Unit 11 · Analysis 1A

Transcendental Functions

The rigorous payoff: define $\ln$ as an integral, $\exp$ as its inverse, and recover every familiar law as a theorem. This is the unit where "what does $2^{\sqrt2}$ even mean?" finally gets an honest answer — built from the FTC of Unit 10 and the inverse function theorem of Unit 8.

By the end you can

Define $\ln$ as an integral and prove the log laws.
Define $\exp$ as the inverse and prove $(e^x)'=e^x$.
Define $a^x$ and $x^\alpha$ for real exponents and differentiate them.
Explain how the trig functions are defined rigorously.

Why it matters & where it comes from

Why it exists

Defining $e^x$ by "repeated multiplication" makes no sense for irrational exponents. Building $\ln$ as $\int_1^x\tfrac1t\,dt$ first gives a rigorous, gap-free construction of all the transcendental functions.

In the real world

$e^x$ governs growth and decay; logarithms scale pH, decibels, and information in bits; trig functions drive every Fourier method in audio, imaging, and signal processing.

Story & open questions

Napier built logarithms in 1614 to ease astronomers' arithmetic; Euler named $e$ and proved it irrational. Hermite and Lindemann later proved $e$ and $\pi$ transcendental, settling "squaring the circle" once and for all.

The logarithm, from the integral up

Definition Natural logarithm

For $x>0$: $\ln x=\displaystyle\int_1^x\frac{1}{t}\,dt$. Since $1/t$ is continuous on $(0,\infty)$, FTC Part 1 gives $(\ln x)'=\tfrac1x$ immediately. Note $\ln1=0$, and $\ln x<0$ for $x<1$ (the integral runs backwards).

Theorem Log laws

For $a,b>0$: $\ln(ab)=\ln a+\ln b$, $\ln\tfrac ab=\ln a-\ln b$, and $\ln(a^r)=r\ln a$ for rational $r$.

Proof of the product law

Fix $a$ and set $g(x)=\ln(ax)$. Chain rule: $g'(x)=\dfrac{a}{ax}=\dfrac1x=(\ln x)'$.

Equal derivatives on an interval force $g(x)-\ln x$ constant (MVT corollary, Unit 7); at $x=1$ the constant is $\ln a$. Set $x=b$. ∎

Example Locating $e$: $2<e<4$

On $[1,2]$, $\tfrac12\le\tfrac1t\le1$, with strict inequality inside; integrating, $\tfrac12<\ln2<1$.

$\ln2<1$ means $2<e$; and $\ln4=2\ln2>1$ means $e<4$. (A finer estimate gives $e\approx2.71828$.)

The exponential, as an inverse

Theorem $\ln$ is a bijection onto $\mathbb{R}$

$(\ln x)'=\tfrac1x>0$, so $\ln$ is strictly increasing on $(0,\infty)$. It is unbounded above ($\ln2^n=n\ln2\to\infty$) and below ($\ln2^{-n}=-n\ln2$), so by the IVT its image is all of $\mathbb{R}$.

Definition exp and $e$

$\exp=\ln^{-1}:\mathbb{R}\to(0,\infty)$, and $e=\exp(1)$ — the unique number with $\ln e=1$. Write $e^x$ for $\exp(x)$.

Proof $(e^x)'=e^x$

The inverse function theorem (Unit 8) applies: $\ln$ is differentiable with nonzero derivative everywhere.

$(\exp)'(x)=\dfrac{1}{(\ln)'(\exp x)} =\dfrac{1}{1/\exp x}=\exp x$. ∎ The exponential is its own derivative — by construction, not by decree.

Now irrational exponents mean something For $a>0$ define $a^x=e^{x\ln a}$, and for $x>0$, $x^\alpha=e^{\alpha\ln x}$. These agree with the old definitions for rational exponents (by the log laws) and finally make sense of $2^{\sqrt2}$. The chain rule gives $(a^x)'=a^x\ln a$ and $(x^\alpha)'=\alpha x^{\alpha-1}$ — the power rule for every real exponent, at last.

Growth rates

Theorem

$\displaystyle\lim_{x\to\infty}\frac{\ln x}{x}=0$ — logarithms grow slower than any positive power, and dually $e^x$ outruns every polynomial.

Proof (a one-line integral estimate)

For $x>1$: $\ln x=\displaystyle\int_1^x\frac{dt}{t}<\int_1^x\frac{dt}{\sqrt t} =2(\sqrt x-1)<2\sqrt x$.

So $0<\dfrac{\ln x}{x}<\dfrac{2}{\sqrt x}\to0$; squeeze. ∎

Trigonometric functions, honestly

"Angle" is not a rigorous notion until you make it one. The clean route (Spivak, ch. 15): define $\pi=2\displaystyle\int_{-1}^{1}\sqrt{1-t^2}\,dt$ (twice the area of the unit half-disc), define $\cos$ on $[0,\pi]$ as the inverse of an explicit area function built from that integral, and set $\sin x=\sqrt{1-\cos^2x}$ there; then extend to all of $\mathbb{R}$ by symmetry and periodicity. The definitions deliver the familiar facts as theorems:

Theorem

$(\sin x)'=\cos x$ and $(\cos x)'=-\sin x$; consequently $\displaystyle\lim_{x\to0}\frac{\sin x}{x}=1$ — it is exactly the statement that $\sin'(0)=\cos0=1$.

Example arctan, revisited

$\tan$ is strictly increasing on $\left(-\tfrac\pi2,\tfrac\pi2\right)$ with derivative $\sec^2x\ge1$, so $\arctan$ exists on all of $\mathbb{R}$ and the inverse rule of Unit 8 gives $(\arctan x)'=\dfrac{1}{1+x^2}$.

Common pitfalls

• $(a^x)'=a^x\ln a$, not $xa^{x-1}$ — the power rule is for a fixed exponent.
• $x^\alpha$ for irrational $\alpha$ is defined as $e^{\alpha\ln x}$ and needs $x>0$.
• Keep domains straight: $\ln$ needs $x>0$; $\ln|x|$ has derivative $1/x$ on each side of $0$ but is undefined at $0$.
• $\lim_{x\to0}\tfrac{\sin x}{x}=1$ is not circular here: it is the derivative of $\sin$ at $0$, proved from the integral-based definitions.

Practice

Original problems written for this guide in the style of Spivak's exercises. Write a full solution before opening the answer.

Practice 1 · Log law

Prove $\ln(x^r)=r\ln x$ for rational $r$ and $x>0$.

Solution

Both sides vanish at $x=1$ and have derivative $\tfrac{1}{x^r}\cdot rx^{r-1}=\tfrac rx=(r\ln x)'$, so they differ by a constant that is $0$.

Practice 2 · Variable exponent

Differentiate $y=x^x$ for $x>0$.

Solution

$x^x=e^{x\ln x}$, so $y'=e^{x\ln x}(\ln x+1)=x^x(\ln x+1)$.

Practice 3 · A limit from a derivative

Compute $\lim_{x\to0}\dfrac{e^x-1}{x}$ without L'Hôpital.

Answer

It is the difference quotient of $e^x$ at $0$, so the limit is $(e^x)'|_{0}=e^0=1$.

Practice 4 · Quotient law

Deduce $\ln\tfrac ab=\ln a-\ln b$ from the product law.

Solution

$\ln a=\ln\big(\tfrac ab\cdot b\big)=\ln\tfrac ab+\ln b$; rearrange.

Practice 5 · Real power rule

From the definition $x^\alpha=e^{\alpha\ln x}$, prove $(x^\alpha)'=\alpha x^{\alpha-1}$ for $x>0$.

Solution

Chain rule: $(e^{\alpha\ln x})'=e^{\alpha\ln x}\cdot\tfrac{\alpha}{x} =\alpha\,\tfrac{x^\alpha}{x}=\alpha x^{\alpha-1}$.

Practice 6 · Growth

Prove $\lim_{x\to\infty}\dfrac{\ln x}{x}=0$ using only an integral estimate.

Solution

For $x>1$, $\ln x=\int_1^x\tfrac{dt}{t}<\int_1^x\tfrac{dt}{\sqrt t}<2\sqrt x$, so $0<\tfrac{\ln x}{x}<\tfrac{2}{\sqrt x}\to0$.

Practice 7 · Pinning down $e$

Show $\tfrac12<\ln2<1$ by estimating the integral, and conclude $2<e<4$.

Solution

On $[1,2]$, $\tfrac12\le\tfrac1t\le1$ with strict inequality on the interior, so integrating over an interval of length 1 gives $\tfrac12<\ln2<1$. Then $\ln2<1=\ln e$ gives $2<e$, and $\ln4=2\ln2>1$ gives $e<4$.

Practice 8 · Both signs

Show that $F(x)=\ln|x|$ satisfies $F'(x)=\tfrac1x$ for every $x\ne0$.

Solution

For $x>0$ this is the FTC. For $x<0$, $F(x)=\ln(-x)$ and the chain rule gives $F'(x)=\tfrac{1}{-x}\cdot(-1)=\tfrac1x$.

Expanded topic checklist

Course coverage. This unit is part of the complete MAT158H1 topic map. It should cover the following items before you move on:

Exponential and logarithm. Construct and differentiate exponential/logarithmic functions rigorously.
Trig functions. Use trigonometric identities and derivative rules with proof awareness.
Inverse functions. Derive inverse trig derivatives from inverse-function theory.
Growth and inequalities. Use MVT and convexity to prove standard exponential/logarithmic inequalities.

Textbook/source alignment

Use the current UofT MAT158H1 calendar topics as the coverage spine, and pair lecture notes with a Spivak-style proof-calculus text for extra reading. The exact assigned textbook should still be checked against the instructor syllabus.

Core theorem map

The transcendental spine is: define carefully, prove monotonicity/invertibility, differentiate, then derive identities and inequalities.

Study routine. Do not treat formulas as isolated facts; connect each formula to FTC, inverse functions, or the differential equation $y\prime=y$.

Deep practice extensions

Use these after the main practice set. They are designed to force definition-level reasoning, theorem selection, and mixed application rather than routine calculation only.

Extension 1 · Log inequality

Prove $\ln x\le x-1$ for $x>0$.

Guide

Let $h(x)=x-1-\ln x$. Then $h\prime(x)=1-1/x$, so $h$ decreases to $0$ on $(0,1]$ and increases from $0$ on $[1,\infty)$.

Extension 2 · Exp inequality

Prove $e^x\ge1+x$.

Guide

Use $h(x)=e^x-1-x$ and derivative sign, or convexity of $e^x$ at $0$.

Extension 3 · Inverse trig

Differentiate $rccos x$.

Guide

Let $y=rccos x$, so $x=\cos y$ and $y\in[0,\pi]$. Then $dy/dx=1/(-\sin y)=-1/\sqrt{1-x^2}$ for $-1

Extension 4 · Growth

Show $\lim_{x\to\infty}\ln x/x=0$ using LHopital if available.

Guide

It is an $\infty/\infty$ form; derivative quotient is $(1/x)/1=1/x\to0$.

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