Learning goals and preparation

We assume algebra, functions and their domains, radians, and elementary exponential and logarithmic laws. This chapter turns an average rate into an instantaneous one using a limit. You will learn to distinguish a limit from a function value, test continuity, calculate and interpret derivatives, and differentiate an implicit engineering constraint.

Calculus is not simply a collection of rules for changing exponents. A derivative exists only when an appropriate limiting slope exists. Its units are output units divided by input units. A mathematical rate also inherits the assumptions and applicable range of the physical model from which it was derived.

1.1 What a limit asserts

The statement $\lim_{x\to a}f(x)=L$ means that $f(x)$ can be made arbitrarily close to $L$ by taking $x$ sufficiently close to, but different from, $a$ within the domain. Formally, for every $\varepsilon>0$ there is $\delta>0$ such that

$$0<|x-a|<\delta\implies |f(x)-L|<\varepsilon.$$

This definition assumes $a$ is an accumulation point of the domain and uses inputs from that domain. The function need not be defined at $a$. When the domain has points arbitrarily close to $a$ on both sides, a two-sided limit exists exactly when the left-hand and right-hand limits both exist and agree. If the domain approaches $a$ from only one side, use the corresponding one-sided or domain-relative limit rather than demanding a nonexistent opposite-side approach.

Worked example 1: a hole in a graph. For $x\ne2$,

$$\frac{x^2-4}{x-2}=x+2.$$

Thus its limit at $2$ is $4$, although the original quotient is undefined there. To verify the limit directly, note that its difference from $4$ has absolute value $|x-2|$. Choosing $\delta=\varepsilon$ proves the defining implication. The missing value and the limiting value are separate facts.

Limit laws allow sums, products, and quotients of limits, with a nonzero limiting denominator for the quotient law. These conditions explain why direct substitution works for polynomials and for rational functions away from denominator zeros. A symbol such as $0/0$ is not a value and is not an answer; it signals that the quotient law cannot be used directly.

Worked example 2: rationalization. Compute

$$\lim_{h\to0}\frac{\sqrt{9+h}-3}{h}.$$

For $h\ne0$ near zero, multiply numerator and denominator by the conjugate to obtain $1/(\sqrt{9+h}+3)$. Its limit is $1/6$. The cancellation was performed only at nonzero $h$, exactly the inputs used in the limit.

The function $|x|/x$ equals $1$ on the positive side and $-1$ on the negative side. Its two-sided limit at zero does not exist. A calculator table containing only positive inputs would miss the obstruction.

1.2 Continuity and attainable values

Definition

Continuity relative to a domain

For $c\in A$, $f:A\to\mathbb R$ is continuous at $c$ if for every $\varepsilon>0$ there is $\delta>0$ such that $x\in A$ and $|x-c|<\delta$ imply $|f(x)-f(c)|<\varepsilon$. At an isolated point of $A$ this is automatic. At an accumulation point it is equivalent to the limit equaling $f(c)$.

Why the real numbers need more than arithmetic

The rational numbers support the familiar ordered arithmetic, yet the set $\{q\in\mathbb Q:q>0,\ q^2<2\}$ has no rational least upper bound. The real number system adds a completeness property: bounded nonempty sets have least upper bounds. This is the source of many existence results later in analysis, not simply another rule for manipulating inequalities.

Definition

Upper bound and supremum

For a nonempty $A\subset\mathbb R$, $u$ is an upper bound if $a\le u$ for every $a\in A$. A number $s$ is $\sup A$ if it is an upper bound and $s\le u$ for every upper bound $u$. A maximum is an element of $A$ that is also an upper bound. The maximum, when it exists, is the supremum; a supremum need not belong to $A$. Lower bounds, infimum and minimum are defined by reversing inequalities.

Theorem

Completeness axiom

Every nonempty subset of $\mathbb R$ that is bounded above has a real supremum. Equivalently, every nonempty subset bounded below has a real infimum, via $\inf A=-\sup(-A)$.

For example, $\sup(0,1)=1$ although $1$ is not a maximum of that set. Conversely, $\sup[0,1]=1$ and it is a maximum. The empty set and an unbounded-above set do not have a real supremum under this definition.

Theorem

Supremum approximation criterion

An upper bound $s$ of a nonempty set $A$ is $\sup A$ if and only if, for every $\varepsilon>0$, there exists $a\in A$ such that $s-\varepsilon<a\le s$.

Proof

Proving both directions

If $s=\sup A$ and no element exceeds $s-\varepsilon$, then $s-\varepsilon$ is a smaller upper bound, contradiction. Conversely, suppose $s$ is an upper bound and the approximation property holds. If another upper bound $u$ were smaller than $s$, choose $\varepsilon=s-u>0$. The property gives $a>u$, contradicting that $u$ is an upper bound. Thus $s$ is the least upper bound.

Worked example

Use both halves of a supremum proof

Let $A=\{n/(n+1):n\in\mathbb N, n\ge1\}$. Every element is less than $1$, so $1$ is an upper bound. Given $\varepsilon>0$, choose $n>1/\varepsilon$ by the Archimedean property below. Then $1-n/(n+1)=1/(n+1)<\varepsilon$. Thus $\sup A=1$. No element equals $1$, so $A$ has no maximum. An upper-bound calculation alone would not prove leastness: $2$ is also an upper bound.

Archimedean property and density

Theorem

Archimedean property

For every real $x$ there exists a positive integer $n>x$. Consequently, for every $\varepsilon>0$ there is $n\in\mathbb N$ such that $1/n<\varepsilon$.

Proof

Completeness forces integers to be unbounded

If the positive integers were bounded above, completeness would give $s=\sup\mathbb N$. Since $s-1$ is not an upper bound, some $n\in\mathbb N$ satisfies $n>s-1$. But $n+1>s$ and $n+1\in\mathbb N$, contradicting that $s$ bounds the set. Apply this to $1/\varepsilon$ to obtain the second statement.

Between any real numbers $x<y$ there is a rational number. Indeed choose $n$ with $n(y-x)>1$; the Archimedean property and the integer-part property yield an integer $m$ with $nx<m<ny$, so $x<m/n<y$. The integer-part property follows from the unboundedness of integers in both directions and the well-ordering of the natural numbers. An irrational also lies between $x$ and $y$: choose a rational $q$ with $x-\sqrt2<q<y-\sqrt2$; then $q+\sqrt2$ is irrational and lies in the interval. Thus density does not say rationals and irrationals are equal sets.

Worked example

An infimum via reflection

For $B=\{2+1/n:n\ge1\}$, every element exceeds $2$. Given $\varepsilon>0$, choose $n>1/\varepsilon$ to obtain $2<2+1/n<2+\varepsilon$. Therefore $2$ is the infimum, but not a minimum. The maximum is $3$, attained at $n=1$.

Theorem

Intermediate value theorem

If $f$ is continuous on $[a,b]$ and $y$ lies between $f(a)$ and $f(b)$, there is $c\in[a,b]$ with $f(c)=y$.

Proof

Completeness locates a crossing

If $y=f(a)$ or $y=f(b)$, take the corresponding endpoint. Otherwise, assume $f(a)<y<f(b)$; the reverse order follows by replacing both $f$ by $-f$ and $y$ by $-y$. The set $S=\{x\in[a,b]:f(x)\le y\}$ is nonempty and bounded above. Let $c=\sup S$. If $f(c)<y$, continuity gives nearby points to the right with value below $y$; since $c<b$, this contradicts the upper-bound property. If $f(c)>y$, continuity gives a neighbourhood of $c$ with value above $y$; the supremum approximation property supplies an element of $S$ in that neighbourhood, a contradiction. Thus $f(c)=y$.

The theorem guarantees existence, not uniqueness or a closed-form root.

Worked example 3: existence before computation. Let $f(x)=x^3+x-1$. It is continuous on $[0,1]$, with $f(0)=-1$ and $f(1)=1$, so a root lies in $(0,1)$. If $0\le u<v$, then

$$f(v)-f(u)=(v-u)(v^2+uv+u^2+1)>0.$$

Thus the function is strictly increasing there, so the root is unique. Continuity and monotonicity play different roles.

1.3 The derivative as a limiting slope

The derivative at an interior point $a$ is

$$f'(a)=\lim_{h\to0}\frac{f(a+h)-f(a)}h,$$

when this finite limit exists. The quotient is the secant slope over an interval of width $h$; the limit is the tangent slope and the instantaneous rate.

Worked example 4: velocity from a position model. Suppose position is $s(t)=2t^2+3t$ metres for times $t\ge0$ in seconds. At an interior time $t>0$,

$$\frac{s(t+h)-s(t)}h=4t+2h+3,$$

so $s'(t)=4t+3$ metres per second. At $t=2$, the instantaneous velocity is $11$ m/s. The average velocity from $2$ to $2+h$ is $11+2h$, which approaches $11$. At the endpoint $t=0$ the model provides a right-hand derivative unless it is explicitly extended to negative time.

Differentiability implies continuity. To see this, write

$$f(a+h)-f(a)=h\frac{f(a+h)-f(a)}h.$$

If the quotient has a finite limit, the product tends to zero. The converse is false: $|x|$ is continuous at zero, but its one-sided difference quotients are $-1$ and $1$, so it has no derivative there.

1.4 Differentiation rules and why products require two terms

At points where the relevant derivatives exist,

$$\begin{aligned}(f+g)'&=f'+g',\\(cf)'&=cf',\\(fg)'&=f'g+fg',\\\left(\frac fg\right)'&=\frac{f'g-fg'}{g^2}\quad(g\ne0),\\(f\circ g)'(x)&=f'(g(x))g'(x).\end{aligned}$$

For the product rule, add and subtract $f(a+h)g(a)$ in the product difference. The quotient becomes

$$f(a+h)\frac{g(a+h)-g(a)}h+g(a)\frac{f(a+h)-f(a)}h.$$

Use continuity of $f$ at $a$ and take limits. This derivation explains why differentiating both factors and multiplying the answers is incorrect.

Useful elementary derivatives include

$$\frac{d}{dx}x^n=nx^{n-1},\quad (e^x)'=e^x,\quad(\ln x)'=\frac1x\ (x>0),\quad(\sin x)'=\cos x,\quad(\cos x)'=-\sin x.$$

The trigonometric formulas use radians. General real powers need an appropriate real domain; for instance the rule for $x^{1/2}$ applies for $x>0$, not as a finite derivative at zero.

The discussion concerns real one-variable functions.

1. Chain and quotient rules

Let $g$ be differentiable at an interior point $a$, and $f$ differentiable at $g(a)$ on a neighborhood of that point. Then $f\circ g$ is differentiable at $a$ and

$$(f\circ g)'(a)=f'(g(a))g'(a).$$

Proof. Differentiability of $f$ gives $f(g(a)+u)-f(g(a))=f'(g(a))u+u\epsilon(u)$, where $\epsilon(u)\to0$ as $u\to0$; define $\epsilon(0)=0$. Put $u=g(a+h)-g(a)$. Differentiability of $g$ implies $u\to0$ and $u/h\to g'(a)$. Thus the difference quotient is $(f'(g(a))+\epsilon(u))(u/h)$ and converges to the claimed product. This argument still works when $u=0$ for some nonzero $h$; dividing by $u$ would not.

If $f,g$ are differentiable at $a$ and $g(a)\ne0$, then

$$\left(\frac fg\right)'(a)=\frac{f'(a)g(a)-f(a)g'(a)}{g(a)^2}.$$

Indeed continuity of $g$ makes it nonzero near $a$. The reciprocal difference quotient equals $-[g(a+h)-g(a)]/[h g(a+h)g(a)]$, whose limit is $-g'(a)/g(a)^2$; apply the product rule to $f(1/g)$.

Example. For every real $x$, $F(x)=\log(1+x^2)$ is differentiable because $1+x^2>0$, and the chain rule gives $F'(x)=2x/(1+x^2)$.

Worked example 5: a composed response. Let

$$v(t)=e^{-2t}\sin(3t).$$

The product and chain rules give

$$v'(t)=e^{-2t}\bigl(3\cos(3t)-2\sin(3t)\bigr).$$

There are two independent contributions: the envelope changes and the oscillatory factor changes. The numerical coefficients are understood with compatible time units so that the exponential and trigonometric arguments are dimensionless.

1.5 Implicit constraints and related rates

Sometimes the inputs and outputs are connected by an equation rather than an explicit formula. If $y=y(x)$ is differentiable and $x^2+y^2=25$, then the chain rule gives $2x+2yy'=0$. At a point with $y\ne0$,

$$y'=-x/y.$$

Dividing by $y$ is legitimate only under that condition. At $(5,0)$ the formula fails, reflecting the vertical tangent of the circle rather than permission to divide by zero.

Worked example 6: a conical vessel. Water occupies a cone whose radius and depth obey $r=h/2$. Assume this geometric relationship holds during filling. Then

$$V=\frac13\pi r^2h=\frac\pi{12}h^3.$$

Differentiating with respect to time yields $dV/dt=(\pi/4)h^2\,dh/dt$. If the inflow is $2$ m$^3$/min and the depth is $2$ m, then $dh/dt=2/\pi$ m/min. The changing surface area explains why a constant volume inflow does not imply constant depth growth.

In related-rates problems, first state the geometric or conservation equation, then identify which quantities vary with time, differentiate, and only afterward substitute the specified instant. Substituting a temporary value such as $h=2$ before differentiation could incorrectly turn a changing variable into a constant.

1.6 Differentials and controlled local approximation

Differentiability says

$$f(a+h)=f(a)+f'(a)h+r(h),\qquad r(h)/h\to0.$$

The differential $df=f'(a)h$ is the linear part of the change. It approximates, but is not generally equal to, the finite change $\Delta f$.

For $f(x)=x^2$, the exact change is $2ah+h^2$. The linear prediction is $2ah$, and the omitted term is $h^2$. At $a=10$, $h=0.01$, this predicts $0.2$, while the exact change is $0.2001$. An explicit remainder is better evidence of accuracy than simply calling a change “small.”

1.7 Graded exercises

Foundation

1. Find $\lim_{x\to3}(x^2-9)/(x-3)$, and state whether the quotient is defined at $3$.
2. Find $\lim_{h\to0}(\sqrt{4+h}-2)/h$ by rationalization.
3. Choose $c$ to make $f(x)=x+1$ for $x<2$, $f(2)=c$, and $f(x)=2x-1$ for $x>2$ continuous at $2$.
4. From the definition, differentiate $f(x)=3x^2-x$.

Core

5. Differentiate $y=(t^2+1)e^{-t}$.
6. Differentiate $y=\ln(1+x^2)$ and state the real domain.
7. If $x^2+xy+y^2=7$ and $y$ is differentiable near $(1,2)$, find $dy/dx$ there.
8. A spherical balloon has radius $r$ centimetres and volume $4\pi r^3/3$. If volume increases at $12\pi$ cm$^3$/s, find $dr/dt$ when $r=3$ cm.

Synthesis

9. Prove that $x^3+2x-2=0$ has exactly one root in $[0,1]$ without using a numerical root finder. State separately the existence and uniqueness arguments.
10. Let $f(x)=1/x$ and $a>0$. Derive the linear approximation to $f(a+h)$ and show that its exact error is $h^2/[a^2(a+h)]$ whenever $a+h\ne0$. If $|h|\le a/2$, bound the error by $2h^2/a^3$.

1.8 Full solutions

1. For $x\ne3$, factor and cancel to obtain $x+3$, so the limit is $6$. The original quotient is not defined at $3$.
2. The quotient equals $1/(\sqrt{4+h}+2)$ for nonzero $h$ near zero. Its limit is $1/4$.
3. Both one-sided limits are $3$. Setting $c=3$ makes the value agree with the common limit and yields continuity.
4. $[3(x+h)^2-(x+h)-(3x^2-x)]/h=6x+3h-1$. Taking $h\to0$ gives $f'(x)=6x-1$.
5. $y'=2te^{-t}-(t^2+1)e^{-t}=e^{-t}(-t^2+2t-1)$.
6. The argument $1+x^2$ is positive for every real $x$, so the domain is $\mathbb R$. The chain rule gives $y'=2x/(1+x^2)$.
7. Differentiating gives $2x+y+xy'+2yy'=0$, so $y'=-(2x+y)/(x+2y)$ where the denominator is nonzero. At $(1,2)$, this is $-4/5$. The point lies on the constraint because $1+2+4=7$.
8. $dV/dt=4\pi r^2dr/dt$. At $r=3$, $12\pi=36\pi\,dr/dt$, giving $dr/dt=1/3$ cm/s.
9. The polynomial is continuous, equals $-2$ at $0$, and equals $1$ at $1$, so the intermediate value theorem gives a root in $(0,1)$. If $0\le u<v\le1$, the difference is $(v-u)(v^2+uv+u^2+2)>0$. Strict increase permits at most one root. Together the two arguments prove exactly one.
10. The derivative at $a$ is $-1/a^2$, so the linear approximation is $1/a-h/a^2$. Subtracting this from $1/(a+h)$ over a common denominator gives $h^2/[a^2(a+h)]$. Under $|h|\le a/2$, the denominator's factor $a+h\ge a/2$, so the nonnegative error is at most $2h^2/a^3$.

Summary and onward map

A limit concerns nearby inputs; continuity additionally concerns the value at the point. A derivative is a two-sided limiting slope at an interior point, and differentiability implies continuity. Product, quotient and chain rules preserve the structure of a composed model. Implicit differentiation and related rates depend on explicit assumptions about which quantities change. A differential is a first-order approximation whose error must be understood.

Sector area and radians

Use these elementary geometric facts: the unit circle has circumference $2\pi$ and enclosed area $\pi$, with the same $\pi$ in both formulas; planar area is invariant under rotations, finitely additive for regions with disjoint interiors, and monotone under inclusion.

Let $0<h<\pi/2$ be measured in radians and put $t=h/(2\pi)$. On the unit circle, $h$ is the arc length, so $t$ is the fraction of the full circumference subtended by the angle. Divide the circle into $n$ equal-angle sectors. Rotations carry each sector onto the next, so all have equal area. Their interiors are disjoint and together they fill the disk; finite additivity gives area $\pi/n$ for each. Thus a sector with rational fraction $t=m/n$ of a turn has area $m\pi/n=\pi t=h/2$.

For arbitrary $t$ in this range, choose rational numbers $0\le r_k\le t\le s_k\le1$ with $r_k,s_k\to t$. Taking all sectors from the same initial ray, the corresponding sectors are nested by inclusion, so area monotonicity and the rational case give

$$\pi r_k\le A(h)\le\pi s_k.$$

Taking $k\to\infty$ yields $A(h)=\pi t=h/2$. This establishes the sector-area formula used in the geometric comparison.

9.5 The fundamental trigonometric limit

All limits in this section use radians. For $0<h<\pi/2$, compare the areas of the inner triangle, unit-circle sector, and outer tangent triangle:

$$\tfrac12\sin h<\tfrac12h<\tfrac12\tan h.$$

The sector area is $h/2$ specifically because $h$ is in radians. Divide through and use positive cosine to obtain

$$\cos h<\frac{\sin h}{h}<1.$$

For $0<h<\pi/2$, the area comparison above also gives $0<\sin h<h$ and $\cos h>0$. Using $\sin^2 h+\cos^2 h=1$,

$$0\le 1-\cos h=\frac{\sin^2 h}{1+\cos h}\le h^2.$$

Thus $\cos h\to1$ as $h\to0^+$. Since cosine is even, the same limit holds from the left.

As $h\to0^+$, cosine tends to one, so the squeeze principle gives $\sin h/h\to1$. The quotient is an even function of $h$, giving the same left limit. The squeeze principle itself follows from the finite-limit definition: if lower and upper functions are both within any prescribed tolerance of $L$, every number between them is too.

Two consequences are

$$\lim_{h\to0}\frac{\tan h}{h}=1,\qquad \lim_{h\to0}\frac{1-\cos h}{h^2}=\frac12.$$

For the first, divide $\sin h/h$ by $\cos h$. For the second, use $1-\cos h=2\sin^2(h/2)$ to get one half times $[\sin(h/2)/(h/2)]^2$. In particular $(\cos h-1)/h\to0$.

10.1 First principles: a method, not a memorized table

For $f$ defined on a neighborhood of $a$,

$$f'(a)=\lim_{h\to0}\frac{f(a+h)-f(a)}h,$$

provided the limit exists as a real number. The numerator must compare the same function at two inputs. Substitution of $h=0$ in the original quotient is never the method.

For a positive integer $n$, the finite binomial theorem gives

$$(a+h)^n=a^n+na^{n-1}h+\sum_{k=2}^{n}\binom nk a^{n-k}h^k.$$

Subtract $a^n$ and divide by nonzero $h$. The remaining sum tends to zero, so $(x^n)'=nx^{n-1}$. This proof applies at $a=0$ as well; for $n=1$ the sum is empty. The constant case is checked directly, and negative integer powers follow by reciprocals on $x\ne0$.

Example 1 — a reciprocal from its quotient. For $a\ne0$,

$$\frac{1/(a+h)-1/a}{h}=-\frac1{a(a+h)}\longrightarrow-\frac1{a^2}.$$

For sufficiently small $h$, $a+h\ne0$, so all stages are defined. For $f(x)=|x|$ at zero, the quotient is $|h|/h$, with one-sided limits $-1,1$; continuity does not rescue differentiability.

10.2 Derivatives of sine and cosine

First make the compound-angle formulas explicit. Rotating the plane through an angle $\beta$ sends the horizontal unit vector to $(\cos\beta,\sin\beta)$ and the vertical unit vector to $(-\sin\beta,\cos\beta)$. A rotation preserves vector addition and real scaling, so the unit vector $(\cos\alpha,\sin\alpha)$ is sent to

$$(\cos\alpha\cos\beta-\sin\alpha\sin\beta, \cos\alpha\sin\beta+\sin\alpha\cos\beta).$$

Its new angle is $\alpha+\beta$; equating coordinates proves both addition formulas. Replacing $\beta$ by $-\beta$ or taking $\beta=\alpha$ supplies difference and double-angle formulas with the same sign conventions. This is a coordinate derivation, not an inference from numerical angle examples.

Now use the sine addition formula to write

$$\frac{\sin(a+h)-\sin a}{h} =\sin a\frac{\cos h-1}{h}+\cos a\frac{\sin h}{h}.$$

Chapter 9's two special limits make this tend to $\cos a$. Similarly,

$$\frac{\cos(a+h)-\cos a}{h} =\cos a\frac{\cos h-1}{h}-\sin a\frac{\sin h}{h}\longrightarrow-\sin a.$$

Thus $(\sin x)'=\cos x$ and $(\cos x)'=-\sin x$ for every real input. The quotient rule then gives

$$(\tan x)'=\sec^2x\quad(\cos x\ne0),\qquad (\cot x)'=-\csc^2x\quad(\sin x\ne0).$$

For example the tangent numerator is $\cos^2x+\sin^2x=1$, not $\cos^2x-\sin^2x$. Reciprocals also give $(\sec x)'=\sec x\tan x$ and $(\csc x)'=-\csc x\cot x$ on their respective domains.

If $d$ is a numerical angle in degrees, its sine is $s(d)=\sin(\pi d/180)$. Therefore $s'(d)=(\pi/180)\cos(\pi d/180)$, not just cosine. The radian convention is what removes the conversion factor.

Learning goals and definitions

Develop the proof behind integrability statements and distinguish boundedness from integrability. Let $a<b$ and $f$ be bounded on $[a,b]$. For a partition $P:a=x_0<\cdots<x_n=b$, put $m_i=\inf_{[x_{i-1},x_i]}f$, $M_i=\sup_{[x_{i-1},x_i]}f$, and define $L(f,P)=\sum m_i\Delta x_i$, $U(f,P)=\sum M_i\Delta x_i$. The lower integral $\underline I$ is the supremum of all lower sums; the upper integral $\overline I$ is the infimum of all upper sums. Refining a partition makes lower sums nondecreasing and upper sums nonincreasing.

For any two partitions $P,Q$, their common refinement gives $L(f,P)\le U(f,Q)$, so $\underline I\le\overline I$. Boundedness ensures these quantities are finite. The function is Riemann integrable when they agree.

Theorem: Darboux criterion

The bounded function is integrable if and only if for every $\varepsilon>0$ there is a partition with $U(f,P)-L(f,P)<\varepsilon$.

Proof. If the two integrals equal $I$, choose a lower sum above $I-\varepsilon/2$ and an upper sum below $I+\varepsilon/2$, using supremum and infimum approximation. Their common refinement has gap less than $\varepsilon$. Conversely the inequalities $0\le\overline I-\underline I\le U(f,P)-L(f,P)$ force the integral gap to be zero when such partitions exist for every positive $\varepsilon$.

Theorem: bounded monotone functions are integrable

For nondecreasing $f$, take a uniform partition with $n$ intervals of width $(b-a)/n$. On interval $i$, the infimum and supremum are the endpoint values. Hence

$$U-L=\frac{b-a}{n}\sum_{i=1}^n[f(x_i)-f(x_{i-1})] =\frac{(b-a)(f(b)-f(a))}{n}\longrightarrow0.$$

The criterion proves integrability even if $f$ has jump discontinuities. For a nonincreasing function apply the result to $-f$.

Bounded proper Riemann integrals: mesh equivalence and elementary properties

Let $a<b$ be finite, and let $f:[a,b]\to\mathbb R$ be bounded. For a partition $P$, write $L(f,P),U(f,P)$ for the lower and upper Darboux sums. We use real completeness and the proved Darboux criterion: equality of lower and upper integrals is equivalent to arbitrarily small $U-L$. This packet does not assume continuity of $f$.

Darboux implies every fine tagged sum converges. Let the Darboux integral be $I$. Given $\varepsilon>0$, choose a fixed partition $P_0$ with gap below $\varepsilon/2$. Choose $M>0$ with $|f|\le M$, and let $k$ be the number of interior partition points of $P_0$. For any other partition $P$, mark as bad every cell whose closure contains one of those $k$ points. Their total length is at most $2k\|P\|$, because each point belongs to at most two cells. Every remaining cell lies within a single $P_0$ cell. The contribution of all good cells to the oscillation sum $U(f,P)-L(f,P)$ is at most the gap for $P_0$; the bad cells contribute at most $2M(2k\|P\|)$. Thus

$$U(f,P)-L(f,P)<\varepsilon/2+4Mk\|P\|.$$

Take $\|P\|<\varepsilon/[8M\max(1,k)]$. Every tagged sum $S(P)$ and $I$ lie between $L(f,P)$ and $U(f,P)$, so $|S(P)-I|<\varepsilon$, independently of tags.

The converse. Suppose all sufficiently fine tagged sums are within $\eta$ of a common finite $I$. Fix one such fine partition. Each cell supremum and infimum can be approximated by actual function values with error below $\eta/(b-a)$. The resulting upper-approximating tagged sum exceeds $U(f,P)-\eta$, and the lower-approximating sum is below $L(f,P)+\eta$. Therefore $U(f,P)<I+2\eta$ and $L(f,P)>I-2\eta$. Since $\eta$ can be arbitrarily small, the Darboux criterion gives integrability. Lower and upper bounds on tagged sums then identify its integral as $I$. Suprema need not be attained for this argument.

Linearity, constants and order. For integrable $f,g$ and real $\alpha,\beta$, on any common tagged partition the exact finite identity is $S(\alpha f+\beta g)=\alpha S(f)+\beta S(g)$. Its limit exists, so the equivalence just proved gives integrability and the corresponding linear integral identity. The constant function $c$ has every sum equal to $c(b-a)$. If $f\le g$, their tagged sums with the same tags obey the same order; their limits do too. Consequently $|\int f|\le M(b-a)$ whenever $|f|\le M$.

Absolute values. The reverse triangle inequality gives $\operatorname{osc}_J|f|\le\operatorname{osc}_J f$ on every cell $J$: apply $||f(x)|-|f(y)||\le|f(x)-f(y)|$ and approximate the suprema/infima. Thus every Darboux gap for $|f|$ is at most the gap for $f$, and $|f|$ is integrable. Since $-|f|\le f\le|f|$, order and linearity give $|\int_a^b f|\le\int_a^b|f|$.

Restriction and interval addition. For $a<c<b$, refine any small-gap partition by $c$. The nonnegative gaps on $[a,c]$ and $[c,b]$ sum to the refined whole gap; each can therefore be arbitrarily small. Both restrictions are integrable. Join partitions of the two pieces with meshes tending to zero. The exact sum on the joined partition is the sum of the two tagged sums, so passage to their three limits gives $\int_a^b f=\int_a^c f+\int_c^b f$. Conversely, if both restrictions are bounded integrable, join two partitions with gaps below $\varepsilon/2$; the resulting whole gap is below $\varepsilon$, proving whole integrability and the same identity. At a shared endpoint both restrictions use the same value of $f$; no endpoint term has positive interval length. Set equal-bound integrals to zero and reversed-bound integrals to the negative. Splitting a finite interval and cancellation now extend addition to every ordering of its three endpoints.

These proofs justify the estimates used in accumulation, continuous FTC, integration by parts and proper substitution. They assert bounded proper Riemann statements only, and do not replace the separate convergence definitions at singularities or infinite endpoints.

From a proper area logarithm to the natural exponential

Use proper integrals of continuous functions on finite closed intervals, their proved additivity/order/linearity, the completeness and Archimedean properties, and the intermediate value theorem. Define

$$L(x)=\int_1^x\frac{dt}{t}\qquad(x>0),$$

with oriented bounds. Every integral here is proper: its interval stays a positive distance from zero.

Continuity and strict increase. For $0<u<v$, additivity and order give $L(v)-L(u)=\int_u^vdt/t\ge(v-u)/v>0$. Near any fixed $a>0$, restrict $x>a/2$. Then $|L(x)-L(a)|\le(2/a)|x-a|$, proving continuity without using a logarithmic derivative.

The product law from actual sums. Fix $u>0$. Scaling a tagged partition of the interval between $1$ and $v>0$ by $u$ scales its mesh by $u$ and its widths by $u$. The corresponding sums satisfy

$$\sum_i\frac{u\Delta s_i}{u\xi_i}=\sum_i\frac{\Delta s_i}{\xi_i}.$$

Both are sums of continuous proper integrands, so their all-mesh limits are equal. For $v<1$ reverse orientation; $v=1$ is zero directly. Hence $\int_u^{uv}dt/t=L(v)$ and

$$L(uv)=L(u)+L(v).$$

In particular $L(1)=0$, $L(1/u)=-L(u)$ and $L(2^n)=nL(2)$. As $L(2)>0$, the latter values are unbounded above by the Archimedean property; the reciprocal values are unbounded below. For any real $y$, choose positive arguments with $L$ below and above $y$. Continuity and the intermediate value theorem give an $x>0$ with $L(x)=y$. Strict increase makes that $x$ unique.

Construct and control the inverse. Define $E:\mathbb R\to(0,\infty)$ as this inverse. It is strictly increasing and continuous. To prove continuity at $y_0=L(a)$, choose $0<d<\min(a,\varepsilon)$. The numbers $L(a-d)<y_0<L(a+d)$ have positive gaps. If $|y-y_0|$ is smaller than both gaps, monotonicity forces $a-d<E(y)<a+d$, so $|E(y)-a|<\varepsilon$. The product law and uniqueness yield $E(s+t)=E(s)E(t)$, $E(0)=1$ and $E(-t)=1/E(t)$. Define $e=E(1)>1$ and $a^t=E(tL(a))$ for $a>0$. It follows that $a^{s+t}=a^sa^t$ and $(a^s)^t=a^{st}$, since $L(E(z))=z$. For integer exponents these equal repeated products; for rational $m/n$, $n>0$, the definition is the unique positive $n$th root of $a^m$. Thus it agrees with the elementary positive rational powers and gives their continuous real extension. In particular $E(t)=e^t$, and the area logarithm is its inverse $\ln=L$.

Derivatives without a cycle. For $a>0$, the average of $1/t$ on the interval between $a$ and $a+h$, with the sign of $h$ and orientation both retained, lies between its smallest and largest values on that interval. Both bounds tend to $1/a$. Thus

$$\frac{L(a+h)-L(a)}h\longrightarrow\frac1a.$$

For $y_0=L(a)$ and $h\ne0$, put $v=E(y_0+h)\ne a$. Inverse continuity gives $v\to a$, and

$$\frac{E(y_0+h)-E(y_0)}h =\left(\frac{L(v)-L(a)}{v-a}\right)^{-1}\longrightarrow a=E(y_0).$$

The denominator limit is nonzero, so the reciprocal is justified. This proves $L'=1/x$ and $E'=E$ from proper area and inverse continuity, with neither l'Hopital nor a previously assumed exponential derivative. The chain rule then gives $(x^r)'=rx^{r-1}$ on $x>0$ by $x^r=E(rL(x))$, and $(b^x)'=(\ln b)b^x$ for $b>0$. The case $b=1$ is the constant one. This supplies the normalization and identity bridge between the area logarithm, the inverse exponential and the real-power notation.

Continuous monotone inverses and principal branches

Let $I$ be a real interval and let $f:I\to\mathbb R$ be continuous and strictly monotone. Its image $J=f(I)$ is an interval: every value between two image values is attained between their preimages by the intermediate value theorem. Strict monotonicity makes $f:I\to J$ bijective. Its inverse is continuous relative to $J$. Indeed fix $a\in I$ and $\varepsilon>0$. If $I$ has a point below $a$, choose $l\in I$ with $a-\varepsilon<l<a$; this is possible by the interval property. If there is no such point, every inverse value is already at least $a$. Similarly, when there are points above $a$, choose $a<r<a+\varepsilon$ in $I$. The positive numbers $|f(l)-f(a)|$ and $|f(r)-f(a)|$ for the sides that exist have a positive minimum $\delta$; if neither side exists, $I=\{a\}$ and continuity is immediate. For $y\in J$ with $|y-f(a)|<\delta$, monotonicity places $f^{-1}(y)$ between the chosen brackets, on each side that exists. Thus $|f^{-1}(y)-a|<\varepsilon$. This argument works for increasing and decreasing functions and for included endpoints.

If $a$ is interior to $I$, $f$ is differentiable at $a$ and $f'(a)\ne0$, continuity and strict monotonicity give an open output interval around $b=f(a)$. For $y\ne b$ in that interval put $x=f^{-1}(y)$. Injectivity gives $x\ne a$, inverse continuity gives $x\to a$, and

$$\frac{f^{-1}(y)-f^{-1}(b)}{y-b} =\left(\frac{f(x)-f(a)}{x-a}\right)^{-1}\longrightarrow\frac1{f'(a)}.$$

The nonzero limit justifies the reciprocal and proves existence of the inverse derivative. No derivative continuity is required once strict monotonicity on the interval has been established.

For $u<v$ in $[-\pi/2,\pi/2]$, the addition formulas give $\sin v-\sin u=2\cos((u+v)/2)\sin((v-u)/2)>0$: the midpoint lies strictly between $-\pi/2$ and $\pi/2$, and $0<(v-u)/2\le\pi/2$. Thus sine is strictly increasing on this closed interval; its endpoint values are $-1,1$. For $0\le u<v\le\pi$, $\cos v-\cos u=-2\sin((u+v)/2)\sin((v-u)/2)<0$, since both sine factors are positive. Cosine is therefore strictly decreasing on $[0,\pi]$, and its endpoint values are $1,-1$. Thus their continuous principal inverses have domains $[-1,1]$ and ranges $[-\pi/2,\pi/2]$ and $[0,\pi]$, respectively. Tangent is continuous on $(-\pi/2,\pi/2)$ and is strictly increasing there: for $u<v$ in that interval, $\tan v-\tan u=\sin(v-u)/(\cos u\cos v)>0$, because $0<v-u<\pi$ and both cosine factors are positive. As its argument tends to the endpoints from within, its sine tends to $-1$ or $1$ and its cosine tends to zero through positive values. Hence tangent tends to $-\infty$ and $+\infty$. IVT gives its range $\mathbb R$ and a continuous inverse into that open interval. The inverse difference quotient therefore applies to all three branches at every interior input. The positive square-root signs in the following formulas follow from cosine being positive on the arcsine interior range and sine being positive on the arccosine interior range.

10.5 Inverse trigonometric derivatives

Suppose a continuous strictly monotone $f$ has a continuous local inverse and $f'(a)\ne0$. Write $y=f(x)$ and $b=f(a)$. For $y\ne b$, injectivity gives $x\ne a$, and

$$\frac{f^{-1}(y)-f^{-1}(b)}{y-b} =\frac{x-a}{f(x)-f(a)}\longrightarrow\frac1{f'(a)}.$$

The inverse's continuity ensures $x\to a$ when $y\to b$. This proves the inverse derivative, rather than assuming its differentiability and using it in a circular chain-rule argument.

For $y=\arcsin x$, the range restriction $-\pi/2\le y\le\pi/2$ makes cosine positive in the interior. Thus, for $|x|<1$,

$$(\arcsin x)'=\frac1{\cos y}=\frac1{\sqrt{1-x^2}}.$$

For arccosine, $0<y<\pi$ makes sine positive, giving $(\arccos x)'=-1/\sqrt{1-x^2}$ on $(-1,1)$. For arctangent, $y\in(-\pi/2,\pi/2)$ and $\sec^2y=1+x^2$, so $(\arctan x)'=1/(1+x^2)$ for every real $x$.

The arcsine and arccosine formulas have no finite extension to inputs $\pm1$; those are endpoints of their domains and the interior slopes become unbounded. For $H(x)=\arcsin(2x)$, the function domain is $[-1/2,1/2]$, whereas its finite derivative formula $2/\sqrt{1-4x^2}$ applies only to $(-1/2,1/2)$.