MATH 145 – Calculus I
October 16, 2025
What functions \(f\) have the property that they are the derivative of themselves?
In other words: \[ \dfrac{d}{dx} f(x) = f(x) \]
The trivial solution is \(f(x) = 0\), but that’s not very exciting…
An exponential function is a function of the form \(f(x) = P_0 a^x\), where \(P_0\) and \(a\) are constants.
This is a family of functions that depend on \(P_0\) and \(a\) We’ve seen other families of functions, such as linear functions (\(f(x) = mx + b)\) or quadratic functions (\(f(x) = ax^2 + bx + c\)).
What is the meaning of \(P_0\)?
\(P_0\) is the initial amount, the value of \(f(x)\) when \(x = 0\).
What is the meaning of a? Consider \(a = 1 + r\).
\(r\) is the percentage rate of growth of an exponential function. It is constant over intervals of the same length.
Let \(f(x) = 1000(1.07)^x\).
Suppose that you start a saving account with $10,000, that gains 1% interest monthly.
The graphs of exponential functions are distinctive because of their monotonic increase/decrease, concavity and asymptotes.
All variables here represent real numbers and all variables in denominators are nonzero.
Example: Simplify \((2 \cdot 3^x \cdot 9^x)^{-2}\), and state it in the form of an exponential function.
Consider an account that starts with $1 and earns an annual interest rate of 100% compounded continuously. How much money will it have in the account?
\[ \begin{aligned} \textrm{Compounding Yearly:} &\, (1 + 1)^1 = 2\\ \textrm{Compounding Every Six Months:} &\, (1 + \frac{1}{2})^2 = 2.25 \\ \textrm{Compounding Every Three Months:} &\, (1 + \frac{1}{4})^4 = 2.4414 \\ \textrm{Compounding Monthly:} &\, (1 + \frac{1}{12})^{12} = 2.6130 \\ \textrm{Compounding Weekly:} &\, (1 + \frac{1}{52})^{52} = 2.6926 \\ \end{aligned} \]
\[ \begin{aligned} \textrm{Compounding Daily:} &\, (1 + \frac{1}{365})^{365} = 2.7146\\ \textrm{Compounding Hourly:} &\, (1 + \frac{1}{8760})^{8760} = 2.7181 \\ \textrm{Compounding Every Minute:} &\, (1 + \frac{1}{525600})^{525600} = 2.718279 \\ \textrm{Compounding Every Second:} &\, (1 + \frac{1}{31536000})^{31536000} = 2.718282\\ \textrm{Compounding Continuously:} &\, \lim\limits_{n \to \infty} (1 + \frac{1}{n})^n = e \approx 2.7182818284\dots \\ \end{aligned} \]
The most important base for the exponential function is Euler’s number, \(e \approx 2.7182818284\dots\).
We call \(f(x) = e^x\) the exponential function.
In an exponential function of the form \(f(x) = P_0 e^{kx}\), \(k\) is called the continuous growth rate.
\[ e^k = \lim\limits_{n \to \infty} \left( 1 + \dfrac{k}{n} \right)^n \]
What is \(\dfrac{d}{dx} \, a^x\)? \[ \begin{aligned} \dfrac{d}{dx} a^x &= \lim\limits_{h \to 0} \frac{a^{x + h} - a^x}{h} \\ &= \lim\limits_{h \to 0} \frac{a^x a^h - a^x}{h} \\ &= \lim\limits_{h \to 0} \frac{a^x (a^h - 1)}{h} \\ &= \lim\limits_{h \to 0} \left( a^x \left( \frac{a^h - 1}{h} \right) \right) \\ &= a^x \left( \lim\limits_{h \to 0} \left( \frac{a^h - 1}{h} \right) \right) \end{aligned} \]
If \(a = e\), then \(\lim\limits_{h \to 0} \left( \frac{a^h - 1}{h} \right) = 1\).
Thus \(\dfrac{d}{dx} \, e^x = e^x\).
Further, \(f(x) = e^x\) is the only function other than the trivial \(f(x) = 0\) for which \(\dfrac{d}{dx} \, f(x) = f(x)\).
Examples: