An exponential function is \(f(x) = b^x\) where \(b\) is any positive constant.
There is a special constant \(e\) which is a natural base for exponential functions. The natural exponential function is \(e^x\), and it has the very important property that its derivative is itself: \[\frac{d}{dx} e^x = e^x.\] In class, we used this property to:
We got the tangent line \(y = x + 1\), which means that \(e^x \approx 1+x\) when \(x\) is close to zero. This led to the following formula, which is the definition of \(e\): \[e = \lim_{n \rightarrow \infty} \left( 1+ \frac{1}{n} \right)^n.\] We used a calculator to approximate this limit with n = 10,000,000 and got \(e \approx 2.71828\).
We solved the following problems:
Find \(\dfrac{d}{dx} x^2 e^x\).
Find \(\dfrac{d}{dx} e^{3x}\).
A boiling cup of water is set out to cool and has temperature \(T(t) = 20 + 80 e^{-0.3t }\) where \(t\) is time in minutes. What is the instantaneous rate of change in the cup’s temperature at time \(t = 0\).
The normal distribution in statistics is given by \(y = e^{-\tfrac{1}{2}x^2}\). Find \(y'\) and \(y''\), and use \(y''\) to find the inflection points.
I tried to finish by explaining the connection between the natural exponential function and compound interest, but I ran out of time. Here’s the idea: if you invest \(P\) dollars (\(P\) for principal) at an interest rate of \(r\) per year, then next year you will have a balance \(B\) given by \[B = P + r P = P(1+r)\] If you wait more than 1 year, your interest will compound, which means you will earn interest on the interest you’ve already earned. So your balance after \(t\) years will be \[B(t) = P(1+r)^t.\] Typically banks will pay out interest quarterly or monthly, so the actual formula will be: \[B(t) = P\left(1 + \frac{r}{n} \right)^{nt}\] where \(n\) is the number of times interest is compounded per year. In the limit as \(n\) goes to infinity, this formula simplifies to the formula for continuously compounded interest:
\[B(t) = P e^{rt}.\] This is because \(\left(1+\frac{r}{n} \right) \approx e^{r/n}\) when \(n\) gets big, so if you raise both sides to the \(nt\) power, you get the result.
A logarithmic scale is a number line with distances that represent multiplying by fixed factors. For example, here is a log scale where each space represents a factor of 2.
A logarithm is a function that does two things.
\(\log_b(x)\) tells you how many steps on a base b log-scale the number \(x\) goes away from 1.
\(\log_b(x)\) tells you the power you need to raise the base to in order to get \(x\).
We calculated the following logarithms right away, without a calculator:
\(\log_5(25)\)
\(\log_{10}(10,000)\)
\(\log_2(\tfrac{1}{2})\)
\(\log_3(\sqrt{3})\)
All logarithms (no matter what base) satisfy three Logarithm Rules
\(\log(ab) = \log(a) + \log(b)\)
\(\log(\tfrac{a}{b}) = \log(a) - \log(b)\)
\(\log(a^p) = p \log(a)\)
We used these rules to simplify the following expressions:
_2(36)-_2(9)
_5(500) - _5(4)
_b(b^x)$
The most important logarithm is the base-e logarithm, which is called the natural logarithm. It is written \(\ln(x)\) instead of \(\log_e(x)\). It is the inverse of the function \(e^x\), which means that \[\ln(e^x) = x ~~~~~ \text{ and } ~~~~~ e^{\ln x} = x.\]
You should also know what the graph of \(\ln x\) looks like.
We did the following natural logarithm problems in class:
Simplify \(\displaystyle \ln \left( \frac{x^2}{\sqrt{y}} \right)\).
Simplify \(\ln(\sqrt{e})\)
Simplify \(\displaystyle \ln \left( \frac{e^2}{\sqrt{e}} \right)\).
Solve \(e^x = 2\).
If I invest in a bond that grows 2% per year, how long will it take for the bond to double? To solve this problem we needed to solve \((1.02)^t = 2\). The trick is to take the natural log of both sides.
The population of the United States is currently 330 million and is growing at 0.5% per year. If this trend continues, how long until the population reaches 400 million?
Today we talked about exponential growth. We started with these problems.
If you invest money at a 5% annual interest rate, how many times bigger will your investment be after \(t\) years?
After 40 years, how many times larger is an investment that grew by 10% every year compared with one that only grew 5% every year (assuming the two investments started with the same initial investment).
Surprisingly 5% growth is not just half of 10% growth. Over time, 10% growth will end up being much more than double 5% growth. Exponential growth can be surprising unintuitive. We watched this video in class: The Most IMPORTANT Video You’ll Ever See which explains a good way to understand exponential growth.
We finished today by looking at why the Rule of 70 from the video is true. It has to do with the linear approximation for the function \(\ln(1+x)\).
We covered several topics today.
\[\log_b(x) = \frac{\ln x}{\ln b}.\]
\[ \dfrac{d}{dx} \ln x = \dfrac{1}{x}.\]
We used this to solve the following problems in class:
Find the derivative of \(y = x \ln x\).
Find the intervals of increase and decrease for \(y = x \ln x\).
Use the chain rule to find \(\dfrac{d}{dx} \ln (x^2+4)\).
Find \(\displaystyle \frac{d}{dx} \ln \left( \frac{\sqrt{x-5}}{x^3}{(x+1)^4} \right)\)$. This problem would be extremely hard with the chain rule, but if you use the rules for logarithms to simplify the function first, it isn’t that bad.
Find \(\displaystyle \frac{d}{dx} \ln \left( \frac{\sqrt{x}}{x+1} \right)\).
The last thing we talked about today was derivatives of exponential functions with other bases.
\[\dfrac{d}{dx} a^x = a^x \ln a.\]
If the population of a town is growing at 5% per year and is currently 3000, use the derivative to find the rate of change of the population 10 years from now.
If the price of an item is currently $3 and is decreasing by 5% per year, find a formula for the price as a function of time (in years) and then find the derivative to estimate how much the price will be decreasing 5 years from now.