Opening the book…
- Note to the Instructor (click to expand)
- A sloppy and superficial treatment of this material would come from a purely mechanical perspective—your students deserve better. Take the time to define \(e\) and derive the theory of continuity and derivatives of the base-\(e\) exponential function.
- (click to expand)
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- Arithmetic
- Simplifying Expressions
The following is a list of learning objectives for this section.
- Learning Objectives (click to expand)
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- Find the derivative of exponential functions, both natural-based and non-natural-based.
- Find the derivative of logarithmic functions, both natural-based and non-natural-based.
- Apply the formulas for derivatives of the hyperbolic functions.
- Explore applications for the derivatives of exponential and hyperbolic functions in the sciences and engineering.
So far, we have learned how to differentiate various functions, including polynomial, radical, rational, and trigonometric functions. In this section, we explore derivatives of exponential and hyperbolic functions.
Discovering \(e\)
In Calculus (and often in Precalculus or College Algebra), a common "beginning" limit to investigate numerically is\[\displaystyle \lim_{k \to \infty}{\left(1 + \frac{1}{k}\right)^k}. \label{etooo}\]Table \(\PageIndex{1}\) shows values of this limit for large values of \(k\).
| \(k\) | \(\left(1 + \frac{1}{k}\right)^k\) |
|---|---|
| \(10\) | 2.5937424601... |
| \(100\) | 2.70481382942153... |
| \(1000\) | 2.71692393223552... |
| \(10000\) | 2.71814592682436... |
| \(100000\) | 2.71826823719753... |
| \(1000000\) | 2.71828046915643... |
| \(10000000\) | 2.71828169398037... |
| \(100000000\) | 2.71828178639580... |
| \(1000000000\) | 2.71828203081451... |
It appears this limit is approaching some value around \(2.71828\ldots\). This exceptional value is a non-terminating, non-repeating number. That is, it is an irrational number—just like \(\pi\). If we could assign importance to the value of this limit, it would be tied with the importance of the numbers \(0\), \(1\), \(\pi\), and \(i = \sqrt{-1}\).
This limit comes up so often in the sciences that we give it a unique name. Rather than calling it "the limit that approaches \(2.718281828459045\ldots\)," we call it \(e\). As was mentioned, it appears that this limit approaches a constant; however, the proof of this fact is reserved for later in the course. For now, it's important to have the limit, \ref{etooo}, in hand for use.
An important alternative form of \ref{etooo} is found by letting \(n = \frac{1}{k}\). As \(k \to \infty\), \(n \to 0\) and we get the form\[\displaystyle \lim_{n \to 0}{\left(1 + n\right)^{1/n}}. \label{eto0}\]Let's formalize this definition before we move on.
The limit\[\displaystyle \lim_{k \to \infty}{\left(1 + \frac{1}{k}\right)^k} = \displaystyle \lim_{n \to 0}{\left(1 + n\right)^{1/n}}\nonumber\]is defined to be the irrational number \(e\).
We will be using this unique number shortly; however, we need to discuss some Calculus before we do.
Derivatives Exponential Functions
Just as when we found the derivatives of other functions, we can find the derivatives of exponential functions using formulas. Recall that all non-transformed exponential functions have the basic form \(B(x) = b^x\), where \(b \gt 0\) and \(b \neq 1\). Our initial goal is to prove that exponential functions are continuous everywhere. Once we have this, the proof of the derivative of an exponential function becomes trivial.
Proving that Exponential Functions are Continuous Everywhere
Believe it or not, we have had a long history with exponential functions. We began in Arithmetic with the definition of \(b^n\), where \(n \in \mathbb{N}\), as the product of \(b\) multiplied by itself \(n\) times. Later, in Algebra, we defined \(b^0=1\), \(b^{-n}=\frac{1}{b^n}\) for \(n \in \mathbb{N}\), and \(b^{s/t}=(\sqrt[t]{b})^s\) for positive integers \(s\) and \(t\). These define exponential functions over the rational numbers but leave open the question of the value of \(b^r\) where \(r\) is an irrational number. That is, we know that \(4^{3.14} = 4^{314/100}\) is a shortcut notation for \(\sqrt[100]{4^{314}}\), but what if that exponent is not a terminating decimal approximation of \(\pi\), but instead is \(\pi\) itself? What does \(4^{\pi}\) mean?
By assuming the continuity of \(B(x)=b^x\), \(b>0\), we may interpret \(b^r\) as \(\displaystyle \lim_{x \to r}b^x\) where the values of \(x\) as we take the limit are rational. For example, we may view \(4^{\pi}\) as the number satisfying\[\begin{array}{rcccl} 4^3 & < & 4^ \pi & < & 4^4 \\ 4^{3.1} & < & 4^ \pi & < & 4^{3.2} \\ 4^{3.14} & < & 4^ \pi & < & 4^{3.15} \\ 4^{3.141} & < & 4^{\pi} & < & 4^{3.142} \\ 4^{3.1415} & < & 4^{\pi} & < & 4^{3.1416} \\ & & \vdots & & \\ \end{array}\nonumber\]As we see in Table \(\PageIndex{2}\), \(4^{\pi} \approx 77.88\).
| \(x\) | \(4^x\) | \(x\) | \(4^x\) |
|---|---|---|---|
| \(4^3\) | \(64\) | \(4^{3.141593}\) | \(77.8802710486\) |
| \(4^{3.1}\) | \(73.5166947198\) | \(4^{3.1416}\) | \(77.8810268071\) |
| \(4^{3.14}\) | \(77.7084726013\) | \(4^{3.142}\) | \(77.9242251944\) |
| \(4^{3.141}\) | \(77.8162741237\) | \(4^{3.15}\) | \(78.7932424541\) |
| \(4^{3.1415}\) | \(77.8702309526\) | \(4^{3.2}\) | \(84.4485062895\) |
| \(4^{3.14159}\) | \(77.8799471543\) | \(4^{4}\) | \(256\) |
While we have stated (via the Generalized Direct Substitution Property) that all our basic functions (including exponentials) are continuous on their domains, we have yet to prove that result for exponential functions officially. To do so, we need to first prove that \(B(x) = b^x\) is continuous at \(x = 0\).
Suppose \(x \gt 0\). \(\forall \epsilon \gt 0\), if we choose \(\delta = \log_b(1 + \epsilon)\), then\[\begin{array}{rccccll} & 0 & \lt & |x - 0| & \lt & \delta & \\ \implies & & & x & \lt & \delta & \left( x \gt 0 \implies |x| = x \right) \\ \implies & & & x & \lt & \log_b(1 + \epsilon) & \\ \implies & & & b^x & \lt & 1 + \epsilon & \\ \implies & & & b^x - 1 & \lt & \epsilon & \\ \implies & & & |b^x - 1| & \lt & \epsilon & \left( x \gt 0 \implies b^x \gt b^0 = 1 \implies b^x - 1 \gt 0 \implies b^x - 1 = |b^x - 1| \right) \\ \end{array}\nonumber\]Therefore, \(\displaystyle \lim_{x \to 0^+} b^x = 1 = b^0\).
Moreover, if \(x \lt 0\), then \(\forall \epsilon \gt 0\), we choose \(\delta = -\log_b(1 - \epsilon)\). Therefore,\[\begin{array}{rccccll} & 0 & \lt & |x - 0| & \lt & \delta & \\ \implies & & & -x & \lt & \delta & \left( x \lt 0 \implies |x| = -x \right) \\ \implies & & & -x & \lt & -\log_b(1 - \epsilon) & \\ \implies & & & x & \gt & \log_b(1 - \epsilon) & \\ \implies & & & b^x & \gt & 1 - \epsilon & \\ \implies & & & b^x - 1 & \gt & -\epsilon & \\ \implies & & & -(b^x - 1) & \lt & \epsilon & \\ \implies & & & |b^x - 1| & \lt & \epsilon & \left( x \lt 0 \implies b^x \lt b^0 = 1 \implies b^x - 1 \lt 0 \implies -(b^x - 1) = |b^x - 1| \right) \\ \end{array}\nonumber\]Therefore, \(\displaystyle \lim_{x \to 0^-} b^x = 1 = b^0\).
Since both one-sided limits are equal to \(b^0\), \(\displaystyle \lim_{x \to 0} b^x = 1 = b^0\). Thus, \(B(x) = b^x\) is continuous at \(x = 0\).
We are now ready to prove that \(B(x) = b^x\) is continuous everywhere.
Let \(B(x) = b^x\) and let \(a\) be arbitrary. Then\[\begin{array}{rclcl} \displaystyle \lim_{x \to a}{b^x} & = & \displaystyle \lim_{h \to 0}{b^{a + h}} & \quad & \left(\text{substituting }x = a + h \implies h = x - a \implies h \to 0\text{ as }x \to a\right) \\[8pt] & = & \displaystyle \lim_{h \to 0}{\left( b^a \cdot b^h \right)} & & \\[8pt] & = & \displaystyle b^a \cdot \lim_{h \to 0}{b^h} & \quad & \left(\text{Constant Multiple Limit Law}\right) \\[8pt] & = & \displaystyle b^a \cdot b^0 & \quad & \left(b^x\text{ is continuous at }0 \right) \\[8pt] & = & \displaystyle b^a & & \\[8pt] \end{array}\nonumber\]Since \(a\) is arbitrary, this shows that the exponential function \(B(x) = b^x\) is continuous everywhere.
The next theorem is a fascinating fact and is the student's favorite derivative. In words, the function \(y = e^x\) is the only function (besides \(y = 0\)) whose derivative is itself!
\[\dfrac{d}{dx}\left(e^x\right) = e^x\nonumber\]
- Proof
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Using the limit definition of the derivative, we get the following:\[\begin{array}{rclcl} \dfrac{d}{dx} \left( e^x \right) & = & \displaystyle \lim_{h \to 0}{\dfrac{e^{x + h} - e^x}{h}} & \quad & \left(\text{definition of a derivative}\right) \\[16pt] & = & \displaystyle \lim_{h \to 0}{\dfrac{e^x \cdot e^h - e^x}{h}} & & \\[16pt] & = & \displaystyle \lim_{h \to 0}{\dfrac{e^x \left(e^h - 1\right)}{h}} & & \\[16pt] & = & \displaystyle e^x \cdot \lim_{h \to 0}{\dfrac{e^h - 1}{h}} & \quad & \left(\text{since }e^x\text{ does not rely on }h\text{, it can be factored out}\right) \\[16pt] \end{array}\nonumber\]Recall that we defined \(e\) to be \(\displaystyle \lim_{n \to 0}{\left(1 + n\right)^{1/n}}\). If we let \(n = e^h - 1\), then \(n + 1 = e^h\). Since exponential functions are continuous everywhere,\[\displaystyle \lim_{h \to 0}{e^h} = e^0 = 1.\nonumber\]This means that \(n \to 0\) as \(h \to 0\). Moreover, solving \(n + 1 = e^h\) for \(h\), we get \(\ln{(n+1)} = h\). Thus,\[\begin{array}{rclcl} \dfrac{d}{dx} \left( e^x \right) & = & \displaystyle e^x \cdot \lim_{h \to 0}{\dfrac{e^h - 1}{h}} & & \\[16pt] & = & \displaystyle e^x \cdot \lim_{n \to 0}{\dfrac{n}{\ln{(1 + n)}}} & & \\[16pt] & = & \displaystyle e^x \cdot \lim_{n \to 0}{\dfrac{1}{\frac{1}{n}\ln{(1 + n)}}} & & \\[16pt] & = & \displaystyle e^x \cdot \lim_{n \to 0}{\dfrac{1}{\ln{(1 + n)^{1/n}}}} & & \\[16pt] & = & e^x \cdot \dfrac{1}{\displaystyle \lim_{n \to 0}{\ln{(1 + n)^{1/n}}}} & \quad & \left(\text{Limit Laws}\right) \\[16pt] & = & e^x \cdot \dfrac{1}{\ln{\left( \displaystyle \lim_{n \to 0}{(1 + n)^{1/n}}\right)}} & \quad & \left(\text{Limit Laws}\right) \\[16pt] & = & e^x \cdot \dfrac{1}{\ln{(e)}} & \quad & \left(\text{definition of }e\right) \\[16pt] & = & e^x \cdot \dfrac{1}{1} & & \\[16pt] & = & e^x & & \\[16pt] \end{array}\nonumber\]
Q.E.D.
The following Interactive Element might help you visualize why the slope of \(y=e^x\) at any point is equal to the height of the curve at that same point.
Interact: Drag the red point along the curve \(y=e^x\).
Observation: The tangent line always crosses the \(x\)-axis exactly one unit to the left of the point, so over a run of \(1\) the rise equals the curve's height. The slope therefore always equals \(e^x\)—the value of the function itself.
Before going into examples, let's add a quick corollary.
\[\dfrac{d}{dx}\left(b^x\right) = b^x \ln{(b)}\nonumber\]
- Proof
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\[\begin{array}{rclcl} \dfrac{d}{dx}\left( b^x \right) & = & \dfrac{d}{dx}\left( e^{\ln{\left(b^x \right)}}\right) & & \\[16pt] & = & \dfrac{d}{dx}\left( e^{x \ln{(b)}}\right) & & \\[16pt] & = & e^{x \ln{(b)}} \cdot \ln{(b)} & \quad & \left(\text{Chain Rule}\right) \\[16pt] & = & e^{\ln{(b^x)}} \cdot \ln{(b)} & & \\[16pt] & = & b^x \cdot \ln{(b)} & & \\[16pt] \end{array}\nonumber\]
Q.E.D.
Find the derivative of \(f(x)=e^{\tan(2x)}\).
- Solution
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Using the derivative formula and the Chain Rule,\[f^{\prime}(x)=e^{\tan(2x)}\frac{d}{dx} \left( \tan(2x) \right) =e^{\tan(2x)}\sec^2(2x) \cdot 2\nonumber\]
Find the derivative of \(y=\frac{e^{x^2}}{x}\).
- Solution
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Use the derivative of the natural exponential function, the Quotient Rule, and the Chain Rule.\[\begin{array}{rclcl} y^{\prime} & = & \dfrac{(e^{x^2} \cdot 2)x \cdot x-1 \cdot e^{x^2}}{x^2} & \quad & \left( \text{Quotient Rule} \right) \\[16pt] & = & \dfrac{e^{x^2}(2x^2-1)}{x^2} & & \\[16pt] \end{array}\nonumber\]
Find the derivative of \(h(x)=xe^{2x}\).
- Answer
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\(h^{\prime}(x)=e^{2x}+2xe^{2x}\)
A colony of mosquitoes has an initial population of 1000. After \(t\) days, the population is given by \(A(t)=1000e^{0.3t}\). Show that the ratio of the rate of change of the population, \(A^{\prime}(t)\), to the population, \(A(t)\) is constant.
- Solution
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First find \(A^{\prime}(t)\). By using the Chain Rule, we have \(A^{\prime}(t)=300e^{0.3t}\). Thus, the ratio of the rate of change of the population to the population is given by\[\frac{A^{\prime}(t)}{A(t)}=\dfrac{300e^{0.3t}}{1000e^{0.3t}}=0.3.\nonumber\]The ratio of the rate of change of the population to the population is the constant 0.3.
If \(A(t)=1000e^{0.3t}\) describes the mosquito population after \(t\) days, as in the preceding example, what is the rate of change of \(A(t)\) after four days?
- Answer
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\(996\)
Find the derivative of \(h(x)=\frac{3^x}{3^x+2}\).
- Solution
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\[\begin{array}{rclcl} h^{\prime}(x) & = & \dfrac{3^x\ln 3(3^x+2)-3^x\ln 3(3^x)}{(3^x+2)^2} & \quad & \left( \text{Quotient Rule} \right) \\[16pt] & = & \dfrac{2 \cdot 3^x\ln 3}{(3x+2)^2} & & \\[16pt] \end{array}\nonumber\]
Find the slope for the line tangent to \(y=3^x\) at \(x=2\).
- Answer
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\(9\ln(3)\)
Derivatives of the Hyperbolic Functions
Recall that the hyperbolic sine and hyperbolic cosine are defined as\[\sinh x=\dfrac{e^x-e^{-x}}{2}\nonumber\]and\[\cosh x=\dfrac{e^x+e^{-x}}{2}.\nonumber\]The other hyperbolic functions are then defined in terms of \(\sinh x\) and \(\cosh x\). The graphs of the hyperbolic functions are shown in Figure \(\PageIndex{1}\).
Developing differentiation formulas for the hyperbolic functions is easy. For example, looking at \(\sinh x\) we have\[\begin{array}{rcl} \dfrac{d}{dx} \left(\sinh x \right) & = & \dfrac{d}{dx} \left(\dfrac{e^x-e^{-x}}{2}\right) \\[16pt] & = & \dfrac{1}{2}\left[\dfrac{d}{dx}(e^x)-\dfrac{d}{dx}(e^{-x})\right] \\[16pt] & = & \dfrac{1}{2}[e^x+e^{-x}] \\[16pt] & = \cosh x. \\[16pt] \end{array}\nonumber\]Similarly,\[\dfrac{d}{dx} \cosh x=\sinh x.\nonumber\]We summarize the differentiation formulas for the hyperbolic functions in Table \(\PageIndex{3}\).
| \(f(x)\) | \(\frac{d}{dx}f(x)\) |
|---|---|
| \(\sinh x\) | \(\cosh x\) |
| \(\cosh x\) | \(\sinh x\) |
| \(\tanh x\) | \(\text{sech}^2 \,x\) |
| \(\text{coth } x\) | \(-\text{csch}^2\, x\) |
| \(\text{sech } x\) | \(-\text{sech}\, x \tanh x\) |
| \(\text{csch } x\) | \(-\text{csch}\, x \coth x\) |
Let's take a moment to compare the derivatives of the hyperbolic functions with the derivatives of the standard trigonometric functions. There are a lot of similarities and differences. For example, the derivatives of the sine functions match:\[\dfrac{d}{dx} \sin x=\cos x\nonumber\]and\[\dfrac{d}{dx} \sinh x=\cosh x.\nonumber\]The derivatives of the cosine functions, however, differ in sign:\[\dfrac{d}{dx} \cos x=-\sin x,\nonumber\]but\[\dfrac{d}{dx} \cosh x=\sinh x.\nonumber\]
Evaluate the following derivatives:
- \(\frac{d}{dx}(\sinh(x^2))\)
- \(\frac{d}{dx}(\cosh x)^2\)
- Solutions
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Using the formulas in Table \(\PageIndex{3}\) and the Chain Rule, we get
\[\dfrac{d}{dx}(\sinh(x^2))=\cosh(x^2) \cdot 2x\nonumber\]
\[\dfrac{d}{dx}(\cosh x)^2=2\cosh x\sinh x\nonumber\]
Evaluate the following derivatives:
- \(\frac{d}{dx}(\tanh(x^2+3x))\)
- \(\frac{d}{dx}\left(\frac{1}{(\sinh x)^2}\right)\)
- Answer a
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\(\frac{d}{dx}(\tanh(x^2+3x))=(\text{sech}^2(x^2+3x))(2x+3)\)
- Answer b
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\(\frac{d}{dx}\left(\frac{1}{(\sinh x)^2}\right)=\frac{d}{dx}(\sinh x)^{-2}=-2(\sinh x)^{-3}\cosh x\)
This page titled 3.5: Derivatives of Exponential and Hyperbolic Functions was last modified on Sun, 29 Dec 2024 21:05:19 GMT and is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by Roy Simpson.
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