Unit 04 · Term 1

Transcendental Functions

Exponential, logarithm, and trigonometric functions, and their inverses — defined carefully and differentiated. The clean way to build them is to define $\ln$ as an integral and recover everything else from there.

By the end you can

Define $\ln$ and $\exp$ rigorously and prove their basic laws.
Differentiate exponential, logarithmic, and trig functions.
Differentiate inverse trig functions.
Use logarithmic differentiation.

Why it matters & where it comes from

Why it exists

Polynomials cannot capture growth that compounds or motion that repeats. Exponentials, logarithms, and trig functions can — they are called "transcendental" because they escape algebra entirely.

In the real world

$e^x$ models compound interest, radioactive decay, and populations; logarithms give us pH, decibels, the Richter scale, and the "bit" of information; trig functions underlie all signal processing — audio, JPEG images, MRI, and GPS all run on Fourier analysis built from sines and cosines.

Story & open questions

Napier invented logarithms in 1614 to shorten astronomers' brutal hand calculations. Euler named $e$ and tied it all together. Later $e$ and $\pi$ were proved transcendental (1873, 1882), which finally settled the 2,000-year-old puzzle of whether you can "square the circle." You cannot.

How the functions are built

The clean MAT137 approach is structural: define $\ln$ by an integral, prove its laws, define $\exp$ as its inverse, then define $a^x$ using $e^{x\ln a}$. This order matters because it keeps irrational exponents and logarithm laws from becoming circular assumptions.

Dependency chain

First: $\ln x=\int_1^x\dfrac1t\,dt$, so $(\ln x)'=1/x$.

Next: log laws follow by comparing derivatives of functions such as $\ln(ax)-\ln x$.

Then: $\exp$ is the inverse of $\ln$, and inverse differentiation gives $(e^x)'=e^x$.

Finally: for $a>0$, define $a^x=e^{x\ln a}$.

Domain discipline

Most mistakes in this unit are not algebra mistakes; they are domain mistakes. Before differentiating, ask where the expression is defined and where the derivative formula is valid.

$\ln x$ requires $x>0$, while $\ln|x|$ is defined for $x\ne0$.
$a^x=e^{x\ln a}$ requires $a>0$ for real-valued functions of all real $x$.
$\arcsin x$ is defined on $[-1,1]$, but its derivative formula only holds on $(-1,1)$.
Logarithmic differentiation requires the quantity inside $\ln$ to be positive, or else an absolute-value version must be justified.

The natural logarithm as an integral

Definition Natural log

For $x>0$, $\displaystyle\ln x=\int_1^x \frac{1}{t}\,dt.$ By the Fundamental Theorem of Calculus, $\dfrac{d}{dx}\ln x=\dfrac1x$.

Theorem Log laws

For $a,b>0$: $\ln(ab)=\ln a+\ln b$ and $\ln(a/b)=\ln a-\ln b$. For rational $r$, $\ln(a^r)=r\ln a$; after real powers are defined by $a^r=e^{r\ln a}$, the same identity holds for every real $r$.

Proof idea of $\ln(ab)=\ln a+\ln b$

Fix $a$ and let $g(x)=\ln(ax)$. Then $g'(x)=\dfrac{a}{ax}= \dfrac1x=(\ln x)'$.

So $g(x)-\ln x$ is constant; at $x=1$ it equals $\ln a$. Hence $\ln(ax)=\ln a+\ln x$; set $x=b$. ∎

The exponential function

Definition exp

$\exp=\ln^{-1}$, and $e=\exp(1)$. Then $\dfrac{d}{dx}e^x=e^x$, and for $a>0$, $a^x=e^{x\ln a}$ with $\dfrac{d}{dx}a^x=a^x\ln a$.

The derivative rule follows from the inverse function theorem (Unit 3 / next units): if $y=e^x$ then $\ln y=x$, so $\dfrac{y'}{y}=1$, giving $y'=y=e^x$.

Trigonometric functions

Key limit

$\displaystyle\lim_{x\to0}\frac{\sin x}{x}=1$ — the fact behind every trig derivative. From it, $\dfrac{d}{dx}\sin x=\cos x$ and $\dfrac{d}{dx}\cos x=-\sin x$.

Example

$\dfrac{d}{dx}\tan x=\dfrac{d}{dx}\dfrac{\sin x}{\cos x}= \dfrac{\cos^2x+\sin^2x}{\cos^2x}=\sec^2x.$

Inverse trigonometric functions

Derivatives

$\dfrac{d}{dx}\arcsin x=\dfrac{1}{\sqrt{1-x^2}}$ for $|x|<1$, $\dfrac{d}{dx}\arctan x=\dfrac{1}{1+x^2}$ for every real $x$.

Example arctan

If $y=\arctan x$ then $\tan y=x$, so $\sec^2y\cdot y'=1$.

$\sec^2y=1+\tan^2y=1+x^2$, hence $y'=\dfrac{1}{1+x^2}$. ∎

Logarithmic differentiation

For products, quotients, and variable exponents, take $\ln$ first.

Example

Differentiate $y=x^x$ ($x>0$).

$\ln y=x\ln x$, so $\dfrac{y'}{y}=\ln x+1$.

$y'=x^x(\ln x+1)$.

Common pitfalls

• $\dfrac{d}{dx}a^x\ne x a^{x-1}$. The power rule is for $x^{n}$, not $a^x$.
• Mixing up $(\ln x)'=1/x$ with $(\log_{10}x)'=1/(x\ln 10)$.
• Forgetting domains: $\arcsin x$ needs $|x|\le1$, and its derivative formula requires $|x|<1$.

Practice

Problems adapted from the MAT137 question bank by Alfonso Gracia-Saz, used under CC BY-SA 4.0. Write full solutions before opening the answer.

Practice 1 · Log + chain

Differentiate $f(x)=\ln(\sec x+\tan x)$ where it is defined.

Answer

$f'(x)=\dfrac{\sec x\tan x+\sec^2x}{\sec x+\tan x}=\sec x.$

Practice 2 · Inverse trig

Show $\dfrac{d}{dx}\arcsin x=\dfrac{1}{\sqrt{1-x^2}}$ for $|x|<1$.

Hint

$y=\arcsin x\Rightarrow\sin y=x\Rightarrow\cos y\,y'=1$; $\cos y=\sqrt{1-x^2}$ since $|y|\le\pi/2$.

Practice 3 · Prove a rule

Obtain and prove the formula for $\dfrac{d}{dx}\arctan x$.

Solution

$y=\arctan x\Rightarrow \tan y=x\Rightarrow\sec^2y\,y'=1$, so $y'=\dfrac{1}{\sec^2y}=\dfrac{1}{1+\tan^2y}=\dfrac{1}{1+x^2}.$

Practice 4 · Absolute value

Find the derivative of $F(x)=\ln|x|$.

Answer

$F'(x)=\dfrac1x$ for $x\ne0$. (For $x<0$, $\ln|x|=\ln(-x)$ and the chain rule gives $\dfrac{-1}{-x}=\dfrac1x$.)

Practice 5 · Product + chain

Differentiate $f(x)=e^{\sin x+\cos x}\ln x$ for $x>0$.

Answer

$f'(x)= e^{\sin x+\cos x}\!\left[(\cos x-\sin x)\ln x+\dfrac1x\right].$

Practice 6 · Log differentiation

Use logarithmic differentiation on $y=\dfrac{(x+1)^2\sqrt{x-3}}{x^4}$.

Hint

$\ln y=2\ln(x+1)+\tfrac12 \ln(x-3)-4\ln x$; differentiate, then multiply by $y$.

Practice 7 · Variable base

Calculate the derivative of $g(x)=x^{\tan x}$ for $x>0$.

Solution

$\ln g=\tan x\,\ln x$, so $\dfrac{g'}{g}=\sec^2x\,\ln x+\dfrac{\tan x}{x}$ and $g'=x^{\tan x}\!\left(\sec^2x\,\ln x+\dfrac{\tan x}{x}\right).$

Practice 8 · Implicit + log

Assume $x,y>0$. Find $y'$ if $x^y=y^x$.

Solution

Take logs: $y\ln x=x\ln y$. Differentiate: $y'\ln x+\dfrac yx=\ln y+\dfrac{xy'}{y}$, so $y'=\dfrac{\ln y-y/x}{\ln x-x/y}.$

Expanded topic checklist

Course coverage. This unit is part of the complete MAT137 topic map. It should cover the following items before you move on:

Logarithm and exponential. Know how $\ln x$ and $e^x$ are defined, why they are inverses, and how their derivative rules follow.
Trigonometric functions. Use standard trig identities and derivative rules without losing domain restrictions.
Inverse trig. Differentiate inverse trigonometric functions through inverse-function reasoning.
Logarithmic differentiation. Use logarithms to differentiate powers, products, quotients, and variable-base expressions.

Textbook/source alignment

Use the Alfonso Gracia-Saz MAT137 materials as the primary local spine, with the current UofT single-variable calculus topic split as the completeness check. The added items are original study prompts, not copied textbook exercises.

Core theorem map

The spine is: define one function rigorously, prove it is monotone and invertible, then use the inverse derivative rule to obtain the companion derivative formula.

Study routine. Build a one-page formula sheet, but next to every formula write its domain and one sentence explaining why the formula is true.

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 derivative

Starting from $\ln x=\int_1^x 1/t\,dt$, explain why $(\ln x)\prime=1/x$ for $x>0$.

Guide

Apply FTC Part 1 to the continuous integrand $1/t$ on $(0,\infty)$.

Extension 2 · Exponential inverse

Use inverse differentiation to show $(e^x)\prime=e^x$ if $e^x$ is the inverse of $\ln x$.

Guide

If $y=e^x$, then $x=\ln y$. The inverse derivative gives $(e^x)\prime=1/(1/y)=y=e^x$.

Extension 3 · Inverse trig

Derive $(rcsin x)\prime=1/\sqrt{1-x^2}$ for $-1<x<1$.

Guide

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

Extension 4 · Log differentiation

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

Guide

Take logs: $\ln y=x\ln x$. Then $y\prime/y=\ln x+1$, so $y\prime=x^x(\ln x+1)$.

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