Derivatives of Exponentials & Logarithmics

We now turn to differentiating exponential and logarithmic functions. Knowing how to compute the derivatives for these functions is especially useful in applied mathematics fields. For example, we frequently see exponential functions with future value calculations in finance, and logarithmic functions for discounting to present value. We also see exponential functions in population growth models. In all of these situations, knowing how to compute various rates of change is invaluabe. And as we've seen, that knowledge is provided through the derivative.

We begin by recalling some of the basic rules of exponents. First, where ${a, b > 0,}$ the following rules hold:

Exponent-product Rule. The exponential of the sum is the exponential of the product:

$$a^{x_1} \cdot a^{x_2} = a^{x_1 + x_2}$$

where ${a \in \uint^{+},}$ ${x_1, x_2 \in \uint.}$

Exponent-quotient Rule. If the bases are the same, keep the base and subtract the exponent in the denominator from the exponent in the numerator.

$$\dfrac{a^{x_1}}{a^{x_2}} = a^{x_1 - x_2} = \dfrac{1}{a^{x_2 - x_1}}$$

where ${a \in \uint^{+},}$ ${x_1, x_2 \in \uint.}$

Raising a Product to a Power. To raise a product to the ${n^{\text{\scriptsize{th}}}}$ power, raise each factor to the ${n^{\text{\scriptsize{th}}}}$ power.

$$(ab)^{x} = a^xb^x$$

where ${a, b \in \uint^{+},}$ ${x \in \uint.}$

Exponent-power Rule. To raise a power to a power, multiply the exponents.

$$(a^{x_1})^{x_2} = a^{x_1 \cdot x_2}$$

where ${a, x_1, x_2 \in \uint^{+}}$

Raising a Quotient to a Power. To raise a quotient to the ${n^{\text{\scriptsize{th}}}}$ power, raise both the numerator and the denominator to the ${n^{\text{\scriptsize{th}}}}$ power.

$$\left(\dfrac{a}{b}\right)^{x} = \dfrac{a^x}{b^x}$$

where ${a, x \in \uint^{+}}$

Exponent of 1. Any number raised to the first power returns the original number.

$$a^1 = a$$

where ${a \in \reals.}$

Exponent of 0. Any number raised to the zero power returns 1.

$$a^0 = 1$$

where ${\{a \in \reals : a \neq 0\};}$ ${0^0 = \text{undefined}.}$

Negative Integer Exponent Rule. Negative exponents are equivalent to an exponent in the denominator.

$$a^{-x} = \dfrac{1}{a^x}$$

where ${a, x \in \uint^{+}.}$

Fractional Exponent Rule. Fractional exponents are equivalent to the ${\text{denom}^{\text{\scriptsize{th}}}}$ root of the ${\text{numer}^{\text{\scriptsize{th}}}}$ power.

$$a^{\frac{x_1}{x_2}} = \sqrt[x_2]{a^{x_1}}$$

where ${a, x_1, x_2 \in \uint^{+}}$

Keeping these rules in mind, let's see some functions. How about ${f(x) = 2^x:}$

To find the derivative of this function, let's think about how Newton's quotient applies. It would look something like this:

What should we do next? Look at the term ${a^{x + \Delta x} -a^x.}$ We can rewrite this term as ${(a^{x})(a^{\Delta x}).}$ Incorporation this rewrite:

Now we see a common factor we can factor out, ${a^x}$:

At this point, we have something to work with conceptually. Let's recall how the limit ${\lim\limits_{\Delta x \to 0}}$ applies here. With this limit, we know that: (1) ${a}$ is fixed, and (2) ${x}$ is fixed. Because of these two facts, it follows that ${a^x}$ is constant. And because ${a^x}$ is constant, we can factor it out of the limit — it has no bearing on our analysis of the limit:

Thus, our analysis is really focused on what the limit term is. Because of how important the term is, we'll write this expression as:

where:

So, what exactly is ${L(a)?}$ To begin, we state an observation. If we plug in the value ${x = 0,}$ we get:

This observation tells us that at ${x = 0,}$ the slope of ${a^x}$ is this limit we're calling ${L(a).}$ More specifically, since ${a = 2,}$ we have a slope of ${L(2)}$ (at ${x = 0.}$) This tells us that if we instead had the function ${f(x) = 3^x,}$ at ${x = 0}$ we'd get ${L(3).}$ For ${f(x) = 13^x,}$ at ${x = 0}$ we'd get ${L(13),}$ and so on. This leads to a broader observation: If we know the slope at ${x = 0}$ for some function ${f(x) = a^x,}$ we can determine the slope everywhere else. We've seen this phenomenon before with sines and cosines: We knew the slope for sine and cosine at ${x = 0,}$ and from the trigonometric formulas, we determined the slope everywhere else.

The difference with exponentials: we're stuck right off the bat. With sine and cosine, we had the benefit of using radians, which allowed us to interpret the limits geometrically. Here, we're dealing directly with a number. There are no clever shortcuts to rely on. So we're going to need a different tactic. In this case, we're going to beg the question.¹

We start by stating a proposition:

There exists a base ${e,}$ where ${e}$ is a unique real number, such that ${M(e) = 1.}$

From the proposition above, we have the following:

This is a very important deduction (so much so that we will encapsulate it in a formal definition later). Furthermore, from our previous analysis, we know that:

As we're all too familiar with in mathematics, we should never take things for granted. Why does ${e}$ exist? How do we know it exists? So we know that ${f(x) = 2^x}$ exists. That is, after all, what we started with. Furthermore, we know that it has the property ${f'(0) = L(2).}$ Now let's say we stretch the function ${f(x) = 2^x.}$ We can stretch this function by multiplying its inputs with a factor ${k}$:

By the exponent power rule, the expression ${2^{kx}}$ is the same as ${(2^k)^x.}$ And since ${2}$ and ${k}$ are both real numbers, ${2^k}$ is a real number, which we can denote as ${b = 2^x.}$ This means we can rewrite the function above as:

Now, what happens when we shrink a function's graph? Well, we essentially shrink the ${x}$-axis, which in turn leads to the graph's slope tilting up. Thus, as ${k \to \texttt{+}\infty}$ (${k}$ gets bigger and bigger), the slope of ${f(kx)}$ gets steeper and steeper. This is numerically confirmed by considering the derivative of ${f(kx):}$

Evaluating at ${0:}$

At this point, we know that ${e}$ exists. When ${k = \dfrac{1}{L(2)},}$ we have:

Substitute ${e}$ into ${b,}$ the equation holds true: ${e^0 = 1.}$ And since the equation holds true, we know that ${e}$ exists.

Knowing that ${e}$ exists, we can continue with our derivative. We must now put all of these observations together. To do so, we rely on a particularly useful operation in mathematics: The natural logarithm. The natural log is defined as such:

The natural logarithm has several useful properties that are worth revisiting:

• ${\ln (x_1 \cdot x_2) = \ln x_1 + \ln x_2}$
• ${\ln 1 = 0}$
• ${\ln e = 1}$

And just to reminder ourselves, the graphs of ${f(x) = e^x}$ and ${g(x) = \ln x:}$

With the graphs above, we can see that ${g(x) = \ln x}$ is indeed the inverse of ${f(x) = e^x.}$ We're switching the roles of ${x}$ and ${y.}$ Notice further that ${g(x) = \ln x}$ is only defined when ${x > 0.}$ This corresponds to the fact that ${f(x) = e^x}$ is always positive.

Since ${g(x) = \ln x}$ is the inverse function of ${f(x) = \ln x,}$ we can use implicit differentiation to find the derivative of the natural logarithm. To do so, we'll write ${w = \ln x.}$ We want to find the derivative of ${w.}$ But we don't know how, so what we'll do is exponentiate:

Given that ${w = e^{\ln x} = x,}$ we can differentiate:

Applying implicit differentiation, we have:

Substitute ${w}$ with ${\ln x}$ (since we defined ${w = \ln x}$):

Hence, we have the following:

We now have two formulas to work with:

• ${(e^x)' = e^x}$
• ${(\ln x)' = \dfrac{1}{x}}$

We now return to our pesky function ${y = a^x.}$ If we can find the derivative for this function, we can find the derivative for any exponential function. Knowing the two formulas above, we can compute the derivative by first rewriting ${a^x}$ in terms of base ${e.}$ I.e., converting ${a^x}$ into base ${e.}$ How do we convert ${a^x}$ to ${e^x?}$ We want to write ${a^x}$ as ${e^{?}.}$ In other words, ${e}$ to some power:

Now we can carry out the differentiation:

The first step is often a point of great confusion, so it's critical that we are comfortable with it as possible. First, we emphasize that ${e}$ and ${\ln a}$ are constants. They're fixed. No movement. The only thing that's moving is ${x.}$ If we were asked to compute the derivative of ${e^{2x,}}$ the derivative would look like:

by the chain rule. The same idea applies to the derivative of ${e^{x \ln a}.}$ Hence:

We know that ${e^{x \ln a} = a^x,}$ so we have the derivative of ${a^x:}$

Now we can go back to our very first problem: ${y = 2^x.}$ Recall we asked, what exactly is ${L(a)?}$ Well now have it:

Accordingly, we have:

Similarly:

This is one of the reasons for why the natural logarithm takes as part of its name the adjective “natural.” No matter what base we choose, the respective exponential function's derivative will contain a natural logarithm. It's a logarithm that appears without any reference.