Calculus I Lesson 12: Exponential Functions

  1. Presentations
  2. Exponential Functions
  3. The number $e$
  4. Exercises
  5. Logarithmic Functions
  6. Other bases: Logarithmic Differentiation
  7. Homework / Upcoming

Presentations

Exponential Functions

Recall that an exponential function is a function of the form $f(x) = b^x$, where $b$ is a constant, positive real number. The most important fact about exponential functions that we need for this section is the rule for addition inside exponents: for any exponential function $b^x$, and any real numbers $x$ and $y$:

\[b^{x+y} = b^x b^y\]

This rule allows us to find a formula for the derivative of exponential functions. Suppose $f(x) = b^x$. To compute $f^\prime(x)$, we compute $\dfrac{b^{x+h}-b^x}{h}$ and then take the limit as $h \rightarrow 0$ of the expression we get. Since $b^{x+h} = b^x b^h$, we see:

\[\frac{b^{x+h}-b^x}{h} = \frac{b^xb^h - b^x}{h} = b^x \frac{b^h - 1}{h}\]

Notice that $b^x$ does not depend on $h$ at all, and so the limit as $h \rightarrow 0$ of the above expression will be $b^x \cdot \lim\limits_{h\rightarrow 0} \dfrac{b^h - 1}{h}$.

This expression on the right, $\dfrac{b^h - 1}{h}$, looks challenging to compute. For example, it’s not obvious what this would be if $b = 2$. Let’s compute $\dfrac{2^h - 1}{h}$ for small, positive values of $h$:

\[\begin{array}{c|c} h & \dfrac{2^h - 1}{h} \\ \hline .1 & .7177 \\ .01 & .6956 \\ .001 & .6934 \\ .0001 & .6932 \end{array}\]

Now let’s compute this same expression for small, negative values of $h$:

\[\begin{array}{c|c} h & \dfrac{2^h - 1}{h} \\ \hline -.1 & .6697 \\ -.01 & .6908 \\ -.001 & .6929 \\ -.0001 & .6931 \end{array}\]

It seems like this does have some limit, but it’s not clear what it is. In fact, this actually turns out to be the derivative of $f(x) = 2^x$ at $x = 0$: recall, to find the derivative at the point $x = 0$, we can use the formula ${\displaystyle \lim_{x \rightarrow 0} \frac{f(x) - f(0)}{x - 0} }$. In the case of $f(x) = 2^x$, we would get

\[\lim_{x \rightarrow 0} \frac{2^x - 1}{x}\]

which is exactly the same as ${\displaystyle \lim_{h \rightarrow 0} \frac{2^h - 1}{h} }$. That is, if $f(x) = 2^x$, then $f^\prime(x) = f(x) \cdot f^\prime(0)$! Recall last time we mentioned that the derivative of an exponential function, at any point, is proportional to the value of the function itself. This is exactly what I mean: $f^\prime(0)$ is some number (a constant, does not depend on $x$), and $f^\prime(x)$ is exactly that same constant multiplied by $f(x)$, no matter what $x$ is.

This same phenomenon happens for every exponential function: if $f(x)$ is an exponential function, then $f^\prime(x) = f(x) \cdot f^\prime(0)$. The derivative of an exponential function is proportional to the value of the function itself.

Now take a look at this video from the “Essence of Calculus” series on exponential functions:

The number $e$

With some work, or just by plugging in numbers or looking at graphs, you can verify, for yourself, that the limit ${\displaystyle \lim_{h \rightarrow 0} \frac{b^h - 1}{h}}$ exists for any positive real number $b$. These limits would be different for different values of $b$, but they all exist. For $b = 2$, for example, we saw that this limit was around 0.693. For $b = 10$, this limit would be about 2.303. That means that if $f(x) = 2^x$, then $f^\prime(x) \approx 0.693 \cdot 2^x$. If $g(x) = 10^x$, then $g^\prime(x) \approx 2.303 \cdot 10^x$. It stands to reason, then, that there is some base, somewhere between $b = 2$ and $b = 10$, such that that limit would be exactly 1. There is such a number: $e \approx 2.71828\ldots$ You’ve likely seen $e$ in the context of studying compound interest: there, the number $e$ was defined as a limit:

\[e = \lim_{n \rightarrow \infty} (1 + \frac{1}{n})^n\]

It turns out that there are many equivalent ways of defining the number $e$. What I mean by this is that you could start with the definition that $e$ is the number $b$ such that ${\displaystyle \lim_{h \rightarrow 0} \frac{b^h - 1}{h}} = 1$, and prove that it’s also equal to the limit as $n \rightarrow \infty$ of $(1 + \dfrac{1}{n})^n$, or you could start with the assumption that $e$ is equal to this limit, and prove that the derivative of $e^x$ at $x = 0$ is exactly $1$.

Either way, we end up with the result that if $f(x) = e^x$, then $f^\prime(x) = e^x$. That is, $e^x$ is its own derivative! This is another really remarkable fact about the number $e$, which we could take as a sort of defining feature.

Using all of our rules, then, we can find more complicated derivatives.

Examples:

  1. Let $y = e^{2x}$. Using the chain rule, let $u = 2x$. Then $\frac{dy}{dx} = \frac{dy}{du} \frac{du}{dx} = e^u \cdot 2 = 2 e^{2x}$.
  2. Let $y = e^{\sin(x)}$. Let $u = \sin(x)$. Then $\frac{dy}{dx} = e^{\sin(x)} \cdot \cos(x)$.

Exercises

Using all the rules you know, find the derivatives of the following functions:

  1. $f(x) = e^{x^2} \cos(x)$
  2. $g(x) = \dfrac{e^{2x}}{\sin(x)}$

Logarithmic Functions

A logarithm is the inverse of an exponential function. That is, if $y = b^x$, then $x = \log_b(y)$, and vice versa: if we are trying to graph $y = \log_b(x)$ for some base $b$, we really should think of this as looking for $(x, y)$ values where $b^y = x$.

That is: $\log_b(x)$ and $b^x$ are inverses. If a point $(x, y)$ is on the graph of one of these functions, then switching $x$ and $y$, we get a point on the other graph. For example, if $y = 2^x$, we know that $(3, 8)$ is on the graph of this function. That means $(8, 3)$ is on the graph of $y = \log_2(x)$. That is: $\log_2(8) = 3$.

The natural logarithm function is $f(x) = \log_e(x)$. It is used so often that it has a special name: $\ln(x)$ means $\log_e(x)$. Again, the same general rule applies: $y = \ln(x)$ means $e^y = x$.

To find the derivative of $y = \ln(x)$, it is easier to reason implicitly using the equation $e^y = x$. Take the derivative of both sides with respect to $x$, and we get $e^y y^\prime = 1$. Then solve for $y^\prime$ and get $y^\prime = \dfrac{1}{e^y}$. Remember, though, that we used the equation $e^y = x$, so that means we can substitute $x$ in for $e^y$: that is, $y^\prime = \dfrac{1}{x}$!

This is covered in the excellent 3Blue1Brown video on Implicit Differentiation. Take a look at that one here, as it may help refresh your memory on this topic.

Other bases: Logarithmic Differentiation

So far, we know a couple of things:

So for other bases $b \neq e$, how can we find that derivative at $0$? To try to tackle this problem, we will use a technique called logarithmic differentiation. Logarithmic differentiation roughly means that, starting with a function $y = f(x)$, we look at $\ln(y) = \ln(f(x))$, use the rules for logarithms to simplify $\ln(f(x))$, take the derivative of both sides (implicitly on the left side), and then solve for $y^\prime$.

I go over this technique in this video:

For example, let’s take a look at $f(x) = 10^x$. Let $y = 10^x$ and then look at the natural log of both sides:

\[\ln(y) = \ln(10^x)\]

Using the rules for logarithms, $\ln(10^x) = x \ln(10)$, or $\ln(10) \cdot x$. So we have $\ln(y) = \ln(10) \cdot x$. Now take the derivative of both sides. On the left, $(\ln(y))^\prime = (\dfrac{1}{y}) \cdot y^\prime = \dfrac{y^\prime}{y}$. As we do more of these logarithmic differentiation problems, you will see this term $\dfrac{y^\prime}{y}$ show up often.

Then on the right, since $\ln(10) \cdot x$ is a constant multiplied by $x$, the derivative is just $\ln(10)$. So:

\[\dfrac{y^prime}{y} = \ln(10)\]

Therefore $y^\prime = \ln(10) \cdot y = \ln(10) \cdot 10^x$, using the fact that $y = 10^x$ is what we started with.

Recall: if $f(x) = b^x$, then $f^\prime(x) = b^x \cdot f^\prime(0)$. That is, $f^\prime(x)$ is just $f(x)$ multiplied by some constant. What is that constant? It turns out, in the case of $b = 10$, that that constant was just $\ln(10)$! In general, for any base $b$, that constant will be $\ln(b)$! Can you figure out how to prove that using the technique of logarithmic differentiation?

Homework / Upcoming

  1. DeltaMath due tonight
  2. Written Homework 5 due on Thursday (10/16):
    • Section 3.5: #180, 192, 201
    • Section 3.6: #230, 232, 248, 254
    • Section 3.8: #302, 316, 325
  3. Topic statement due Monday (10/20)
  4. DeltaMath due Monday (10/20)
  5. Quiz next week?