ and $G(x, y, u, v) = 0$ at a point where $\dfrac{\partial(F, G)}{\partial(u, v)} = F_uG_v - F_vG_u \ne 0$ , then near that point they define $u = u(x, y)$ , $v = v(x, y)$ with

u_x = -\frac{\partial(F, G)/\partial(x, v)}{\partial(F, G)/\partial(u, v)}, \qquad v_x = -\frac{\partial(F, G)/\partial(u, x)}{\partial(F, G)/\partial(u, v)},

and similarly for $u_y$ , $v_y$ with $y$ in place of $x$ .

In one lineUnder the hypotheses of the implicit function theorem: $\dfrac{dy}{dx} = -\dfrac{F_x}{F_y}$ ; for $F(x, y, z) = 0$ , $z_x = -\dfrac{F_x}{F_z}$ and $z_y = -\dfrac{F_y}{F_z}$ ; for two equations in $u, v$ , $u_x = -\dfrac{\partial(F, G)/\partial(x, v)}{\partial(F, G)/\partial(u, v)}$ .

Why it is trueThese are the formulas used in every implicit differentiation problem, with the exact condition (the denominator not zero) that makes them valid.

ProofApostol 9.6, 9.7

The idea: Everything is the matrix formula Dg = -(D_yF)^(-1) D_xF in the three cases (one unknown, one unknown with two inputs, two unknowns). For the second derivative, differentiate the first formula again, remembering that y depends on x.

(a)

g'(x) = -F_y^{-1}F_x

The implicit function theorem, with

m = 1

and the single input

x

D_xF = F_x

D_yF = F_y

are numbers.

g'' = -\dfrac{(F_{xx} + F_{xy}g')F_y - F_x(F_{yx} + F_{yy}g')}{F_y^2}

Quotient rule and the chain rule:

F_x

and

F_y

are functions of

x

and

y = g(x)

, so

\frac d{dx}F_x(x, g(x)) = F_{xx} + F_{xy}\,g'

and the same for

F_y

with

F_{yx}

F_{yy}

This simplifies to the formula for

g''

Put

g' = -F_x/F_y

and use

F_{xy} = F_{yx}

(5.5). The numerator becomes

F_{xx}F_y - 2F_xF_{xy} + F_{yy}F_x^2/F_y

; multiply top and bottom by

F_y

(b)

(z_x, z_y) = -F_z^{-1}(F_x, F_y)

Implicit function theorem with the single unknown

z

and the two inputs

(x, y)

D_xF = (F_x, F_y)

and

D_yF = F_z

(c)

u_x = \dfrac{\det\begin{bmatrix}-F_x & F_v\\ -G_x & G_v\end{bmatrix}}{\partial(F, G)/\partial(u, v)}

and

v_x = \dfrac{\det\begin{bmatrix}F_u & -F_x\\ G_u & -G_x\end{bmatrix}}{\partial(F, G)/\partial(u, v)}

Here the unknowns are

(u, v)

, so

D_{(u,v)}F = \begin{bmatrix}F_u & F_v\\ G_u & G_v\end{bmatrix}

with determinant

\partial(F, G)/\partial(u, v)

. Differentiating the two identities in

x

gives the linear system

F_uu_x + F_vv_x = -F_x

G_uu_x + G_vv_x = -G_x

; solve it by Cramer's rule (2.7).

These are the stated formulas.

Pull the

-1

out of the first column of the first determinant and out of the second column of the second determinant.

\blacksquare

Worked example MATH 376 Worksheet 1 #1

Let

x^2 + y^2 = 1

. Find

\dfrac{dx}{dy}

, the tangent vector at

\big(\frac{\sqrt2}2, \frac{\sqrt2}2\big)

, and

\dfrac{d^2x}{dy^2}

Solution.

Choose the dependent variable. The worksheet treats $x$ as a function of $y$ , so the hypothesis it needs is $F_x = 2x \ne 0$ . The theorem solves for the variable whose partial derivative you divide by. Near the points $(\pm1, 0)$ the circle is a graph over the $y$ -axis; near $(0, \pm1)$ it is not (there it is a graph over the $x$ -axis).
Differentiate the identity. $x(y)^2 + y^2 = 1$ gives $2x\,\dfrac{dx}{dy} + 2y = 0$ , so $\dfrac{dx}{dy} = -\dfrac yx$ . Chain rule on $x(y)^2$ ; the formula $-F_y/F_x$ of the theorem.
Tangent vector. A curve $(x(y), y)$ has velocity $\big(\frac{dx}{dy}, 1\big)$ . At the point, $\frac{dx}{dy} = -1$ , so the velocity is $\langle-1, 1\rangle$ . Velocity of $y \mapsto (x(y), y)$ ; it is perpendicular to $\nabla F = (\sqrt2, \sqrt2)$ , as the gradient theorem of 5.3 says.
Second derivative. Differentiate $\dfrac{dx}{dy} = -\dfrac yx$ in $y$ with the quotient rule, remembering that $x$ depends on $y$ : $\dfrac{d^2x}{dy^2} = -\dfrac{x - y\,\frac{dx}{dy}}{x^2} = -\dfrac{x + y^2/x}{x^2} = -\dfrac{x^2 + y^2}{x^3} = -\dfrac1{x^3}$ . Replace $\frac{dx}{dy}$ by $-\frac yx$ , then use the equation of the circle, $x^2 + y^2 = 1$ .
Evaluate. At $x = \frac{\sqrt2}2$ : $-\dfrac1{x^3} = -\dfrac{1}{\sqrt2/4} = -2\sqrt2$ . $\big(\frac{\sqrt2}2\big)^3 = \frac{2\sqrt2}8 = \frac{\sqrt2}4$ .

Answer. $\dfrac{dx}{dy} = -\dfrac yx$ , tangent vector $\langle-1, 1\rangle$ , $\dfrac{d^2x}{dy^2} = -\dfrac1{x^3} = -2\sqrt2$ at the point. It is negative: the right half of the circle bends to the left as a function of $y$ (the curve is concave as seen from the $y$ -axis).

Your answerOn the circle

x^2 + y^2 = 25

, what is

\dfrac{dy}{dx}

(3, 4)

Your answerFor

x^2 + y^2 = 1

, find

\dfrac{d^2x}{dy^2}

\big(\frac{\sqrt2}2, \frac{\sqrt2}2\big)

Worked example MATH 376 Worksheet 1 #2

The equation

x^2 + y^2 + z^2 = 1

defines

z = z(x, y)

where

z \ne 0

. Find

z_x

z_y

z_{xx}

z_{yy}

, and show

z_{xx} + z_{yy} = -\dfrac{1 + z^2}{z^3}

Solution.

State the assumption. Write $z = z(x, y)$ , so $x^2 + y^2 + z(x, y)^2 - 1 = 0$ is an identity in $(x, y)$ near the point. The identity may be differentiated; this is the implicit function theorem with $F_z = 2z \ne 0$ .
First derivatives. In $x$ with $y$ fixed: $2x + 2z\,z_x = 0$ , so $z_x = -\dfrac xz$ . In $y$ : $z_y = -\dfrac yz$ . Chain rule on $z^2$ : $\frac{d}{dx}z^2 = 2z\,z_x$ because $z$ depends on $x$ . Dividing by $z$ needs $z \ne 0$ .
Second derivative in $x$ . Quotient rule on $-x/z$ with $z$ depending on $x$ : $z_{xx} = -\dfrac{z - x\,z_x}{z^2} = -\dfrac{z + x^2/z}{z^2} = -\dfrac{z^2 + x^2}{z^3}$ . Replace $z_x$ by $-x/z$ so no derivative is left, then multiply top and bottom by $z$ .
Second derivative in $y$ . The same computation gives $z_{yy} = -\dfrac{z^2 + y^2}{z^3}$ .
Add and simplify. $z_{xx} + z_{yy} = -\dfrac{2z^2 + x^2 + y^2}{z^3}$ , and $x^2 + y^2 = 1 - z^2$ , so the numerator is $2z^2 + 1 - z^2 = 1 + z^2$ . Use the original equation to eliminate $x^2 + y^2$ .

Answer. $z_x = -\dfrac xz$ , $z_y = -\dfrac yz$ , $z_{xx} + z_{yy} = -\dfrac{1 + z^2}{z^3}$ . Check at the north pole $z = 1$ : the upper hemisphere $z = \sqrt{1 - x^2 - y^2}$ has $z_{xx} = z_{yy} = -1$ there, and the formula gives $-\frac{1 + 1}{1} = -2$ . ✓.

Worked example MATH 376 Worksheet 1 #3

Differentiate

y = \arcsin x

by implicit differentiation.

Solution.

Turn the inverse into an equation. $y = \arcsin x$ means $\sin y = x$ with $y \in [-\frac\pi2, \frac\pi2]$ . Put $F(x, y) = \sin y - x$ . An inverse function is the implicit function of an equation.
Check the hypothesis. $F_y = \cos y \ne 0$ exactly when $y \ne \pm\frac\pi2$ , that is $x \ne \pm1$ . This is where the inverse function is allowed to be differentiated.
Differentiate. $\dfrac{dy}{dx} = -\dfrac{F_x}{F_y} = \dfrac1{\cos y}$ . $F_x = -1$ .
Express in $x$ . On $[-\frac\pi2, \frac\pi2]$ , $\cos y \ge 0$ , so $\cos y = \sqrt{1 - \sin^2y} = \sqrt{1 - x^2}$ . The sign of the square root is fixed by the range of $\arcsin$ .

Answer. $\dfrac{d}{dx}\arcsin x = \dfrac1{\sqrt{1 - x^2}}$ , valid for $-1 \lt x \lt 1$ . At $x = \pm1$ the hypothesis fails ( $F_y = 0$ ): the graph of $\arcsin$ has a vertical tangent there. The same method gives $\frac{d}{dx}\arctan x = \cos^2 y = \frac1{1 + x^2}$ .

Worked example MATH 376 Worksheet 1 #4

The equations

x^2 + y^2 + z^2 - w^2 = 1

and

x + y + z - w = 0

define

z = z(x, y)

and

w = w(x, y)

. Find

z_x

w_x

, and say where the answer is valid.

Solution.

Name the two equations. $F = x^2 + y^2 + z^2 - w^2 - 1$ and $G = x + y + z - w$ . Two unknowns $(z, w)$ , two equations: the theorem applies with $m = 2$ .
The condition. $\dfrac{\partial(F, G)}{\partial(z, w)} = \det\begin{bmatrix}F_z & F_w\\ G_z & G_w\end{bmatrix} = \det\begin{bmatrix}2z & -2w\\ 1 & -1\end{bmatrix} = -2z + 2w = 2(w - z)$ . It must be nonzero. The determinant of the matrix you invert, which is the one you divide by below.
Differentiate both equations in $x$ . $2x + 2z\,z_x - 2w\,w_x = 0$ and $1 + z_x - w_x = 0$ . Chain rule; $z$ and $w$ both depend on $x$ , and $y$ is held fixed.
Solve the linear system. From the second, $w_x = 1 + z_x$ . Substituting into the first: $2x + 2z\,z_x - 2w - 2w\,z_x = 0$ , so $2(z - w)z_x = 2(w - x)$ and $z_x = \dfrac{w - x}{z - w}$ . Then $w_x = 1 + z_x = \dfrac{z - x}{z - w}$ . Divide by $2(z - w)$ , which is allowed exactly when the determinant of step 2 is not zero.

Answer. $z_x = \dfrac{w - x}{z - w}$ , $w_x = \dfrac{z - x}{z - w}$ , valid where $z \ne w$ . (The same method gives $z_y = \frac{w - y}{z - w}$ , $w_y = \frac{z - y}{z - w}$ . The matrix formula $Dg = -D_yF^{-1}D_xF$ gives all four at once.)

Worked example Apostol 9.6, Example

Assume

y^2 + xz + z^2 - e^z - c = 0

defines

z = f(x, y)

. Find

c

so that

f(0, e) = 2

, and then

\dfrac{\partial f}{\partial x}

and

\dfrac{\partial f}{\partial y}

(0, e)

Solution.

Find $c$ . At $(x, y, z) = (0, e, 2)$ : $e^2 + 0 + 4 - e^2 - c = 0$ , so $c = 4$ . The point must lie on the surface.
Gradient of $F = y^2 + xz + z^2 - e^z - 4$ . $F_x = z$ , $F_y = 2y$ , $F_z = x + 2z - e^z$ . Ordinary partial derivatives.
Check the hypothesis. $F_z(0, e, 2) = 0 + 4 - e^2 \approx -3.39 \ne 0$ . The theorem applies: $z$ is a differentiable function of $(x, y)$ near the point, although we cannot find a formula for it.
Apply the formulas. $f_x = -\dfrac{F_x}{F_z} = -\dfrac{2}{4 - e^2} = \dfrac2{e^2 - 4}$ and $f_y = -\dfrac{F_y}{F_z} = \dfrac{2e}{e^2 - 4}$ . $z_x = -F_x/F_z$ and $z_y = -F_y/F_z$ .

Answer. $c = 4$ , $f_x = \dfrac2{e^2 - 4} \approx 0.590$ , $f_y = \dfrac{2e}{e^2 - 4} \approx 1.604$ . We found the derivatives using the value of $f$ at a single point.

Your answerIn the example above, what is

\partial f/\partial x

(0, e)

, to three decimal places?

Worked example new (a classic implicit-differentiation example; compare Apostol 9.8)

Study the folium of Descartes

x^3 + y^3 = 3xy

: the slope

\dfrac{dy}{dx}

, the slope at

\big(\frac32, \frac32\big)

, the points of horizontal tangent, and what happens at the origin.

Solution.

Write $F$ . $F(x, y) = x^3 + y^3 - 3xy$ , with $F_x = 3x^2 - 3y$ and $F_y = 3y^2 - 3x$ . Partial derivatives.
Slope. $\dfrac{dy}{dx} = -\dfrac{F_x}{F_y} = \dfrac{y - x^2}{y^2 - x}$ , valid where $y^2 \ne x$ . The formula of the theorem; the hypothesis is $F_y \ne 0$ .
At $\big(\frac32, \frac32\big)$ . It is on the curve: $2\cdot\frac{27}8 = \frac{27}4 = 3\cdot\frac94$ . The slope is $\dfrac{\frac32 - \frac94}{\frac94 - \frac32} = \dfrac{-3/4}{3/4} = -1$ . Substitute; the tangent is perpendicular to the diagonal, since the curve is symmetric under $x \leftrightarrow y$ .
Horizontal tangents. $F_x = 0$ means $y = x^2$ . Substituting in the equation: $x^3 + x^6 = 3x^3$ , so $x^6 = 2x^3$ , giving $x = 0$ or $x^3 = 2$ . For $x = 2^{1/3}$ we get $y = 2^{2/3}$ , the top of the loop. A horizontal tangent has $\frac{dy}{dx} = 0$ , which needs $F_x = 0$ and $F_y \ne 0$ . Check: $F_y = 3(2^{4/3}) - 3(2^{1/3}) \ne 0$ .
The origin. $F_x(0, 0) = 0$ and $F_y(0, 0) = 0$ , so the theorem is silent. The curve passes through the origin twice: near $(0, 0)$ the cubic terms are small and the equation is about $3xy = 0$ , so the two branches run along the two axes. The point $x = 0$ found in step 4 has $y = 0$ , where $F_y = 0$ too, so the slope formula is $0/0$ there. The two branches have tangents $y = 0$ and $x = 0$ .

Answer. $\dfrac{dy}{dx} = \dfrac{y - x^2}{y^2 - x}$ ; slope $-1$ at $\big(\frac32, \frac32\big)$ ; horizontal tangent at $\big(2^{1/3}, 2^{2/3}\big)$ (and by symmetry vertical tangent at $\big(2^{2/3}, 2^{1/3}\big)$ ); a double point at the origin where $\nabla F = O$ and the curve is not a graph.

Your answerWhat is the slope of the folium

x^3 + y^3 = 3xy

\big(\frac32, \frac32\big)

Quick checkAt which point of the circle

x^2 + y^2 = 1

does the implicit function theorem fail to give

y

as a function of

x

near the point?

Quick check

F(x, y) = x - y^3

has

F_y(0, 0) = 0

. Does the equation

F = 0

define

y

as a function of

x

near the origin?

Quick checkThe two equations

F = 0

and

G = 0

in the unknowns

u, v

(with

x, y

as parameters) can be solved for

u, v

near a solution when:

Inverse maps in practice

Worked example new (compare Apostol 8.22 and Practice Final 2013 #10)

Where is

F(x, y) = (x^2 - y^2,\ 2xy)

locally invertible? Find the derivative of the local inverse at

F(1, 1) = (0, 2)

without finding the inverse.

Solution.

The Jacobian. $DF = \begin{bmatrix}2x & -2y\\ 2y & 2x\end{bmatrix}$ and $\det DF = 4x^2 + 4y^2$ . Rows are the gradients of $x^2 - y^2$ and $2xy$ .
Where the theorem applies. $F$ is continuously differentiable and $\det DF \ne 0$ at every point except the origin. So $F$ has a differentiable local inverse near every point except $(0, 0)$ . Inverse function theorem.
At the origin. There is no local inverse: $F(-x, -y) = F(x, y)$ , so every value near $(0, 0)$ (other than $(0, 0)$ ) has exactly two preimages arbitrarily close to $(0, 0)$ . This is $z \mapsto z^2$ on the complex numbers, which folds the plane in half by doubling angles.
Derivative of the inverse at $(0, 2)$ . $DF(1, 1) = \begin{bmatrix}2 & -2\\ 2 & 2\end{bmatrix}$ , determinant $8$ , inverse $\dfrac18\begin{bmatrix}2 & 2\\ -2 & 2\end{bmatrix} = \begin{bmatrix}\frac14 & \frac14\\ -\frac14 & \frac14\end{bmatrix}$ . $DG(F(x)) = DF(x)^{-1}$ ; the $2 \times 2$ inverse swaps the diagonal, negates the off-diagonal and divides by the determinant.
Check with the real inverse. Near $(1, 1)$ , $G(X, Y)$ is the principal square root of $X + iY$ . Its derivative at $X + iY = 2i$ is $\frac1{2\sqrt{2i}} = \frac1{2(1 + i)}= \frac{1 - i}4$ , which as a matrix acts as $\begin{bmatrix}\frac14 & \frac14\\ -\frac14 & \frac14\end{bmatrix}$ on $(X, Y)$ (real part $\frac{X + Y}4$ , imaginary part $\frac{Y - X}4$ ). Complex derivative of the square root, $\frac{1}{2z}$ , with $z = 1 + i$ ; multiplying by $\frac{1 - i}4$ is the matrix above.

Answer. Locally invertible everywhere except the origin; $DG(0, 2) = \begin{bmatrix}\frac14 & \frac14\\ -\frac14 & \frac14\end{bmatrix}$ . Note that $F$ is not one-to-one on the plane minus the origin, so the inverse exists only locally.

Your answerIn the example above, what is the entry in row 1, column 1 of the derivative of the local inverse at

F(1, 1) = (0, 2)

Worked example Apostol 9.7, Example 3

The equations

2x = v^2 - u^2

and

y = uv

define

u, v

as functions of

x, y

. Find

u_x

u_y

v_x

v_y

Solution.

Write the system. $F = v^2 - u^2 - 2x$ , $G = uv - y$ , unknowns $(u, v)$ . Move everything to one side.
The condition. $\dfrac{\partial(F, G)}{\partial(u, v)} = \det\begin{bmatrix}-2u & 2v\\ v & u\end{bmatrix} = -2u^2 - 2v^2 = -2(u^2 + v^2)$ , nonzero except at $u = v = 0$ . The matrix to invert.
Differentiate in $x$ . $-2u\,u_x + 2v\,v_x = 2$ and $v\,u_x + u\,v_x = 0$ . Chain rule with $y$ held fixed; $\frac{\partial}{\partial x}(2x) = 2$ .
Solve. From the second, $v_x = -\dfrac{v}{u}u_x$ (if $u \ne 0$ ). Then $-2u\,u_x - \dfrac{2v^2}{u}u_x = 2$ , so $u_x\Big(-\dfrac{2(u^2 + v^2)}{u}\Big) = 2$ and $u_x = -\dfrac{u}{u^2 + v^2}$ , $v_x = \dfrac{v}{u^2 + v^2}$ . Multiply through by $u$ and factor.
Differentiate in $y$ . $-2u\,u_y + 2v\,v_y = 0$ and $v\,u_y + u\,v_y = 1$ . The solution is $u_y = \dfrac{v}{u^2 + v^2}$ , $v_y = \dfrac{u}{u^2 + v^2}$ . Same steps with the right-hand sides $(0, 1)$ . Check with the matrix formula: $-D_{(u,v)}F^{-1}D_{(x,y)}F$ .

Answer. $u_x = -\dfrac{u}{u^2 + v^2}$ , $v_x = \dfrac{v}{u^2 + v^2}$ , $u_y = \dfrac{v}{u^2 + v^2}$ , $v_y = \dfrac{u}{u^2 + v^2}$ , the same as Apostol’s printed answers.

Python labUndoing a map by iteration: the proof's contraction against Newton

import numpy as np

def F(p):  return np.array([p[0]**2 - p[1]**2, 2 * p[0] * p[1]])
def DF(p): return np.array([[2 * p[0], -2 * p[1]], [2 * p[1], 2 * p[0]]])

target = np.array([0.0, 2.8])                   # solve F(x) = target near a = (1, 1)
z = np.sqrt(complex(*target)); exact = np.array([z.real, z.imag])   # z -> z^2 has a complex square root
a = np.array([1.0, 1.0]); Ainv = np.linalg.inv(DF(a))
p = a.copy(); q = a.copy()
print(" k   contraction error   Newton error     ratio of contraction errors")
prev = np.linalg.norm(p - exact)
for k in range(1, 9):
    p = p + Ainv @ (target - F(p))              # fixed matrix A^(-1): the proof's contraction
    q = q + np.linalg.solve(DF(q), target - F(q))   # Newton: new Jacobian every step
    e = np.linalg.norm(p - exact)
    print(f"{k:2d}   {e:.3e}          {np.linalg.norm(q - exact):.3e}      {e/prev:.3f}")
    prev = e
print("solution:", exact, " F(solution) =", F(exact))

Try this. Run it. The iteration x to x + A^(-1)(y - F(x)) of the proof uses the Jacobian A at the starting point once and for all. Its error shrinks by a roughly constant factor each step (about 0.18 here), the contraction constant of the theorem. Newton's iteration recomputes the Jacobian at every step, and its error is roughly squared each time (2.4e-2, 1.7e-4, 8e-9, then machine precision). Change the target to (0, 2.2), closer to F(1, 1) = (0, 2): the contraction factor drops to about 0.05. Now change the start a to (0.2, 0.2), where the fixed matrix is a poor model of F at the solution: the contraction iteration blows up (the proof needs a small enough ball), while Newton wanders and then converges.

Level sets are smooth where the gradient is not zero

5.3 defined the tangent plane to a level surface as $\{x : \nabla f(a)\cdot(x - a) = 0\}$ and admitted that nothing yet showed that the surface really is a smooth surface with that plane. The implicit function theorem pays that debt, in any dimension.

Definition B

Regular point

Let $F = (F_1, \dots, F_m): \mathbb R^N \to \mathbb R^m$ be continuously differentiable, $m \lt N$ . A point $a$ is a regular point of $F$ if the $m \times N$ matrix $DF(a)$ has rank $m$ , that is, its rows $\nabla F_1(a), \dots, \nabla F_m(a)$ are linearly independent (2.2). For a single equation ( $m = 1$ ) it means $\nabla F(a) \ne O$ . A point that is not regular is singular: the folium’s origin, the cusp’s origin.

In one lineFor $F: \mathbb R^N \to \mathbb R^m$ with $m \lt N$ , a point $a$ where the Jacobian $DF(a)$ has rank $m$ : the $m$ gradients $\nabla F_1(a), \dots, \nabla F_m(a)$ are linearly independent.

Why it existsIt is the condition under which $m$ equations cut out a surface of the expected dimension $N - m$ with a tangent plane, and the condition Lagrange multipliers need in 5.8.

Corollary B Rudin Thm 9.28; Spivak Thm 5-1

Level sets near a regular point are graphs

Let $F$ be continuously differentiable near $a$ , $F(a) = c$ , and $DF(a)$ of rank $m$ . Then (a) after renaming the coordinates, $N - m$ of them $x$ and $m$ of them $y$ , there is a $C^1$ function $g$ with $\{F = c\} = \{(x, g(x))\}$ near $a$ ; (b) the set of velocity vectors $r'(0)$ of $C^1$ curves $r$ in $\{F = c\}$ with $r(0) = a$ is exactly the null space $\ker DF(a)$ , a subspace of dimension $N - m$ .

In one lineIf $a$ is a regular point of $F: \mathbb R^N \to \mathbb R^m$ with $F(a) = c$ , then near $a$ the level set $\{F = c\}$ is the graph of a $C^1$ function of $N - m$ of the coordinates, and its tangent vectors at $a$ are exactly the vectors in the null space of $DF(a)$ .

Why it is trueIt proves that level sets are genuine smooth curves and surfaces away from singular points, and it identifies the tangent space, which is the fact the proof of the Lagrange multiplier theorem uses.

ProofSpivak Thm 5-1

The idea: Rank m means some m columns of the Jacobian are independent, so the m by m block for those variables is invertible and the implicit function theorem solves for them. For the tangent space, a curve in the level set has F of it constant, so its velocity is killed by DF; conversely any vector killed by DF is the velocity of the curve built from the implicit function.

Rename the coordinates so that

D_yF(a)

is invertible.

The rank is the number of independent columns (2.6). Reorder the variables so that the last

m

columns, for the variables

y

, are independent; they then form an invertible

m \times m

matrix

D_yF(a)

There is a

C^1

function

g

near

a_x

with

F(x, g(x)) = c

, and these are the only solutions near

a

. This is (a).

Implicit function theorem for

F(x, y) - c = O

r

is a curve in the level set through

a

, then

DF(a)\,r'(0) = O

, so

r'(0) \in \ker DF(a)

Chain rule along the curve:

F(r(t)) = c

for all

t

, so

\frac d{dt}F(r(t)) = DF(r(t))\,r'(t) = O

Conversely, let

v \in \ker DF(a)

. Define the curve

r(t) = \big(a_x + tv_x,\ g(a_x + tv_x)\big)

, which lies in the level set.

Write

v = (v_x, v_y)

DF(a)v = D_xF\,v_x + D_yF\,v_y = O

means

v_y = -D_yF^{-1}D_xF\,v_x = Dg(a_x)\,v_x

(the implicit function theorem formula).

Its velocity at

t = 0

v

r'(0) = (v_x, Dg(a_x)v_x) = (v_x, v_y) = v

by the previous step. Rank–nullity (2.3) gives

\dim\ker DF(a) = N - m

\blacksquare

The corollary closes the gap in 5.3: at a regular point of a level surface in $\mathbb R^3$ the vectors perpendicular to $\nabla f(a)$ fill the whole plane, and every one of them is the velocity of a curve in the surface. The same holds for a curve cut out by two equations in $\mathbb R^3$ (as in Apostol 8.17 #9 and 9.8 #4, the next example): its tangent line is the null space of the $2 \times 3$ Jacobian, that is, the direction perpendicular to both gradients.

Worked example Apostol 9.8 #4(a)

The surfaces

2x^2 + 3y^2 - z^2 = 25

and

x^2 + y^2 = z^2

meet in a curve through

P = (\sqrt7, 3, 4)

. Find a unit tangent vector to the curve at

P

without parametrising it.

Solution.

Check the point. $14 + 27 - 16 = 25$ and $7 + 9 = 16$ . $P$ is on both surfaces.
Gradients at $P$ . $\nabla F = (4x, 6y, -2z) = (4\sqrt7, 18, -8)$ and $\nabla G = (2x, 2y, -2z) = (2\sqrt7, 6, -8)$ , for $F = 2x^2 + 3y^2 - z^2$ and $G = x^2 + y^2 - z^2$ . The level surfaces $F = 25$ and $G = 0$ .
Regular point. The two gradients are not parallel (the ratio of the first entries is $2$ , of the second $3$ ). So $P$ is a regular point of $(F, G)$ and the intersection is a smooth curve near $P$ .
Tangent direction. The tangent is perpendicular to both gradients, so it is parallel to their cross product: $\nabla F\times\nabla G = \big(18(-8) - (-8)(6),\ (-8)(2\sqrt7) - 4\sqrt7(-8),\ 4\sqrt7\cdot6 - 18\cdot2\sqrt7\big) = (-96,\ 16\sqrt7,\ -12\sqrt7)$ . The null space of a $2 \times 3$ matrix of rank $2$ is the line spanned by the cross product of its rows.
Normalise. Divide by $4$ : $(-24, 4\sqrt7, -3\sqrt7)$ has length $\sqrt{576 + 112 + 63} = \sqrt{751}$ . $4\sqrt7$ squared is $112$ and $3\sqrt7$ squared is $63$ .
Cross-check by the formula. Solving for $x, y$ in terms of $z$ : $x^2 = 2z^2 - 25$ and $y^2 = 25 - z^2$ , so $\frac{dx}{dz} = \frac{2z}{x} = \frac{8}{\sqrt7}$ and $\frac{dy}{dz} = -\frac zy = -\frac43$ at $P$ . The velocity vector $\big(\frac8{\sqrt7}, -\frac43, 1\big)$ of the curve $z \mapsto (x(z), y(z), z)$ equals $\dfrac{-1}{3\sqrt7}\cdot(-24,\ 4\sqrt7,\ -3\sqrt7)$ , a multiple of the cross product. Both descriptions give the same line: the null space of the Jacobian is the tangent line.

Answer. $T = \pm\dfrac{(24,\ -4\sqrt7,\ 3\sqrt7)}{\sqrt{751}}$ .

Where it is used

Used inside maths· Inside maths: smooth surfaces, moving roots and differentiable eigenvalues

The corollary above is why the sphere, the torus and every level surface $\{f = c\}$ with $\nabla f \ne O$ are smooth surfaces: it is the starting point of differential geometry and the definition of a manifold in MATH 551 and 561. Roots of polynomials depend differentiably on the coefficients wherever the root is simple. For $p(x, a) = x^3 - 3x + a$ at $a = -2$ the root $x = 2$ has $\partial p/\partial x = 3x^2 - 3 = 9 \ne 0$ , so $\frac{dx}{da} = -\frac{p_a}{p_x} = -\frac19$ : raise $a$ by $0.1$ and the root moves by about $-0.011$ (the true root of $x^3 - 3x - 1.9 = 0$ is $1.9888$ ). At the double root $x = 1$ , $a = 2$ , where $p_x = 0$ , the roots collide and then become complex: the theorem fails there, and the roots move like $\pm\sqrt{(2 - a)/3}$ , not linearly. The same argument shows that a simple eigenvalue of a matrix is a differentiable function of the entries (problem below), which is how the perturbation theory of 3.1 and 3.5 is justified. Finally, the open mapping property: a $C^1$ map $\mathbb R^n \to \mathbb R^n$ with $\det DF \ne 0$ everywhere sends open sets to open sets (problem below).

Your answerThe root of

x^3 - 3x + a = 0

near

x = 2

(at

a = -2

) moves with

a

. What is

dx/da

a = -2

Used in the real world· Robot arms and thermodynamics: solving F(x) = y in practice

Robot inverse kinematics. The two-link arm of 5.3 (links of length 1 m, joint angles $\theta_1, \theta_2$ ) puts the hand at $F(\theta) = (\cos\theta_1 + \cos(\theta_1 + \theta_2),\ \sin\theta_1 + \sin(\theta_1 + \theta_2))$ , with $\det DF = \sin\theta_2$ . The controller needs the opposite: given a target, find the angles. Set the target to $(1, 1)$ and start at $\theta = (0.3, 1.2)$ , hand at $(1.026, 1.293)$ . Newton’s iteration $\theta \leftarrow \theta + DF(\theta)^{-1}\big(\text{target} - F(\theta)\big)$ gives $\theta = (-0.0156, 1.6352)$ (hand $(0.951, 0.983)$ ), then $(-0.0012, 1.5720)$ (hand $(0.99994, 0.99881)$ ), then $(-1.4\times10^{-9}, 1.570797)$ : three steps from $0.3$ m away to a millionth of a metre. The answer is $\theta = (0, \frac\pi2)$ , the elbow bent a right angle. A second solution, $(\frac\pi2, -\frac\pi2)$ , has $\det DF = \sin(-\frac\pi2) = -1$ : the elbow bent the other way. The inverse function theorem says each is isolated and depends differentiably on the target, because $\det DF = \pm1 \ne 0$ ; near full stretch ( $\theta_2 \to 0$ ) the determinant tends to $0$ and the joint speeds needed blow up.

Thermodynamics. An ideal gas has three state variables with one relation, $PV = nRT$ . One mole ( $n = 1$ , $R = 8.314$ J/(mol K)) at $T = 300$ K in $V = 0.025$ m $^3$ has $P = 99{,}768$ Pa. Physicists write $\big(\frac{\partial P}{\partial V}\big)_T$ for the derivative with $T$ held fixed, which is an implicit derivative of $F(P, V, T) = PV - nRT$ . Here it is $-\frac{nRT}{V^2} = -3.99\times10^6$ Pa/m $^3$ . The three partial derivatives $\big(\frac{\partial P}{\partial V}\big)_T = -\frac{nRT}{V^2}$ , $\big(\frac{\partial V}{\partial T}\big)_P = \frac{nR}{P}$ , $\big(\frac{\partial T}{\partial P}\big)_V = \frac{V}{nR}$ multiply to $-\frac{nRT}{VP} = -1$ (because $PV = nRT$ ), not $+1$ as the naive cancellation of ‘fractions’ would suggest. The problem below proves this for any equation $F(x, y, z) = 0$ .

Traps

Watch out· The minus sign, and the cancellation that is not a cancellation

$\dfrac{dy}{dx} = -\dfrac{F_x}{F_y}$ has a minus sign. And $\dfrac{\partial x}{\partial y}\dfrac{\partial y}{\partial z}\dfrac{\partial z}{\partial x} = -1$ , not $+1$ : partial derivatives are not fractions, and each one holds a different variable fixed. Check the sign against the circle: at the top right $F_x, F_y \gt 0$ and the curve is falling.

Watch out· Divide by the right partial derivative

The hypothesis names the variable you solve for: $F_y \ne 0$ for $y = g(x)$ , $F_x \ne 0$ for $x = h(y)$ , $F_z \ne 0$ for $z = z(x, y)$ . The worksheet that treats $x$ as a function of $y$ needs $F_x \ne 0$ . At $(1, 0)$ on the circle the first choice fails and the second works. For a system, the condition is a determinant, not separate partial derivatives.

Watch out· Remember the chain rule when differentiating a second time

In $z_x = -\dfrac xz$ , the $z$ in the denominator depends on $x$ too. Forgetting to differentiate it (writing $z_{xx} = -\frac1z$ ) gives a wrong answer (Worksheet 1 #2). The cure is to substitute the first derivative back before simplifying.

Watch out· Nonzero Jacobian is local, and the theorem needs C¹

$\det DF \ne 0$ at every point does not make $F$ one-to-one (the exponential map of 5.4, $z \mapsto z^2$ away from the origin). And the derivative must be continuous: the function $\frac x2 + x^2\sin\frac1x$ has $f'(0) = \frac12$ yet is not one-to-one near $0$ . On the Fall 2025 final the same lesson came as a true or false question: smooth level sets need $\nabla f \ne O$ , and a smooth function can have level sets that end or have corners where the gradient vanishes.

Watch out· The theorem gives existence, not a formula

It says $g$ exists and tells you $g'$ ; it does not give $g$ . To compute values of $g$ you iterate (Newton’s method, or the contraction in the proof), as in the lab above. Series for $g$ come from differentiating repeatedly, as the second derivative formula did.

Problems

checkMATH 376 Worksheet 1 style (hypothesis check)

Does $x^2 + y^2 + z^2 = 1$ define $z$ as a differentiable function of $(x, y)$ near $(0, 0, 1)$ ? Near $(1, 0, 0)$ ? If not, which variable can be solved for near $(1, 0, 0)$ ?

Solution

$F_z = 2z$ . At $(0, 0, 1)$ , $F_z = 2 \ne 0$ : yes, $z = \sqrt{1 - x^2 - y^2}$ . At $(1, 0, 0)$ , $F_z = 0$ : the hypothesis fails; indeed for $(x, y)$ near $(1, 0)$ with $x^2 + y^2 \gt 1$ there is no $z$ at all. But $F_x = 2 \ne 0$ , so $x = \sqrt{1 - y^2 - z^2}$ is a differentiable function of $(y, z)$ near $(0, 0)$ .

check

Find $\dfrac{dy}{dx}$ at $(0, 1)$ for the curve $e^{xy} + y = 2$ .

Solution

The point is on the curve: $e^0 + 1 = 2$ . $F = e^{xy} + y - 2$ , $F_x = ye^{xy} = 1$ and $F_y = xe^{xy} + 1 = 1$ at $(0, 1)$ . So $\frac{dy}{dx} = -\frac{F_x}{F_y} = -1$ .

check

Is $F(x, y) = (x^2 + y,\ x + y^2)$ locally invertible near $(1, 1)$ ? Near $\big(\frac12, \frac12\big)$ ?

Solution

$DF = \begin{bmatrix}2x & 1\\ 1 & 2y\end{bmatrix}$ , $\det DF = 4xy - 1$ . At $(1, 1)$ it is $3 \ne 0$ : yes. At $\big(\frac12, \frac12\big)$ it is $0$ : the theorem gives no conclusion. In this case the news is bad: along the line $y = 1 - x$ the first component is $x^2 + 1 - x$ and the second is $x + (1 - x)^2 = x^2 - x + 1$ , the same number. So the line $x + y = 1$ is sent into the diagonal, and $x$ and $1 - x$ have the same image, so $F$ is not one-to-one on any ball around $\big(\frac12, \frac12\big)$ , as the vanishing determinant warned.

check

$F(x, y) = x - y^3$ has $F_y(0, 0) = 0$ . Does $F = 0$ define $y$ as a function of $x$ near $(0, 0)$ ? Is the function differentiable at $0$ ?

Solution

The solution is $y = x^{1/3}$ , a function, defined for all real $x$ . It is not differentiable at $0$ (its difference quotient $h^{-2/3}$ is unbounded). So the conclusion of the implicit function theorem fails without the hypothesis, although a function exists.

useApostol 9.8 #1

The equations $x + y = uv$ and $xy = u - v$ define $x$ and $y$ as functions of $u$ and $v$ . Show that $\dfrac{\partial x}{\partial u} = \dfrac{xv - 1}{x - y}$ when $x \ne y$ , and say what the hypothesis is.

Solution

Unknowns $(x, y)$ , with $F = x + y - uv$ and $G = xy - u + v$ . The determinant $\frac{\partial(F, G)}{\partial(x, y)} = \det\begin{bmatrix}1 & 1\\ y & x\end{bmatrix} = x - y$ , so the hypothesis is $x \ne y$ . Differentiate in $u$ with $v$ fixed: $x_u + y_u = v$ and $y\,x_u + x\,y_u = 1$ . From the first, $y_u = v - x_u$ ; substituting, $y\,x_u + xv - x\,x_u = 1$ , so $(y - x)x_u = 1 - xv$ and $x_u = \dfrac{xv - 1}{x - y}$ .

useApostol 9.7, Example 5

Eliminating $u$ from $x = u + v$ and $y = uv^2$ gives an equation defining $v = h(x, y)$ . Show that $\dfrac{\partial h}{\partial x} = \dfrac{h}{3h - 2x}$ and find $\dfrac{\partial h}{\partial y}$ .

Solution

$u = x - v$ , so $y = (x - v)v^2$ , that is $F(x, y, v) = xv^2 - v^3 - y = 0$ . $F_x = v^2$ , $F_y = -1$ , $F_v = 2xv - 3v^2$ . So $h_x = -\dfrac{v^2}{2xv - 3v^2} = \dfrac{v}{3v - 2x} = \dfrac{h}{3h - 2x}$ , and $h_y = -\dfrac{-1}{2xv - 3v^2} = \dfrac1{2xh - 3h^2} = -\dfrac{1}{h(3h - 2x)}$ .

usenew (thermodynamics)

For $F(P, V, T) = PV - nRT$ , compute $\big(\frac{\partial P}{\partial V}\big)_T$ , $\big(\frac{\partial V}{\partial T}\big)_P$ , $\big(\frac{\partial T}{\partial P}\big)_V$ with the implicit formulas, and verify that the product is $-1$ .

Solution

$F_P = V$ , $F_V = P$ , $F_T = -nR$ . $\big(\frac{\partial P}{\partial V}\big)_T = -\frac{F_V}{F_P} = -\frac PV$ ; $\big(\frac{\partial V}{\partial T}\big)_P = -\frac{F_T}{F_V} = \frac{nR}{P}$ ; $\big(\frac{\partial T}{\partial P}\big)_V = -\frac{F_P}{F_T} = \frac{V}{nR}$ . The product is $-\frac PV\cdot\frac{nR}P\cdot\frac V{nR} = -1$ .

useMATH 376 Worksheet 1 #3 style

Use implicit differentiation to find $\dfrac{d}{dx}\operatorname{arctan}x$ and state where the hypothesis fails (if anywhere).

Solution

$y = \arctan x$ means $\tan y = x$ , $F = \tan y - x$ , $F_y = \sec^2y \ge 1$ , never zero. So $y' = \frac{1}{\sec^2y} = \cos^2y = \frac1{1 + \tan^2y} = \frac1{1 + x^2}$ for all real $x$ . The hypothesis never fails, so $\arctan$ is differentiable everywhere, unlike $\arcsin$ at $\pm1$ .

prove

Let $F(x, y)$ be $C^1$ , $F(a, b) = 0$ and $F_y(a, b) \ne 0$ . Prove that the tangent line to the level curve $F = 0$ at $(a, b)$ is $F_x(a, b)(x - a) + F_y(a, b)(y - b) = 0$ .

Solution

Plan. The implicit function $g$ has slope $-F_x/F_y$ ; write the tangent line to its graph.

By the theorem, near $a$ the curve is the graph of $g$ with $g(a) = b$ and $g'(a) = -F_x(a, b)/F_y(a, b)$ . The tangent line to a graph at $(a, b)$ is $y - b = g'(a)(x - a)$ , that is $y - b = -\frac{F_x}{F_y}(x - a)$ . Multiply by $F_y \ne 0$ : $F_y(y - b) + F_x(x - a) = 0$ . $\blacksquare$ This is the tangent line $\nabla F(a, b)\cdot((x, y) - (a, b)) = 0$ of 5.3, now justified.

prove

Let $F(x, y, z) = 0$ define each variable as a differentiable function of the other two, with all partial derivatives of $F$ nonzero. Prove that $\dfrac{\partial x}{\partial y}\dfrac{\partial y}{\partial z}\dfrac{\partial z}{\partial x} = -1$ .

Solution

Plan. Write each factor with the formula $-F_{\text{other}}/F_{\text{self}}$ .

Solving for $x$ with $y$ and $z$ as inputs: $\frac{\partial x}{\partial y} = -\frac{F_y}{F_x}$ . Solving for $y$ with $z$ and $x$ as inputs: $\frac{\partial y}{\partial z} = -\frac{F_z}{F_y}$ . Solving for $z$ with $x$ and $y$ as inputs: $\frac{\partial z}{\partial x} = -\frac{F_x}{F_z}$ . Each is valid because the relevant partial derivative of $F$ is nonzero. The product is $(-1)^3\frac{F_yF_zF_x}{F_xF_yF_z} = -1$ . $\blacksquare$

provenew (a global contraction)

Let $h: \mathbb R^n \to \mathbb R^n$ satisfy $\lVert h(x) - h(x')\rVert \le q\lVert x - x'\rVert$ for all $x, x'$ , with $q \lt 1$ . Prove that $F(x) = x + h(x)$ is a bijection of $\mathbb R^n$ onto $\mathbb R^n$ . (Compare with the local theorem: here nothing is local.)

Solution

Plan. Solving $F(x) = y$ means $x = y - h(x)$ ; the right side is a contraction of the complete space $\mathbb R^n$ .

Fix $y$ and put $T_y(x) = y - h(x)$ . Then $\lVert T_y(x) - T_y(x')\rVert = \lVert h(x) - h(x')\rVert \le q\lVert x - x'\rVert$ , so $T_y$ is a contraction of $\mathbb R^n$ , which is complete (4.2). By the Banach fixed point theorem $T_y$ has exactly one fixed point $x$ . And $T_y(x) = x$ means $y - h(x) = x$ , that is $F(x) = y$ . So every $y$ has exactly one preimage: $F$ is a bijection. $\blacksquare$

prove

Let $F: \mathbb R^n \to \mathbb R^n$ be continuously differentiable with $\det DF(x) \ne 0$ for every $x$ . Prove that $F$ maps open sets to open sets.

Solution

Plan. Apply the inverse function theorem at each point of the open set.

Let $O$ be open and $y \in F(O)$ , say $y = F(a)$ with $a \in O$ . By the inverse function theorem there are open sets $U' \ni a$ and $V \ni y$ with $F: U' \to V$ a bijection. Shrink: $U'' = U' \cap O$ is open and contains $a$ . The image $F(U'')$ is the set of $y' \in V$ with $G(y') \in U''$ , that is the preimage of the open set $U''$ under the continuous map $G: V \to U'$ ; so it is open in $V$ , hence open in $\mathbb R^n$ . So $F(U'')$ is an open set containing $y$ and contained in $F(O)$ . Every point of $F(O)$ has such a ball, so $F(O)$ is open. $\blacksquare$

stretch

For $f(x) = \dfrac x2 + x^2\sin\dfrac1x$ ( $f(0) = 0$ ), show that $f'(0) = \frac12$ but $f$ is not one-to-one on any interval around $0$ .

Solution

Difference quotient: $\frac{f(h)}{h} = \frac12 + h\sin\frac1h \to \frac12$ , so $f'(0) = \frac12$ . For $x \ne 0$ , $f'(x) = \frac12 + 2x\sin\frac1x - \cos\frac1x$ , and at $x_k = \frac1{2\pi k}$ this is $\frac12 + 0 - 1 = -\frac12$ . Suppose $f$ were one-to-one on an interval $I \ni 0$ . It is continuous, and a continuous one-to-one function on an interval is strictly monotone (by the intermediate value theorem of 1.3). If it were increasing then $f' \ge 0$ on $I$ , contradicting $f'(x_k) = -\frac12$ for large $k$ ; if decreasing then $f' \le 0$ , contradicting $f'(0) = \frac12$ . So $f$ is not one-to-one near $0$ , although $f'(0) \ne 0$ : the continuity of $f'$ cannot be dropped from the inverse function theorem.

stretchnew (links to 3.1 and 3.5)

The matrix $A_\varepsilon = \begin{bmatrix}2 + \varepsilon & 1\\ 1 & 2\end{bmatrix}$ has the eigenvalue $\lambda = 3$ at $\varepsilon = 0$ . Use the implicit function theorem on the characteristic polynomial to find $\dfrac{d\lambda}{d\varepsilon}$ at $\varepsilon = 0$ , and check it against the eigenvector formula $\frac{d\lambda}{d\varepsilon} = u^{\mathsf T}\big(\frac{dA}{d\varepsilon}\big)u$ for a unit eigenvector $u$ .

Solution

$p(\lambda, \varepsilon) = \det(A_\varepsilon - \lambda I) = (2 + \varepsilon - \lambda)(2 - \lambda) - 1$ . At $(\lambda, \varepsilon) = (3, 0)$ : $p = (-1)(-1) - 1 = 0$ . $p_\lambda = -(2 - \lambda) - (2 + \varepsilon - \lambda) = 1 + 1 = 2 \ne 0$ (so $\lambda = 3$ is a simple root), and $p_\varepsilon = 2 - \lambda = -1$ . So $\frac{d\lambda}{d\varepsilon} = -\frac{p_\varepsilon}{p_\lambda} = \frac12$ . Eigenvector formula: $u = \frac{(1, 1)}{\sqrt2}$ and $\frac{dA}{d\varepsilon} = \begin{bmatrix}1 & 0\\ 0 & 0\end{bmatrix}$ , so $u^{\mathsf T}\frac{dA}{d\varepsilon}u = \frac12$ . Exact check: the larger eigenvalue is $\frac{4 + \varepsilon + \sqrt{4 + \varepsilon^2}}2$ , whose derivative at $0$ is $\frac12$ .

Summary

The inverse function theorem: if $F$ is continuously differentiable and $\det DF(a) \ne 0$ , then $F$ has a continuously differentiable local inverse $G$ near $a$ with $DG = (DF\circ G)^{-1}$ . It is local only, and it needs the derivative to be continuous.
Proof idea (B): rewrite $F(x) = y$ as the fixed point problem $x = x + A^{-1}(y - F(x))$ with $A = DF(a)$ . The mean value inequality shows this map shrinks distances by $\frac12$ on a small ball, and the Banach fixed point theorem (4.2) gives the solution.
The implicit function theorem: if $F(a, b) = O$ and the block $D_yF(a, b)$ is invertible, then $F(x, y) = O$ defines a unique $C^1$ function $y = g(x)$ near $(a, b)$ with $Dg = -D_yF^{-1}D_xF$ . It is the inverse function theorem applied to $(x, y) \mapsto (x, F(x, y))$ .
Implicit differentiation is legal exactly when the denominator is nonzero: $\frac{dy}{dx} = -\frac{F_x}{F_y}$ , $z_x = -\frac{F_x}{F_z}$ , and for systems the ratio of Jacobian determinants; second derivatives need the chain rule applied once more.
Failure points are where the relevant partial derivative or determinant is zero: the circle at $(\pm1, 0)$ (a vertical tangent, fixed by solving for the other variable), the folium’s origin and the cusp (where $\nabla F = O$ ). The condition is sufficient, not necessary ( $y = x^{1/3}$ ).
A regular point (Jacobian of full rank $m$ ) of $F: \mathbb R^N \to \mathbb R^m$ has a level set that is locally a graph, with tangent space $\ker DF(a)$ . This is why level surfaces are smooth and why the Lagrange multiplier theorem of 5.8 holds.
In practice, inverses are computed by iteration: the proof’s contraction converges linearly, and Newton’s method (which recomputes the Jacobian) converges quadratically (5.10).

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The picture: an equation is a curve, a function is a graph

The idea, in words

Definitions and theorems

Local inverse

Mean value inequality for fields

Inverse function theorem

What the hypotheses are for

Implicit function

Implicit function theorem

Implicit differentiation, with permission

Implicit differentiation formulas

Inverse maps in practice

Level sets are smooth where the gradient is not zero

Regular point

Level sets near a regular point are graphs

Where it is used

Traps

Problems

Summary

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