Section 8.8 The Derivative of Exponential Functions
¶
Subsection 8.8.1 Elementary Exponential Functions
We learned that the elementary exponential functions are of the form
\begin{equation*}
\exp_b(x) = b^x
\end{equation*}
for positive real numbers \(b\text{.}\) We now explore the derivative rule associated with these functions. It is important to note that the exponential functions are fundamentally different from the power functions and therefore will have a different rule of differentiation.
We use the definition of the derivative. The key property necessary for the exponential function is a consequence of the properties of exponents,
\begin{equation*}
\exp_b(x+y) = b^{x+y} = b^x \cdot b^y = \exp_b(x) \cdot \exp_b(y).
\end{equation*}
Using these properties
\begin{align*}
\exp_b'(x) &= \lim_{h \to 0} \frac{\exp_b(x+h)  \exp_b(x)}{h} \\
&= \lim_{h \to 0} \frac{\exp_b(x) \cdot \exp_b(h)  \exp_b(x)}{h}
= \lim_{h \to 0} \exp_b(x) \cdot \frac{\exp_b(h)1}{h} \\
&= \exp_b(x) \cdot \lim_{h \to 0} \frac{b^{h}1}{h}.
\end{align*}
This means that the derivative of \(b^x\) is just \(b^x\) times some number that depends on \(b\) that is calculated as the limit,
\begin{equation*}
L(b) = \lim_{h \to 0}\frac{b^h1}{h}.
\end{equation*}
The following table illustrates the value of this limit for a variety of bases.
\(h\) 
\(\displaystyle \frac{2^h1}{h}\) 
\(\displaystyle \frac{3^h1}{h}\) 
\(\displaystyle \frac{5^h1}{h}\) 
\(\displaystyle \frac{10^h1}{h}\) 
\(h=0.1\) 
0.71773 
1.16123 
1.74619 
2.58925 
\(h=0.01\) 
0.69555 
1.10467 
1.62246 
2.32930 
\(h=0.001\) 
0.69339 
1.09922 
1.61073 
2.30524 
\(h=0.0001\) 
0.69317 
1.09867 
1.60957 
2.30285 
\(h \to 0\) 
0.69315 
1.09861 
1.60944 
2.30259 
Every positive real number \(b\) has such a limit, \(L(b)\text{.}\) This limit corresponds to the slope of the elementary exponential function \(y=b^x\) at the point \(x=0\text{,}\)
\begin{equation*}
L(b) = \exp_b'(0) = \lim_{h \to 0} \frac{b^{0+h}  b^{0}}{h} = \lim_{h \to 0} \frac{b^h1}{h}.
\end{equation*}
What this calculation shows is that
\begin{equation*}
\frac{d}{dx}[b^x] = L(b) \cdot b^x.
\end{equation*}
For the four bases used above, this corresponds to the following derivatives:
\begin{align*}
\frac{d}{dx}[2^x] &= L(2) \cdot 2^x \approx 0.69315 \cdot 2^x \\
\frac{d}{dx}[3^x] &= L(3) \cdot 3^x \approx 1.09861 \cdot 3^x \\
\frac{d}{dx}[5^x] &= L(5) \cdot 5^x \approx 1.60944 \cdot 5^x \\
\frac{d}{dx}[10^x] &= L(10) \cdot 10^x \approx 2.30259 \cdot 10^x
\end{align*}
There is a particular base \(b\) between 2 and 3 such that \(L(b)=1\text{,}\) which has a value of \(b \approx 2.71828183\text{,}\) such that for this special base, the derivative and the function are the same. This base is called the natural base and is given the special symbol \(e\text{.}\) Consequently,
\begin{equation*}
\frac{d}{dx}[e^x] = e^x.
\end{equation*}
This is why the function \(e^x\) is so significant—it is the unique elementary exponential function that is equal to its own derivative. Every other exponential function is proportional to its derivative, with the proportionality constant given by the value \(L(b)\text{,}\) which we will determine below as a consequence of the chain rule.
Subsection 8.8.2 The Chain Rule with Exponentials
Because the exponential function \(\exp(x) = e^x\) is defined with the natural base, \(\exp'(x) = e^x\) is the same as the original function, \(\exp'=\exp\text{.}\) Combining this with the chain rule, we get a generalized derivative of compositions with exponentials,
\begin{equation*}
\frac{d}{dx}[e^{u(x)}] = e^{u(x)} \cdot u'(x) = e^{u} \cdot \frac{du}{dx}.
\end{equation*}
Example 8.8.1
Find the derivatives of the following functions.
 \(f(x) = e^{5x}\)
 \(g(x) = e^{x^3}\)
 \(h(x) = e^{x^24x}\)
Solution
In each case, we will identify the formula \(u(x)\) and then apply the chain rule.

For \(f(x) = e^{5x}\text{,}\) we have \(u(x)=5x\) so that \(f(x) = \exp(u(x))\text{.}\) We will use \(u'(x)=5\text{.}\) The chain rule gives:
\begin{align*}
f'(x) &= \frac{d}{dx}[e^{5x}] \\
& \underset{(u=5x)}{=} \frac{d}{dx}[e^u] \underset{(u=5x)}{=} e^u \frac{du}{dx} \\
&= e^{5x} \cdot 5 = 5e^{5x}
\end{align*}
So \(f'(x) = 5e^{5x}\text{.}\)

The function \(g(x) = e^{x^3}\) involves a composition with \(u(x)=x^3\) such that \(u'(x)=3x^2\text{.}\)
\begin{align*}
g'(x) &= \frac{d}{dx}[e^{x^3}] \\
&\underset{(u=x^3)}{=} \frac{d}{dx}[e^u] \underset{(u=x^3)}{=} e^u \frac{du}{dx} \\
&= e^{x^3} \cdot (3x^2) = 3x^2 e^{x^3}
\end{align*}
Thus \(g'(x) = 3x^2e^{x^3}\text{.}\)

The function \(h(x) = e^{x^24x}\) involves a composition with \(u(x)=x^24x\) such that \(u'(x)=2x4\text{.}\)
\begin{align*}
h'(x) &= \frac{d}{dx}[e^{x^24x}] \\
&\underset{(u=x^24x)}{=} \frac{d}{dx}[e^u] \underset{(u=x^24x)}{=} e^u \frac{du}{dx} \\
&= e^{x^24x} \cdot (2x4) = (2x4) e^{x^24x}
\end{align*}
Thus \(h'(x) = (2x4)e^{x^24x}\text{.}\)
Now that we have the exponential function in our toolbox of elementary functions with known derivatives, we can expand the class of functions we know how to differentiate.
Subsection 8.8.3 Power Function or Exponential Function
One of the challenges for a calculus novice is identifying which rule applies. It is essential that you can distinguish between an exponential function and a power function.
Recall that a power function has a constant power while an exponential function has a constant as the base. Furthermore, don't be fooled by numbers that look like other symbols. For example, \(x^e\) is a power function since the power is the number \(e\) (the natural exponential base):
\begin{equation*}
\frac{d}{dx}[x^e] = e x^{e1}.
\end{equation*}
Similarly, \(x^{\sqrt{2}}\) is a power function because the power \(\sqrt{2}\) is a number (even though it is represented as a formula, it does not have any variables):
\begin{equation*}
\frac{d}{dx}[x^{\sqrt{2}}] = \sqrt{2} x^{\sqrt{2}1}.
\end{equation*}
Furthermore, you must look at a formula and determine which operation determines the differentiation rule that is required (constant multiple, sum, product, quotient or chain). This is always based on the last operation to be applied under the rules of order of operations. When the number of steps is small, you should be able to write the derivative down directly. When the number of steps is large, you might need to use the differentiation operator to allow yourself the chance to work part of the way and indicate that there are still steps remaining.
Example 8.8.2
Find the derivative \(\displaystyle \frac{d}{dx}[ (x^2+4)^5 e^{5x^3} ]\text{.}\)
Solution
Start by identifying the final operation. In this problem, the function \(f(x) = (x^2+4)^5 e^{5x^3}\) is a product of \((x^2+4)^5\) and \(e^{5x^3}\text{.}\) So we begin by using the product rule. If you want to emphasize this without having to write down the derivatives of the factors, then we use the differentiation operator:
\begin{equation*}
f'(x) = \frac{d}{dx}[(x^2+4)^5] \cdot e^{5x^3} + (x^2+4)^5 \cdot \frac{d}{dx}[e^{5x^3}].
\end{equation*}
Notice that the differentiation operator is pointing out where we still need to find derivatives in order to complete the problem.
The first term \((x^2+4)^5\) should be recognized as a composition with a power function \(u^5\) where \(u=x^2+4\text{.}\) We will use the chain rule:
\begin{equation*}
\frac{d}{dx}[(x^2+4)^5] \underset{(u=x^2+4)}{=} \frac{d}{du}[u^5] \cdot \frac{du}{dx}
= 5(x^2+4)^4 \cdot (2x).
\end{equation*}
The second term \(e^{5x^3}\) should be recognized as a composition with the exponential function \(e^u\) where \(u=5x^3\text{.}\) Again, the chain rule guides us:
\begin{equation*}
\frac{d}{dx}[e^{5x^3}] \underset{(u=5x^3)}{=} \frac{d}{du}[e^u] \cdot \frac{du}{dx}
= e^{5x^3} \cdot (15x^2).
\end{equation*}
The previous paragraph represents work that you either think through mentally or write out as scratch work. Putting the pieces together gives us the overall answer.
\begin{align*}
f'(x) = \frac{d}{dx}[(x^2+4)^5 e^{5x^3}]
& = \frac{d}{dx}[(x^2+4)^5] \cdot e^{5x^3} + (x^2+4)^5 \cdot \frac{d}{dx}[e^{5x^3}]\\
& = 5(x^2+4)^4 (2x) \cdot e^{5x^3} + (x^2+4)^5 \cdot e^{5x^3} \cdot (15x^2) \\
& = 10x (x^2+4)^4 e^{5x^3} + 15x^2(x^2+4)^5 e^{5x^3}
\end{align*}
As we start to use our derivatives in applications, we will often need to factor our formulas. To illustrate this principle, we identify all of the common factors of the terms.
\begin{align*}
f'(x) &= 5x(x^2+4)^4e^{5x^3} \cdot(2 + 3x(x^2+4)) \\
& = 5x(x^2+4)^4 e^{5x^3} (2+12x+3x^2)
\end{align*}
Subsection 8.8.4 Other Bases
Every function involving an exponential can be rewritten using the natural exponential function \(\exp(x)=e^x\text{.}\) Recall that the exponential and the logarithm are inverse functions so that for every number \(u \gt 0\) we have
\begin{equation*}
\exp(\ln(u)) = u.
\end{equation*}
This means that any power \(u^w\) where \(u \gt 0\) can be rewritten as
\begin{equation*}
u^w = \exp(\ln(u^w)) = e^{\ln(u^w)} = e^{w \cdot \ln(u) \cdot w} = \exp(\ln(u) \cdot w).
\end{equation*}
Example 8.8.3
Rewrite each of the following in terms of the natural exponential function.
 \(f(x) = 2^x\)
 \(g(x) = 5^x\)
 \(p(x) = x^{3/4}\)
 \(r(x) = x^x\)
Solution
 \(f(x) = 2^x = \exp(\ln(2^x)) = e^{\ln(2^x)} = e^{x \ln(2)}\)
 \(g(x) = 5^x = \exp(\ln(5^x)) = e^{\ln(5^x)} = e^{x \ln(5)}\)
 \(p(x) = x^{3/4} = \exp(\ln(x^{3/4})) = e^{\ln(x^{3/4})} = e^{\frac{3}{4}\ln(x)}\)
 \(r(x) = x^x = \exp(\ln(x^x)) = e^{\ln(x^x)} = e^{x \ln(x)}\)
Because every exponential can be rewritten in terms of the natural exponential, we have a special result for all other exponential functions,
\begin{equation*}
b^x = e^{x \cdot \ln(b)}.
\end{equation*}
Using the chain rule with \(u=x \cdot \ln(b)\) and \(\frac{du}{dx}=\ln(b)\text{,}\) we discover
\begin{equation*}
\frac{d}{dx}[b^x] = \frac{d}{dx}[e^{x \ln(b)}] = e^{x \ln(b)} \cdot \ln(b) = b^x \cdot \ln(b).
\end{equation*}
In other words, the scaling factor for the derivative of an exponential is the natural logarithm of the base:
\begin{equation*}
L(b) = \lim_{h \to 0} \frac{b^h1}{h} = \ln(b).
\end{equation*}