Exponential derivatives

Section 2.12 Exponential derivatives

Objectives

Introduce the base \(e\) in the context of exponential derivatives.
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Rewrite any exponential function with a base of \(e\text{.}\)
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Establish a derivative formula for exponential functions.
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Compute derivatives of functions build from exponential functions and algebraic operations.
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Subsection The natural exponential function and its derivative

Our previous investigation into exponential functions yields some valuable qualitative information about their derivatives, as we now summarize. Let \(f(x)=b^x\text{.}\)

Exponential growth: \(b> 1\).
If \(b> 1\text{,}\) then \(f\) is an increasing function on \(\R\text{,}\) and hence \(f'(x)> 0\) for all \(x\in \R\text{.}\)
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Moreover, since \(f\) is concave up on \(\R\text{,}\) we have \(f''(x)> 0\) for all \(x\in \R\text{,}\) and thus \(f'\) is positive and gets more positive as \(x\) increases. This last quality can be also be seen as a consequence of the fact that exponential growth functions model percent growth: for each input \(x\text{,}\) the rate at which \(f\) is growing is proportional to its current value \(f(x)=b^x\text{.}\)
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Exponential decay: \(0 < b < 1\).
Similar reasoning shows that if \(0 < b < 1\text{,}\) then because \(f\) is a decreasing and concave up function on \(\R\text{,}\) we have \(f'(x)< 0\) and \(f''(x)> 0\) for all \(x\in \R\text{,}\) which means \(f'\) is negative but becomes less negative as \(x\) increases.
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Thus, qualitatively we have a good grip on \(f'(x)\) for an exponential function \(f(x)=b^x\text{.}\) But can we derive an actual derivative formula? Recall that exponential functions \(f(x)=b^x\text{,}\) where the input variable appears in the exponent, are not to be confused with power functions \(g(x)=x^r\text{,}\) where the input appears in the base. In particular, we cannot make use of the power rule \(g'(x)=rx^{r-1}\) to differentiate an exponential function \(f(x)=b^x\text{.}\) As such we must rely on the limit definition of the derivative. Let’s see how far we get with this. We have

\begin{align*} f'(x) \amp =\lim\limits_{h\to 0}\frac{f(x+h)-f(x)}{h}\\ \amp = \lim\limits_{h\to 0}\frac{b^{x+h}-b^x}{h}\\ \amp = \lim\limits_{h\to 0}\frac{b^x\cdot b^h - b^x}{h}\\ \amp = \lim\limits_{h\to 0} b^x\cdot \frac{b^h - 1}{h}\\ \amp = b^x \lim\limits_{h\to 0}\frac{b^h-b^0}{h} \text{.} \end{align*}

Interesting! Note that \(b^x\) is just \(f(x)\text{,}\) and the remaining limit expression in the last line is precisely \(f'(0)\text{,}\) assuming the limit actually exists. As such, assuming the derivative exists, we conclude that

\begin{align*} f'(x) \amp =f'(0)f(x)\text{.} \end{align*}

Let’s reflect a moment on what this discovery means. Setting \(f'(0)=c\) for some \(c\in \R\text{,}\) we see that \(f'(x)=cf(x)\text{.}\) In terms of rate of change language, this means that the rate of change \(f'\) is proportional to the current value \(f(x)\text{.}\) This is exactly as we expected from our percent growth interpretation of the exponential function!

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We are certainly getting somewhere, but we are still significantly short of an actual formula. As it turns out, the remaining limit

\begin{align*} \lim\limits_{h\to 0}\frac{b^h-1}{h} \amp = \lim\limits_{h\to 0}\frac{b^h-b^0}{h} \end{align*}

is not one we can analytically compute with tools currently at our disposal. We can, however, estimate its value for various choices of base \(b\text{.}\) We do so below for the choices \(b=2\) and \(b=3\text{.}\)

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Table 2.12.1. Approximating \(\lim\limits_{h\to 0}\frac{2^h-1}{h}\)

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\(h\)	\(0.5\)	\(0.1\)	\(0.01\)	\(0.001\)	\(0.0001\)
\(\frac{2^h-1}{h}\)	\(0.8284\)	\(0.7177\)	\(0.6956\)	\(0.6934\)	\(0.6932\)

Table 2.12.2. Approximating \(\lim\limits_{h\to 0}\frac{3^h-1}{h}\)

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\(h\)	\(0.5\)	\(0.1\)	\(0.01\)	\(0.001\)	\(0.0001\)
\(\frac{3^h-1}{h}\)	\(1.4641\)	\(1.1612\)	\(1.1047\)	\(1.0992\)	\(1.0987\)

From these tables, we see that

\begin{align} \lim_{h\to 0}\frac{2^h-1}{h} \amp \approx 0.693 \amp \tag{2.40}\\ \lim_{h\to 0}\frac{3^h-1}{h} \amp \approx 1.099 \amp \text{.}\tag{2.41} \end{align}

So far this is pretty mysterious. Recall that, assuming that the limits exist, we have \(\lim\limits_{h\to 0}\frac{b^h-1}{h}=f'(0)\text{,}\) where \(f(x)=b^x\text{.}\) Graphically speaking, this value represents the slope of the tangent line to the graph of \(f\) at the point \((0,1)\text{.}\) For \(b=2\) this slope is approximately \(0.693\text{,}\) a little less than one. It is no surprise that for \(b=3\) the slope at \(x=0\) is larger, approximately \(1.099\text{,}\) since we have seen already that as we increase \(b\text{,}\) the graph of the exponential function \(f(x)=b^x\) gets steeper at each point. This suggests that perhaps there is a base \(b\) between \(2\) and \(3\) for which the corresponding exponential function \(f(x)=b^x\) satisfies \(f'(0)=1\text{.}\) And indeed there is such a base \(b\text{:}\) this mysterious number is called \(e\) and has approximate value \(e\approx 2.71828\text{.}\)

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Graph of base-two exponential function with tangent at x equals 0 — (a) \(f(x)=2^x\)
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Graph of base-three exponential function with tangent at x equals 0 — (a) \(f(x)=2^x\)
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Definition 2.12.4. Natural exponential function and \(e\).

The number \(e\) is defined as the unique positive real number satisfying

\begin{align*} \lim\limits_{h\to 0}\frac{e^h-1}{h} \amp = 1\text{.} \end{align*}

The function \(f(x)=e^x\) is called the natural exponential function.

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The real number \(e\) is named in honor of the great Swiss mathematician Leonard Euler (1707-1783). Like \(\pi\text{,}\) \(e\) is an irrational number that cannot be written as a ratio of integers, and whose decimal expansion never repeats. Of its many remarkable properties, the one that is of most value to us is its connection to exponential functions \(f(x)=b^x\text{.}\) Namely, since we showed earlier that

\begin{align*} f'(x) \amp = f'(0)f(x) \amp f'(0)\amp=\lim\limits_{h\to 0}\frac{b^h-1}{h} \text{,} \end{align*}

and since \(e\) is the unique choice of \(b\) for which

\begin{align*} f'(0) \amp =\lim\limits_{h\to 0}\frac{b^h-1}{h}=1\text{,} \end{align*}

we conclude that the natural exponential function \(f(x)=e^x\) satisfies

\begin{align*} f'(x) \amp =f(x)\text{.} \end{align*}

In other words, \(f\) is its own derivative!

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Theorem 2.12.5. Derivative of natural exponential function.

Let \(f(x)=e^x\text{.}\) We have \(f'=f\text{.}\) Equivalently,

\begin{align*} \frac{d}{dx}(e^x) \amp = e^x \end{align*}

for all \(x\in \R\text{.}\)

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Example 2.12.6. Derivatives with natural exponential.

Compute the derivative of the given function.

\(\displaystyle f(x)=e^{2x}\)
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\(\displaystyle g(s)=\frac{1}{e^s+e^{-s}}\)
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\(\displaystyle h(t)=\sqrt{e^{\sqrt{t}}}\)
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Solution.

For the first computation, we will make explicit our use of the chain rule. We have

\begin{align*} f'(x) \amp = \frac{d}{du}(e^u)\, \frac{d}{dx} u \amp (u=2x)\\ \amp = e^u \cdot 2\\ \amp = 2e^{2x}\text{.} \end{align*}

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We have

\begin{align*} \frac{d}{dx}(1/(e^x+e^{-x})) \amp = \frac{d}{dx}(e^x+e^{-x})^{-1} \\ \amp = -(e^x+e^{-x})^{-2}(e^x+e^{-x})'\\ \amp = -\frac{e^x-e^{-x}}{(e^x+e^{-x})^2}\text{.} \end{align*}

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We have

\begin{align*} h'(t) \amp =\frac{d}{dt}((e^{\sqrt{t}})^{1/2})\\ \amp = \frac{1}{2}(e^{\sqrt{t}})^{-1/2}(e^{\sqrt{t}})'\\ \amp = \frac{e^{\sqrt{t}}(\sqrt{t})'}{2\sqrt{e^{\sqrt{t}}}} \\ \amp = \frac{\tfrac{1}{2\sqrt{t}}e^{\sqrt{t}}}{\sqrt{e^{\sqrt{t}}}}\\ \amp =\frac{e^{\sqrt{t}}}{4\sqrt{t}\sqrt{e^{\sqrt{t}}}}\text{.} \end{align*}

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Interestingly, the derivative formula for \(f(x)=e^x\) is so simple, that sometimes the need to use the chain rule is obscured. Accordingly, we add a chain rule-enhanced version of the exponential derivative formula.

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Corollary 2.12.7. Exponential derivative with chain rule.

Let \(u(x)\) be a differentiable function on the interval \(I\text{.}\) We have

\begin{align} \frac{d}{dx}(e^{u(x)})\amp = u'(x)\, e^{u(x)} \tag{2.42} \end{align}

for all \(x\in I\text{.}\)

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Subsection General derivative formula for \(\)

It is not difficult to generalize our base-\(e\) exponential derivative formula to arbitrary bases \(b\text{.}\) This is done by first observing that for any \(b> 0\text{,}\) we have \(b=e^{k}\text{,}\) where \(k=\log_e b\text{,}\) and thus

\begin{align*} b^x \amp = (e^k)^x=e^{kx} \text{.} \end{align*}

Thus, using the chain rule, we have

\begin{align*} \frac{d}{dx}(b^x) \amp = \frac{d}{dx}(e^{kx}) \\ \amp = \frac{d}{dx}e^{kx} \amp (u=kx)\\ \amp = e^{kx} (kx)' \\ \amp = e^{kx} \cdot k \amp \\ \amp = \log_e b\, e^{kx} \\ \amp = \log_e b\, b^x\text{.} \end{align*}

Before making this result official, we give a special name to the base-\(e\) logarithm \(\log_e\text{.}\)

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Definition 2.12.8. Natural logarithm.

The base-\(e\) logarithmic function \(\log_e\) is called the natural logarithm, and is denoted \(\ln\text{.}\)

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Remark 2.12.9. Natural logarithm.

Because the notation for the natural logarithm is significantly different from the standard \(\log_b\) notation, we will do well to make note of what the various logarithmic identities mean in terms of \(\ln=\log_e\text{.}\) We collect them all below.

\begin{align*} \ln r=s \amp\iff e^s=r \\ \ln(e^s) \amp = s\\ e^{\ln r} \amp = r \end{align*}

for all \(r\in (0,\infty)\) and \(s\in \R\text{.}\) Additionally, we have

\begin{align*} \ln 1 \amp =0\\ \ln e \amp =1\text{.} \end{align*}

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Theorem 2.12.10. Exponential derivatives.

Let \(b\) be a positive real number, and let \(f(x)=b^x\text{.}\)

We have \(b^x=e^{kx}\) for all \(x\in \R\text{,}\) where \(k=\ln b\text{.}\)
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We have \(f'(x)=\ln b\, b^x\text{.}\) Equivalently,

\begin{align} \frac{d}{dx}(b^x) \amp = \ln b\, b^x\tag{2.43} \end{align}

for all \(x\in \R\text{.}\)

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More generally, if \(u(x)\) is a differentiable function on the interval \(I\text{,}\) then

\begin{align} \frac{d}{dx} (b^{u(x)}) \amp = u'(x)\, \ln b\, b^{u(x)} \tag{2.44} \end{align}

for all \(x\in I\text{.}\)

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Example 2.12.11. General exponential derivatives.

Compute the derivative of the given function.

\(\displaystyle h(x)=3^{x^2+x+1}\)
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\(\displaystyle s(t)=\frac{5^{3x}}{x+1}\)
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Solution.

We compute

\begin{align*} h'(x) \amp =(x^2+x+1)'3^{x^2+x+1}\\ \amp =(2x+1)3^{x^2+x+1}\text{.} \end{align*}

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We compute

\begin{align*} s'(t) \amp =\frac{(5^{3x})'(x+1)-5^{3x}(x+1)'}{(x+1)^2}\\ \amp =\frac{\ln (5)\, 5^{3x}(3x)'(x+1)-5^{3x}}{(x+1)^2}\\ \amp =\frac{3\ln (5)\, 5^{3x}(x+1)-5^{3x}}{(x+1)^2}\\ \amp =\frac{(3\ln (5)\, (x+1)-1)5^{3x}}{(x+1)^2}\text{.} \end{align*}

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Example 2.12.12. Tangent line to exponential function.

Let \(f(x)=2^x\text{.}\) Find the input \(a\) where the tangent line to the point \(P=(a,f(a))\) on the graph of \(f\) has slope \(3\text{.}\)

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Solution.

The slope of the tangent line to the graph of \(f\) at a point \((a,f(a))\) is given by \(f'(a)=\ln(2) 2^a\text{.}\) To find \(a\) where this slope is equal to \(3\text{,}\) we thus solve \(\ln(2)2^{a}=3\text{:}\)

\begin{align*} \ln(2) 2^a \amp = 3\\ 2^a \amp =\frac{3}{\ln(2)}\\ \ln(2^a) \amp = \ln\left(\frac{3}{\ln(2)}\right) \\ a \ln(2) \amp = \ln(3)-\ln(\ln(2))\\ a \amp = \frac{\ln(3)-\ln(\ln(2))}{\ln(2)} \text{.} \end{align*}

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Exercises Exercises

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Certain pieces of antique furniture increased very rapidly in price in the 1970s and 1980s. For example, the value of a particular rocking chair is well approximated by

\begin{equation*} V = 115 (1.35)^t, \end{equation*}

where \(V\) is in dollars and \(t\) is the number of years since 1975. Find the rate, in dollars per year, at which the price is increasing.

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rate = dollars/yr

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Answer.

\(115\ln\mathopen{}\left(1.35\right)\cdot 1.35^{t}\)

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Solution.

\begin{equation*} \hbox{rate} = 115 \ln(1.35) 1.35^t \quad\hbox{dollars/year}. \end{equation*}

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