Learning goals and preparation

Prior calculus is not assumed. We assume fractions, equations, powers, and elementary coordinate graphs. We shall define a limit and a derivative, then use them to describe a changing quantity. At the end you should be able to calculate an instantaneous rate, state its units, and tell when a simple optimization argument needs endpoint or model checks.

The examples use hypothetical scientific quantities.

1.1 A function includes its allowed inputs

A function $f:D\to\mathbb R$ assigns exactly one real output to every input in its domain $D$. A graph consists of points $(x,f(x))$. The formula and the domain together determine the function.

For a rational expression, denominators must be nonzero. For a real square root, the radicand must be nonnegative. For $\ln x$, the input must be positive. Model conditions can restrict inputs further: time may be nonnegative, or a population may need to remain nonnegative.

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.

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 $(\cos h-1)/h\to0$.

Learning goals and allowed earlier results

Construct rather than assume an elementary function, prove its algebraic laws, and justify the logarithm as an inverse. We use the previously proved ratio test, uniform convergence of power and derivative series on compact intervals, the mean value theorem, and the intermediate value theorem.

Define

$$E(x)=\sum_{n=0}^{\infty}\frac{x^n}{n!},\qquad x\in\mathbb R.$$

For fixed $x\ne0$, the ratio of successive absolute terms is $|x|/(n+1)\to0$, so the series converges absolutely; at $x=0$ it equals one directly. On any fixed interval $|x|\le M$, both its series and derivative series have summable factorial majorants. The uniform derivative theorem therefore applies to polynomial partial sums and gives $E'(x)=E(x)$ everywhere, with $E(0)=1$.

Theorem: positivity and exponential law

For every real $x,y$, $E(x)>0$, $E(-x)=1/E(x)$, and $E(x+y)=E(x)E(y)$.

Proof. Differentiate $H(x)=E(x)E(-x)$ to obtain $H'(x)=E(x)E(-x)-E(x)E(-x)=0$. The mean value theorem makes $H$ constant; at zero it equals one. Thus $E$ never vanishes. Since it is continuous and $E(0)>0$, an intermediate-value argument prevents it from becoming negative anywhere, proving positivity and the reciprocal identity. For fixed $y$, differentiate $K(x)=E(x+y)E(-x)$; the same cancellation gives zero derivative. At zero $K(0)=E(y)$, so $E(x+y)E(-x)=E(y)$. Multiplication by $E(x)$ yields the addition law. No rearrangement of a double infinite series was needed.

Since $E'=E>0$, it is strictly increasing. For $x\ge0$, all terms are nonnegative and $E(x)\ge1+x$, so $E(x)\to\infty$ at positive infinity. The reciprocal identity gives $E(x)\to0$ at negative infinity. Continuity and monotonicity therefore make $E:\mathbb R\to(0,\infty)$ a bijection.

Define $\exp=E$, $e=E(1)$, and $\ln:(0,\infty)\to\mathbb R$ as its inverse. The inverse is continuous: if $x_0=E(t_0)$, then for any $\varepsilon>0$, $E(t_0-\varepsilon)<x_0<E(t_0+\varepsilon)$; taking $x$ within the smaller output gap confines $\ln x$ between those inputs.

Theorem: logarithmic derivative and laws

For $x>0$, $\ln'(x)=1/x$. For positive $x,y$ and real $r$, $\ln(xy)=\ln x+\ln y$ and $\ln(x^r)=r\ln x$, where $x^r$ is defined as $E(r\ln x)$.

Proof. Write $x_0=E(t_0)$, $x=E(t)$ and use inverse continuity to ensure $t\to t_0$ as $x\to x_0$. Then

$$\frac{\ln x-\ln x_0}{x-x_0}=\frac{t-t_0}{E(t)-E(t_0)}\longrightarrow\frac1{E'(t_0)}=\frac1{x_0}.$$

The exponential law gives $E(\ln x+\ln y)=xy$, and uniqueness of inverse values proves the product law. The real-power definition and inverse identity give the other formula. Differentiating that definition yields $(x^r)'=rx^{r-1}$ for $x>0$. This derivation did not first assume the real-power derivative to define the logarithm.

Worked example 1: a saturating response. Let $R(c)=6c/(2+c)$ model a response to concentration $c\ge0$. The formula's natural domain excludes $-2$, but the model domain is $[0,\infty)$. Also $R(c)=6-12/(2+c)$, so the response starts at zero and stays below $6$. Algebraically meaningful negative concentrations do not belong to this model.

A composition $(g\circ f)(x)=g(f(x))$ is allowed only when $x$ belongs to the domain of $f$ and $f(x)$ belongs to the domain of $g$. Domain restrictions persist even if a formula simplifies.

Elementary trigonometric and inverse functions

Angles in radians are arc length divided by radius, with one full turn $2\pi$. The unit-circle point is $(\cos x,\sin x)$, so $\sin^2x+\cos^2x=1$. Sine and cosine are defined on all real inputs and have period $2\pi$; tangent is $\sin x/\cos x$ and excludes $x=\pi/2+k\pi,\ k\in\mathbb Z$. For a periodic model $m+A\cos(\omega t+\phi)$, the amplitude is $|A|$ and period $2\pi/|\omega|$ for nonzero $A,\omega$. Its argument is dimensionless, so time and frequency must use compatible units.

A function is injective when equal outputs force equal inputs. Its inverse goes from its range back to its domain. Sine is inverted only after restriction to $[-\pi/2,\pi/2]$, giving $\arcsin:[-1,1]\to[-\pi/2,\pi/2]$; cosine's principal inverse has range $[0,\pi]$, and tangent's has range $(-\pi/2,\pi/2)$. For example $\arcsin(\sin(3\pi/4))=\pi/4$, not $3\pi/4$. An inverse value is not the full periodic solution set: $\sin x=1/2$ has both $\pi/6$ and $5\pi/6$ in $[0,2\pi)$.

1.2 Limits separate nearby behaviour from a point value

Let $a$ be an accumulation point of $D$. The assertion $\lim_{x\to a}f(x)=L$ means that outputs approach $L$ when $x\in D$ tends to $a$ with $x\ne a$. Formally, for every $\varepsilon>0$ there is $\delta>0$ such that

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

When $D$ has points approaching $a$ from both sides, a two-sided limit exists exactly when both one-sided limits exist and agree. A right-hand limit approaches through domain points with $x>a$, as at the start of a nonnegative time domain.

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 $n=1$.

Quantifiers in convergence

Definition

Convergence of a real sequence

A sequence $(x_n)$ converges to $L\in\mathbb R$ if for every $\varepsilon>0$ there exists $N\in\mathbb N$ such that for every $n\ge N$, $|x_n-L|<\varepsilon$. Write $x_n\to L$. The index $N$ may depend on $\varepsilon$ but must work for all later indices.

To negate convergence to $L$, there must be some $\varepsilon_0>0$ such that for every $N$ there is an $n\ge N$ with $|x_n-L|\ge\varepsilon_0$. A few early terms do not affect convergence; a recurring failure arbitrarily late does.

Worked example

Construct an epsilon–N proof

For $x_n=(3n+1)/(n+2)$, guess $L=3$. Since

$$\left|\frac{3n+1}{n+2}-3\right|=\frac5{n+2}<\frac5n,$$

choose $N>5/\varepsilon$. Then every $n\ge N$ gives $|x_n-3|<\varepsilon$. The guess is motivation; the inequality and quantified choice establish the proof.

Theorem

Uniqueness, boundedness and limit laws

A convergent real sequence has a unique limit and is bounded. If $x_n\to x$ and $y_n\to y$, then $x_n+y_n\to x+y$, $x_ny_n\to xy$, and $x_n/y_n\to x/y$ when $y\ne0$ and the quotient is defined eventually. If $x_n\le z_n\le y_n$ eventually and both outer sequences converge to $L$, then $z_n\to L$.

Proof

Uniqueness and boundedness

If $x_n\to L$ and $x_n\to M$, then for any $\varepsilon>0$, sufficiently late terms satisfy both $|x_n-L|<\varepsilon/2$ and $|x_n-M|<\varepsilon/2$. The triangle inequality yields $|L-M|<\varepsilon$. Since this holds for every positive $\varepsilon$, $L=M$. For boundedness choose $N$ so that $|x_n-L|<1$ for $n\ge N$. The finite initial terms and the bound $|x_n|\le|L|+1$ for the tail together give a finite bound for the whole sequence.

The converse of boundedness is false: $(-1)^n$ is bounded but does not converge. If it converged to $L$, its even subsequence would converge to $1$ and its odd subsequence to $-1$, violating uniqueness.

Proof

Product law mechanism

A convergent sequence $(y_n)$ is bounded, say $|y_n|\le M$. Write

$$|x_ny_n-xy|\le|y_n||x_n-x|+|x||y_n-y|.$$

Both terms tend to zero. For the quotient, $y_n\to y\ne0$ implies $|y_n|\ge|y|/2$ eventually; this keeps the denominator away from zero and gives $1/y_n\to1/y$.

Subsequences

Definition

Subsequence

A subsequence of $(x_n)$ is $(x_{n_k})$ where $n_1<n_2<\cdots$ are positive integers. If $x_n\to L$, then every subsequence converges to $L$ because $n_k\ge k$ and therefore eventually lies beyond any prescribed index.

Theorem

Bolzano–Weierstrass

Every bounded real sequence has a convergent subsequence.

Proof

Nested-interval proof

Place every term in a closed bounded interval $I_1$. Bisect it. At least one half contains infinitely many terms; choose that half as $I_2$. Continue, obtaining nested closed intervals $I_k$ of lengths tending to zero, each containing infinitely many terms. Choose indices $n_k$ strictly increasing with $x_{n_k}\in I_k$; this is possible because each interval contains infinitely many terms. Completeness gives a common point $L$ in all intervals: the supremum of their left endpoints lies in each interval because every left endpoint is at most every right endpoint. Since $x_{n_k},L\in I_k$, their distance is at most the interval length and tends to zero.

The conclusion is about a subsequence, not necessarily the whole sequence. For example, $(-1)^n$ has constant subsequences converging to $1$ and $-1$, but does not converge.

Real closed sets: the sequence criterion and the open complement

Work on the real line in the usual classical set-theoretic setting, with countable witness selections available. A set $U$ is open when every $y$ in $U$ has some $r>0$ with $(y-r,y+r)$ contained in $U$. We use the Archimedean consequence that $1/n$ tends to zero. For every $K$ subset of the real line, the following conditions are equivalent:

1. Every sequence of points of $K$ that converges in the real line has its limit in $K$.
2. The complement of $K$ is open.

Proof of 2 implies 1. Suppose $x_n$ belongs to $K$ and $x_n$ tends to $x$. If $x$ were outside $K$, openness would give $r>0$ with $(x-r,x+r)$ disjoint from $K$. Convergence gives a threshold after which $|x_n-x|<r$, contradicting $x_n$ in $K$. Therefore $x$ belongs to $K$.

Proof of 1 implies 2. Fix $y$ outside $K$. If no positive-radius interval around $y$ avoided $K$, then for every positive integer $n$ there would be a point $x_n$ in $K$ with $|x_n-y|<1/n$. Choose these witnesses as a sequence. Given $\varepsilon>0$, choose $N$ with $1/N<\varepsilon$. For $n\ge N$, $|x_n-y|<1/n\le1/N<\varepsilon$, so $x_n$ tends to $y$. Condition 1 would force $y$ into $K$, a contradiction. Hence some interval around $y$ avoids $K$. This holds for every point of the complement, proving it open. The witness-sequence step uses the stated classical setting; this proof is not offered as a theorem about choice-free foundations. If $K$ is empty there is no sequence in $K$, and its complement is the whole open line. If $K$ is the whole line, its empty complement is open vacuously. Both edge cases also satisfy the equivalence.

Learning goals and definitions

A set $K\subseteq\mathbb R$ is compact if every family of open sets whose union contains $K$ has a finite subfamily still covering $K$. A set is closed if it contains all limits of convergent sequences of its points, equivalently if its complement is open. Sequential compactness means every sequence in $K$ has a subsequence converging to a member of $K$.

The empty set is compact, closed and bounded, and sequentially compact vacuously. In proofs using a selected point or an infinite sequence, treat that case separately. We may assume the real completeness property and its nested-interval and Bolzano–Weierstrass consequences from the earlier sequence units.

Theorem: Heine–Borel in the real line

For $K\subseteq\mathbb R$, the following are equivalent: compactness by open covers; closedness and boundedness; sequential compactness.

Compact implies closed and bounded. The intervals $(-n,n)$, $n\ge1$, cover $\mathbb R$, hence $K$. A finite subcover gives a bound. To show closedness, take $y\notin K$. For each $x\in K$, the interval $B(x,|x-y|/2)$ avoids $y$. A finite subfamily covers $K$. Let $\delta$ be the minimum of their positive half-distances to $y$. Every point in the chosen interval centered at $x$ has distance from $y$ strictly greater than $|x-y|/2\ge\delta$. Thus $B(y,\delta)$ misses $K$, showing its complement open. If $K$ is empty this is immediate.

Closed and bounded implies compact. Enclose nonempty $K$ in a closed bounded interval $I_0$. Suppose an open cover of $K$ has no finite subcover. Bisect $I_0$. At least one closed half meets $K$ in a set not finitely coverable; otherwise two finite collections would cover $K$. Choose such a half and repeat. We obtain nested closed intervals $I_n$ with lengths tending to zero, and $K\cap I_n$ not finitely coverable. Each intersection is nonempty; choose $x_n\in K\cap I_n$. The nested-interval property gives a unique common point $p$, and $x_n\to p$ because both lie in $I_n$. Closedness gives $p\in K$. Some member $O$ of the cover contains $p$, and openness gives a ball around $p$ inside $O$. For large $n$, the whole interval $I_n$ lies in that ball, so $O$ alone covers $K\cap I_n$, a contradiction.

Closed and bounded implies sequentially compact. Every sequence in $K$ is bounded, so Bolzano–Weierstrass supplies a convergent subsequence. Closedness retains its limit in $K$.

Sequentially compact implies closed and bounded. If unbounded, choose $x_n\in K$ with $|x_n|>n$; no subsequence can converge to a finite real number, a contradiction. If $p$ is a limit of points in $K$, choose $x_n\in K$ with $|x_n-p|<1/n$. A subsequence converges to a point of $K$ by sequential compactness, but it also converges to $p$; uniqueness of real limits gives $p\in K$. Thus $K$ is closed. This completes all implications.

Compact subsets of the real line

A set $K\subset\mathbb R$ is compact if every open cover has a finite subcover. In $\mathbb R$, Heine–Borel says this is equivalent to being closed and bounded; it is also equivalent to every sequence in $K$ having a subsequence converging to a point of $K$. The closedness requirement keeps the subsequential limit in the set; boundedness permits extraction. We use the sequential form in the proofs below.

Theorem

Extreme value theorem

If $K\subset\mathbb R$ is nonempty and compact and $f:K\to\mathbb R$ is continuous, then $f$ is bounded and attains both a maximum and a minimum on $K$.

Proof

Extract a maximizing subsequence

If $f$ were unbounded above, choose $x_n\in K$ with $f(x_n)>n$. A subsequence $x_{n_k}\to x\in K$ exists; continuity would give $f(x_{n_k})\to f(x)$, contradicting divergence to infinity. The same reasoning applied to $-f$ gives boundedness below. Let $s=\sup f(K)$. Choose $x_n\in K$ with $s-1/n<f(x_n)\le s$. A convergent subsequence tends to some $x\in K$, and continuity implies $f(x)=s$. Apply the argument to $-f$ for a minimum.

Worked example

Which hypothesis fails?

The continuous function $f(x)=x$ on $(0,1)$ is bounded but has no maximum or minimum: $0$ and $1$ are not in the domain. The set is bounded but not closed, hence not compact. On $[0,1]$, both endpoints are present and the extrema are attained. Compactness is a sufficient global hypothesis, not a condition required for every individual function to attain extrema.

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 $f(c)$.

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 $f(c)=y$.

Worked example 2: an undefined formula with a limit. For $x\ne1$, $(x^2-1)/(x-1)=x+1$. Therefore its limit at $1$ is $2$. The original quotient is still undefined at $1$. To verify the limit, the absolute difference from $2$ is $|x-1|$; choose $\delta=\varepsilon$.

For finite limits, sums and products pass to the corresponding sum and product. Quotients do so when the denominator's limit is nonzero. If direct substitution produces $0/0$, it does not produce a number: another argument is needed.

For $f:D\to\mathbb R$ and $a\in D$, continuity relative to $D$ means that for every $\varepsilon>0$ there is $\delta>0$ such that

$$x\in D,\ |x-a|<\delta\implies |f(x)-f(a)|<\varepsilon.$$

At an accumulation point this is equivalent to $\lim_{x\to a}f(x)=f(a)$. An isolated domain point is continuous under this definition. Polynomials are continuous everywhere; rational functions are continuous where their denominators are nonzero. A continuous function on $[a,b]$ takes every value between its endpoint values, by the intermediate value theorem.

Worked example 3: guarantee a solution. The equation $x^2+x=1$ has a root between zero and one because $f(x)=x^2+x-1$ is continuous, $f(0)=-1$ and $f(1)=1$. This argument guarantees existence without calculating the root. It does not by itself prove uniqueness.

1.3 Average and instantaneous rates

Let $a$ be an interior point of $D$. The average rate of $f$ from $a$ to $a+h$ is $[f(a+h)-f(a)]/h$ for $h\ne0$ with $a+h\in D$. It has output units divided by input units. The derivative is its finite limit as the interval shrinks:

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

This definition is two-sided at an interior input. A boundary input requires a one-sided convention or a specified extension. A function can be continuous without being differentiable: the left and right limiting slopes of $|x|$ at zero are $-1$ and $1$.

The derivative is a limit

Definition

Derivative at an interior point

For $f$ defined near $a$, $f$ is differentiable at $a$ if the finite limit

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

exists. Equivalently $f(a+h)=f(a)+f'(a)h+o(h)$, where $o(h)/h\to0$ as $h\to0$.

Differentiability at $a$ implies continuity there: write $f(a+h)-f(a)=h[(f(a+h)-f(a))/h]$, and let $h\to0$. The converse fails: $|x|$ is continuous but not differentiable at zero because the left and right difference quotients tend to $-1$ and $1$.

Theorem

Fermat, Rolle and the mean value theorem

If a differentiable function has a local extremum at an interior point, its derivative there is zero. If $a<b$, $f$ is continuous on $[a,b]$, differentiable on $(a,b)$, and $f(a)=f(b)$, then there is $c\in(a,b)$ with $f'(c)=0$ (Rolle). Consequently, under the same continuity and differentiability assumptions but without equal endpoints, some $c\in(a,b)$ satisfies

$$f'(c)=\frac{f(b)-f(a)}{b-a}.$$

Proof

From extrema to mean value

At a local maximum, the difference quotient for positive increments is nonpositive and for negative increments is nonnegative. If both converge to the same derivative, that derivative is zero. The minimum case is analogous. For Rolle, the extreme value theorem guarantees a maximum and minimum on $[a,b]$. If the function is constant, any interior point works; otherwise at least one extremum is interior because the endpoint values agree, and Fermat applies. For the mean value theorem, subtract the affine function joining $(a,f(a))$ to $(b,f(b))$ and apply Rolle.

Worked example

A derivative bound controls a function

If $|f'(x)|\le4$ on an interval and $f$ is continuous on any closed segment within it, then for $x<y$ the mean value theorem yields $|f(y)-f(x)|\le4|y-x|$. Thus $f$ is Lipschitz and uniformly continuous there. The bound gives a sufficient condition, not a necessary condition for uniform continuity.

The theorem's interval hypotheses matter: $f(x)=|x|$ on $[-1,1]$ has equal endpoint values but is not differentiable at zero; no point in the open interval has derivative zero. A local derivative bound is also not automatically a single global bound.

Worked example 4: a nonlinear temperature model. Let $T(t)=20+3t+t^2$ for $t\ge0$, with time in minutes and temperature in degrees. For an interior time,

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

so $T'(t)=3+2t$. At $t=2$, the rate is $7$ degrees per minute. The exact change over a step $h$ is $7h+h^2$, while the derivative predicts the linear part $7h$.

The tangent line at $(a,f(a))$ is $y=f(a)+f'(a)(x-a)$. It is a local approximation, not a formula guaranteed to match the original function at distant inputs.

1.4 Calculating derivatives

The derivative rules below apply where the functions and derivatives involved exist:

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

Useful elementary derivatives are $(x^n)'=nx^{n-1}$ for integer powers with $n\ne0$ on their domains, $(e^x)'=e^x$, and $(\ln x)'=1/x$ for $x>0$. For $n=0$, define $x^0=1$ as a constant function; its derivative is zero, including at $x=0$. The chain rule supplies coefficients for inner expressions.

The usual radian derivatives are $(\sin x)'=\cos x$, $(\cos x)'=-\sin x$, and $(\tan x)'=\sec^2x$ on tangent's domain. The sine derivative follows from the angle-addition identity and the limits $\sin h/h\to1$, $(\cos h-1)/h\to0$; the first limit is squeezed between $\cos h$ and one using unit-circle geometry for small positive $h$, and symmetry gives the other side. For inverse branches, differentiate $\sin y=x$ to get $(\arcsin x)'=1/\sqrt{1-x^2}$ for $|x|<1$, because cosine is nonnegative on the principal branch. Likewise $(\arctan x)'=1/(1+x^2)$ on all real inputs. The inverse remains defined at some endpoints where its finite derivative does not.

For the product rule, write the change as

$$f(a+h)g(a+h)-f(a)g(a) =f(a+h)(g(a+h)-g(a))+g(a)(f(a+h)-f(a)).$$

Divide by $h$ and take limits; differentiability implies continuity, producing the stated two-term formula. The chain rule similarly preserves both the outer response and the inner input rate.

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: differentiating saturation. For $R(c)=6c/(2+c)$,

$$R'(c)=\frac{6(2+c)-6c}{(2+c)^2}=\frac{12}{(2+c)^2}>0\quad(c\ge0).$$

At $c=0$, this means the right derivative on the model domain; it equals the ordinary derivative of the same rational formula on $c>-2$.

Thus the model increases with concentration, but its marginal response becomes small as concentration grows. The derivative is not the same as the response-per-concentration ratio $R(c)/c=6/(2+c)$ for $c>0$.

For a positive function $N(t)$, its relative rate is $N'(t)/N(t)$. In $N=N_0e^{kt}$ this equals the constant $k$. A constant absolute rate instead produces an affine model, not an exponential one.

1.5 From derivative signs to optimization

For $a<b$, the mean value theorem says that if $f$ is continuous on $[a,b]$ and differentiable inside, then $f(b)-f(a)=f'(c)(b-a)$ for some interior $c$. Therefore a strictly positive derivative on an interval implies strict increase; a negative derivative implies decrease.

An interior differentiable local maximum or minimum must have $f'=0$. The condition is not sufficient: $x^3$ has zero derivative at zero but no extremum there. On a closed interval, a continuous function attains extrema; check endpoints as well as interior stationary or nondifferentiable points.

Worked example 6: a bounded treatment model. Suppose a hypothetical benefit model is $B(d)=8d-d^2$ on $0\le d\le5$. Its derivative $8-2d$ is positive below $4$ and negative above, so the maximum is at $d=4$, with value $16$. Compare endpoints: $B(0)=0$, $B(5)=15$. If the admissible range were $0\le d\le3$, the maximum would instead occur at the endpoint $3$, with value $15$.

1.6 Graded exercises

Foundation

1. Find the natural real domain of $f(x)=\sqrt{x+2}/(x-1)$.
2. Calculate $\lim_{x\to2}(x^2-4)/(x-2)$ and explain why the quotient is not defined at the limiting input.
3. From the derivative definition, find the derivative of $f(x)=2x^2+1$.

Core

4. For $N(t)=40e^{0.3t}$, calculate its absolute and relative rates at $t=0$.
5. Differentiate $f(x)=(1+x^2)e^{-x}$.
6. Find the tangent-line approximation to $f(x)=x^2$ at $x=3$ and use it to approximate $f(3.02)$. Compare with the exact value.
7. Maximize $f(x)=12x-x^2$ on $[0,4]$, with an endpoint check.

Synthesis

8. Show that $x^3+x-2=0$ has exactly one root in $[0,2]$, using separate existence and uniqueness arguments.
9. Let $f(x)=|x-1|$. Prove that it is continuous but not differentiable at $1$ by the relevant limits.
10. For $R(c)=Ac/(K+c)$ with $A,K>0$ and $c\ge0$, find $R'$ and $R''$. At $c=0$, evaluate these derivatives using the same rational formula extended to $c>-K$. Prove $R$ increases toward but never reaches $A$ at finite $c$. Explain how the second derivative quantifies diminishing marginal response.

Additional elementary-function checks

11. State amplitude and period of $3+2\cos(4t)$ and calculate its value at zero.
12. Differentiate $\sin(2t)+\cos t$ using radian conventions.
13. Find the derivative and real domain of $\arctan(3x)$; distinguish the domain of $\arcsin x$ from that of its finite derivative.

1.7 Full solutions

1. Require $x+2\ge0$ and $x\ne1$. The domain is $[-2,1)\cup(1,\infty)$.

2. For $x\ne2$, the quotient equals $x+2$, whose limit is $4$. At $x=2$ the original denominator is zero; cancellation does not supply a value to the original function.

3. $[2(x+h)^2+1-(2x^2+1)]/h=4x+2h$, whose limit is $4x$.

4. $N'=12e^{0.3t}$, so $N'(0)=12$ quantity units per time unit. The relative rate $N'/N=0.3$ per time unit is constant, including at zero.

5. The product rule gives $f'=2xe^{-x}-(1+x^2)e^{-x}=(-x^2+2x-1)e^{-x}$.

6. The tangent line is $L(x)=9+6(x-3)$. At $3.02$, it gives $9.12$. Exactly, $(3.02)^2=9.1204$, so the approximation is low by $0.0004$, the squared input change.

7. $f'=12-2x>0$ throughout $[0,4]$. Thus the maximum is at $4$, with value $32$. The other endpoint gives zero; the unconstrained stationary point $6$ is infeasible.

8. The polynomial is continuous, equals $-2$ at zero and $8$ at two, so a root exists. Its derivative is $3x^2+1>0$, so the mean value theorem proves strict increase and hence at most one root. Together these establish exactly one (indeed $x=1$).

9. $f(1)=0$ and $|x-1|\to0$ as $x\to1$, so it is continuous. Its difference quotient is $|h|/h$, equal to $1$ for $h>0$ and $-1$ for $h<0$, so no two-sided derivative exists.

10. $R'=AK/(K+c)^2>0$ and $R''=-2AK/(K+c)^3<0$. Also $R=A-AK/(K+c)<A$ for finite $c\ge0$, while the subtracted term tends to zero as $c\to\infty$. The negative second derivative means the positive first derivative decreases as concentration rises; increments continue to increase response but have diminishing marginal effect.

11. Amplitude is two, period is $\pi/2$, and the initial value is five.

12. The derivative is $2\cos(2t)-\sin t$; the inner factor two comes from the chain rule.

13. The derivative is $3/(1+9x^2)$, defined for every real $x$, as is the function. The function $\arcsin x$ is defined on $[-1,1]$, but its finite derivative formula applies only on $(-1,1)$; the endpoints are not silently added to the derivative's domain.

Summary and onward map

Domains are part of functions. Limits concern nearby values; continuity is relative to the domain, and at an accumulation point matches the nearby limit to the point value. Derivatives turn shrinking-interval average rates into instantaneous rates. Derivative rules preserve the structure of a composed model, while the mean value theorem connects derivative signs to changes over intervals. Optimization still requires feasibility and endpoint analysis.

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

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

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 $(-1/2,1/2)$.