The Timpson Lyceum · Calculus 1 (2024)

Inverse Functions, Logarithms, and Definition of 'e' - Lecture 19 (Calculus 1)

Overview

This lecture opens the study of transcendental functions — functions that cannot be built from ordinary algebraic operations. It develops inverse functions and a rule for differentiating them, then builds logarithms from the ground up as the inverses of exponential functions, proving the standard log laws. It introduces the number $e$, both through the compound-interest limit $e = \lim_{n\to\infty}\left(1+\tfrac{1}{n}\right)^n$ and as the natural base, and derives the derivative of $\log_a x$ and $\ln x$. Finally it applies natural logs to evaluate integrals that produce $\ln$ (including $\tan$, $\cot$, $\sec$, $\csc$) and uses logarithmic differentiation to tame messy products and quotients. These tools are central to the rest of calculus.

Key Concepts

One-to-one functions

• A function pairs every input with a unique output (passes the vertical line test).
• A function is one-to-one if every output comes from a unique input: if $f(a)=f(b)$ then $a=b$. Equivalently it passes the horizontal line test.
• Example: $f(x)=x^2+1$ is not one-to-one since $f(2)=f(-2)=5$. Restricting the domain to $[0,\infty)$ makes it one-to-one.

Inverse functions

• Only one-to-one functions have inverses. The inverse is written $f^{-1}(x)$ — this is notation, not a reciprocal: $f^{-1}(x)\neq \frac{1}{f(x)}$.
• Definition: $f^{-1}(b)=a$ if and only if $f(a)=b$. The inverse maps outputs back to inputs.
• Cancellation: $f\!\left(f^{-1}(x)\right)=x$ and $f^{-1}\!\left(f(x)\right)=x$.
• To find an inverse: write $y=f(x)$, swap $x$ and $y$, and solve for $y$.
• Graphically: the graph of $f^{-1}$ is the reflection of $f$ across the line $y=x$; if $(a,b)$ lies on $f$, then $(b,a)$ lies on $f^{-1}$.
• Inverse trig functions invert the trig functions: e.g. $\sin^{-1}\!\left(\tfrac{\sqrt{3}}{2}\right)=\tfrac{\pi}{3}$ because $\sin\tfrac{\pi}{3}=\tfrac{\sqrt 3}{2}$.

Derivative of an inverse function

• The slope of $f^{-1}$ at a point is the reciprocal of the slope of $f$ at the corresponding point: $\left(f^{-1}\right)'(b)=\frac{1}{f'\!\left(f^{-1}(b)\right)}.$
• This lets you find the slope of $f^{-1}$ at a point without ever computing $f^{-1}$, as long as you can identify which input maps to the given output.

Exponents and exponential functions

• An exponential function has the variable in the exponent, e.g. $f(x)=2^x$.
• Key rules: $a^0=1$, $a^1=a$, $a^{-1}=\tfrac{1}{a}$, $a^m a^n=a^{m+n}$, $\left(a^m\right)^n=a^{mn}$, $\dfrac{a^m}{a^n}=a^{m-n}$.

Logarithms

• Definition: $y=\log_a x$ means exactly $x=a^y$. A log takes a "product" and returns the exponent; it is the inverse of the exponential $a^x$. The base $a$ and the argument $x$ must be positive.
• Examples: $\log_2 32 = 5$ (since $2^5=32$); $\log_{10}1000=3$; $\log_{\pi}\pi^2 = 2$.
• Properties:
  • $\log_a 1 = 0$, $\quad \log_a a = 1$, $\quad \log_a\!\left(a^x\right)=x$.
  • $\log_a(xy)=\log_a x + \log_a y$
  • $\log_a\!\left(\tfrac{x}{y}\right)=\log_a x - \log_a y$
  • $\log_a\!\left(x^k\right)=k\log_a x$
  • Change of base: $\log_a x = \dfrac{\log_b x}{\log_b a}$.

The number e and the natural log

• $e = \displaystyle\lim_{n\to\infty}\left(1+\frac{1}{n}\right)^n \approx 2.718$.
• It arises from continuously compounded interest: discrete compounding $A=P\left(1+\tfrac{r}{n}\right)^{nt}$ becomes $A=Pe^{rt}$ as $n\to\infty$.
• The natural log is $\ln x = \log_e x$. It satisfies all log laws and is the most convenient base in calculus.

Derivatives of logarithms

• General rule: $\dfrac{d}{dx}\log_a x = \dfrac{1}{x}\log_a e$.
• For the natural log the extra factor $\log_e e = 1$ disappears, which is why $\ln$ is preferred: $\frac{d}{dx}\ln x = \frac{1}{x},\qquad \frac{d}{dx}\ln u = \frac{1}{u}\,\frac{du}{dx}.$

Integrals producing natural logs

• $\displaystyle\int \frac{1}{x}\,dx = \ln|x| + C$ — the missing case of the power rule (you cannot add $1$ to the exponent $-1$ and divide by $0$).
• $\ln x$ also satisfies the textbook definition $\displaystyle\ln x = \int_1^x \frac{1}{t}\,dt$, and $\ln 1 = 0$.
• Standard antiderivatives obtained by $u$-substitution:
  • $\displaystyle\int \tan\theta\,d\theta = \ln|\sec\theta| + C$
  • $\displaystyle\int \cot\theta\,d\theta = \ln|\sin\theta| + C$
  • $\displaystyle\int \sec\theta\,d\theta = \ln|\sec\theta + \tan\theta| + C$
  • $\displaystyle\int \csc\theta\,d\theta = -\ln|\csc\theta + \cot\theta| + C$

Logarithmic differentiation

• To differentiate a complicated product/quotient/power, take $\ln$ of both sides, expand with log laws into bite-sized pieces, differentiate (using $\tfrac{d}{dx}\ln y = \tfrac{y'}{y}$), then solve for $y'$ and substitute $y$ back in.

Worked Example

Derivative of an inverse without finding the inverse. Given $f(x)=x^3-2$, find $\left(f^{-1}\right)'(6)$.

The rule is $\left(f^{-1}\right)'(b)=\dfrac{1}{f'\!\left(f^{-1}(b)\right)}$. First note $f(2)=2^3-2=6$, so $f^{-1}(6)=2$, and $f'(x)=3x^2$.

Logarithmic differentiation. Find $y'$ for $y=\dfrac{\left(x^2+1\right)(x+3)^{1/2}}{x-1}$ on $x>1$.

Take $\ln$ of both sides and expand: $\ln y = \ln(x^2+1) + \tfrac{1}{2}\ln(x+3) - \ln(x-1)$. Differentiate term by term and solve for $y'$.

Summary

• A function has an inverse only if it is one-to-one; find $f^{-1}$ by swapping $x$ and $y$ and solving, and graph it as the reflection of $f$ across $y=x$.
• The derivative of an inverse is the reciprocal of the original's derivative: $\left(f^{-1}\right)'(b)=\dfrac{1}{f'\left(f^{-1}(b)\right)}$ — usable without computing $f^{-1}$.
• Logarithms invert exponentials: $\log_a x = y \iff a^y = x$, with the laws for products, quotients, powers, and change of base following directly from the exponent rules.
• The number $e=\lim_{n\to\infty}\left(1+\tfrac1n\right)^n\approx 2.718$ is the natural base, and $\ln x = \log_e x$.
• Differentiation is simplest in base $e$: $\dfrac{d}{dx}\ln x = \dfrac{1}{x}$ and $\dfrac{d}{dx}\ln u = \dfrac{u'}{u}$, while $\int \tfrac{1}{x}\,dx = \ln|x|+C$ fills the gap in the power rule.
• Natural logs evaluate integrals of $\tan,\cot,\sec,\csc$ and, via logarithmic differentiation, turn ugly products and quotients into easily differentiated sums.