4.8 Exponential and Logarithmic Functions

4.8.1 Basic Properties of Exponential Functions

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Definition 1.
The exponential function with base \(\pmb{a}>0\), \(a\neq 1\), is defined by \(a^x\), for \(x\in \mathbb{R}\).
The domain and range of the exponential function are
\[D_{a^x}=\mathbb{R} \;\;\;\; R_{a^x} = (0,\infty).\]
Definition 2. A function \(f\) defined on an interval \((a,b)\) is said to be
increasing, if \(a < x_1 < x_2 < b \;\; \Longrightarrow \;\; f(x_1) > f(x_2)\).
decreasing, if \(a < x_1 < x_2 < b \;\; \Longrightarrow \;\; f(x_1) > f(x_2)\).

It turns out that (1) if \(a > 1\), then \(y = a^x\) is increasing; (2) if \( 0 < a < 1 \), \(y = a^x\) is decreasing (see Figure 4.14).

Figure 4.14 (a)

Figure 4.14 (b)

Note that the \(x\)-axis is a horizontal asymptote.
Laws of exponents. Let \(a>0\), \(b>0\), \(a\neq 1\), \(b \neq 1\), and \(x, \; y \in \mathbb{R}\), then
\[a^xa^y=a^{x+y},\;\;\;\; (a^x)^y=a^{xy},\;\;\;\;(ab)^x=a^xb^x,\;\;\;\; \left(\frac{a}{b}\right)^x=\frac{a^x}{b^x}, \;\;\;\; \frac{a^x}{a^y}=a^{x-y}.\]

The exponential function has base \(e\), that is, \(y = e^x\). Another notation is \(e^x = \exp x\).

Exponential functions are one-to-one (because they are increasing/decreasing). This means
\[a^x=a^y \Longleftrightarrow x = y.\]
In particular, they have inverses. The inverse of an exponential function is a logarithmic function.

4.8.2 Basic Properties of Logarithmic Functions

Definition 3.
Let \(f(x) =a^x\), be an exponential function. Its inverse function, \(f^{-1}(x) = \log_ax\), is the logarithmic function of base \(\pmb{a}\).

The domain and range of a logarithmic function are
\[D_{\log_a}= (0,\infty)\;\;\;\; R_{\log_a} =\mathbb{R}\]
(see Figure 4.15).

In addition
\[ y =\log_a x\Longleftrightarrow a^x=y.\]
This can be interpreted by saying that

“the logarithm of base \(a\) of a positive number \(x\) is the exponent to which \(a\) must be raised to yield \(x\).”

(1) if \(a > 1\), then \(y = \log_ax\) is increasing; (2) if \( 0 < a < 1 \), \(y = \log_ax\) is decreasing (see Figure 4.15).

Figure 4.15 (a)

Figure 4.15 (b)

Note that the \(y\)-axis is a vertical asymptote of the graph.Laws of logarithms. Let \(a>0\), \(a\neq 1\), \(x>0\), and
\(y>0\).
\[\log_a(x y )= \log_ax+\log_a y,\;\;\;\; \log_a\frac{x}{y}=\log_a x – \log_ay,
\;\;\;\; \log_ax^y=y\log_ax.\]

The natural logarithm is the logarithm of base \(e\), \(\ln x = \log_ex\), that is, the inverse of the exponential function, \(y = e^x\). The logarithm of base 10 of a number \(x\) is denoted by \(\log x\) (no subindex 10).

4.8.3 Derivative of the Exponential Function

Theorem 1.
\begin{equation}\label{ch04-08-eq01}
\lim_{h\to 0}\frac{e^h-1}{h} =1\tag{4.9}
\end{equation}

We will not prove this Theorem. Rather, we will use it to obtain the derivative of the exponential function.

Proposition 1.
\begin{equation}\label{ch04-08-eq02}
\exp'(x) = e^x\;\;\;\forall \, x\in \mathbb{R}.\tag{4.10}
\end{equation}

Proof.
\begin{align*}
\exp'(x) = \lim_{h\to 0} \frac{e^{x+h}-e^x}{h} =
\lim_{h\to 0} e^x\frac{e^{h}-1}{h} = e^x
\end{align*}

Example 1. Compute the derivative of the function.    (a) \(f(t) = e^{at}\);    (b) \(g(x) = x^2e^{\sin \pi x}\);

(c) \(y = \arctan e^{x^2}\);    (d) \(\displaystyle y = \frac{x e^{-2x}}{\sqrt{1+x^2}}\).

Solution.
(a) Applying the chain rule, \(y’ = a e^{ax}\).

(b) Applying the Product and chain rules
\[g'(x) = 2xe^{\sin \pi x} + x^2e^{\sin \pi x}\pi \cos \pi x.\]

(c) Let \(g(x)= e^{x^2}\). Applying the chain rule to \(\arctan g(x)\) and then to \(g(x)\), we have
\[y’ = \frac{1}{1+(g(x))^2}\cdot g'(x) = \frac{1}{1+(e^{x^2})^2}\cdot e^{x^2}2x = \frac{2xe^{x^2}}{1+e^{2x^2} }.\]

(d) This example applies the Product, Quotient, and chain rules.
\begin{align*}
y’ &=
\frac{\sqrt{1+x^2}\cdot \left (x e^{-2x}\right )’ – x e^{-2x}\left( \sqrt{1+x^2}\right)’ }{1+x^2} =
\frac{\displaystyle \sqrt{1+x^2}\cdot \left (e^{-2x} -2x e^{-2x}\right )- x e^{-2x}\frac{x}{\sqrt{1+x^2}}}{1+x^2} =\\
& \\
& =
\frac{(1+x^2)\cdot \left (e^{-2x} -2x e^{-2x}\right )- x^2 e^{-2x}}{(1+x^2)\sqrt{1+x^2}} =
\frac{e^{-2x} \, \left ( 1 – 2 (x + x^3)\right )}{(1+x^2)\sqrt{1+x^2}}.
\end{align*}

Example 2. Given \(\displaystyle f(x) = \frac{\sqrt[3]{x} \, e^{x}}{2x+5}\), find the \(x\)-coordinate(s) of the point(s) on the graph of \(f\) whose tangent line is horizontal.

Solution.
\begin{align*}
f'(x)& = \frac{\displaystyle (2x+5)\left (\frac{1}{3\,\sqrt[3]{x^2} }\, e^x + \sqrt[3]{x} \, e^x\right )- 2\sqrt[3]{x} \, e^x}{(2x+5)^2}
= \frac{(2x+5)\left (e^x + 3x e^x\right )- 6x \, e^x}{3\,\sqrt[3]{x^2}(2x+5)^2} =\\
&\underset{?}{=} \frac{(6 x^2+11 x+5) e^x}{3\,\sqrt[3]{x^2}\,(2x+5)^2}
\end{align*}
The graph has a horizontal tangent line at \(x\) if and only if (the curly brace \(“\{“\) stands for the connective “and”)
\[\left \{
\begin{array}{l}
x\in D_f\\
f'(x)=0
\end{array}
\right.
\Longleftrightarrow
\left \{
\begin{array}{l}
x \neq -5/2\\
6 x^2+11 x+5=0
\end{array}
\right.
\Longleftrightarrow
\left \{
\begin{array}{l}
(6 x+5)(x+1)=0
\end{array}
\right.
\Longleftrightarrow x\in \left \{-1, \frac{5}{6}\right \}.
\]

4.8.4 Derivative of the Natural Logarithm Function

The formula to compute the derivative of \(y = \ln x\) is obtained in an analogous way to the procedure to compute the derivatives of inverse trigonometric functions.

Proposition 2.
\begin{equation}\label{ch04-08-eq03}
\frac{d}{dx} \ln |x|= \frac{1}{x}\;\;\;\forall \, x \in (-\infty, 0)\cup (0, \infty)\tag{4.11}.
\end{equation}
Proof. Assume \(x>0\). Then
\[y = \ln |x| = \ln x \;\; \Longrightarrow \;\; e^{ y} = x \;\; \Longrightarrow \]
\[\Longrightarrow\frac{d}{dx} e^{ y} = 1 \;\; \Longrightarrow \;\; e^{ y} \frac{dy}{dx} = 1 \;\; \Longrightarrow \;\; \frac{d y}{dx}= \frac{1}{e^y} \Longrightarrow y’= \frac{1}{x}.\]
Let now \(x <0\). Then \[y = \ln |x| = \ln (- x) \;\; \Longrightarrow y’ = \frac{d}{dx} \ln (-x)\; \underset{\text{chain rule}}{=} \; \frac{1}{-x}\cdot (-1) = \frac{1}{x}.\]

Example 1. Find the domain and compute the derivative of the function.   (a) \(f(t) =\ln (\ln x ) \);    (b) \(\displaystyle g(x) = \ln \left (\frac{1}{2} – \frac{2}{\pi}\arcsin x\right )\);    (c) \(\displaystyle y = \ln^2 \frac{1}{\sqrt[4]{x^3}}\);    (d) \(y =\sin \left[ \exp \left (\ln^3 x^2\right )\right]\).
Solution. (a) \(x\in D_f \Longleftrightarrow \left \{\begin{array}{l} x > 0\\ \ln x > 0\end{array} \right .\Longleftrightarrow x > 1\). Applying the chain rule
\[
\frac{d}{dx}
\ln (\ln x) = \frac{1}{\ln x}\frac{1}{x} = \frac{1}{x \ln x}.\]
(b)
\[x\in D_f \;\; \Longleftrightarrow \;\;
\left \{ \renewcommand{\arraystretch}{2}
\begin{array}{l} -1 \leq x \leq 1\\\displaystyle \frac{1}{2} – \frac{2}{\pi}\arcsin x > 0\end{array} \right .
\;\; \underset{?}{\Longleftrightarrow} \;\; \arcsin x < \frac{\pi}{4}
\;\; \underset{?}{\Longleftrightarrow} \;\; -1 \leq x < \frac{\sqrt{2}}{2} .\] Applying the chain rule \begin{align*} \frac{d}{dx} \ln\left (\frac{1}{2} – \frac{2}{\pi}\arcsin x\right ) &= \frac{1}{\displaystyle \frac{1}{2} – \frac{2}{\pi}\arcsin x} \cdot \frac{d}{dx}\left (\frac{1}{2} – \frac{2}{\pi}\arcsin x\right )=\\ &= \frac{2\pi}{\displaystyle \pi – 4\arcsin x} \cdot \left ( – \frac{2}{\pi}\frac{1}{\sqrt{1-x^2}}\right ) -\frac{4}{\displaystyle \left (\pi – 4\arcsin x\right )\sqrt{1-x^2}} . \end{align*} (c) \(x\in D_y \Longleftrightarrow x > 0\). Before computing the derivative, it is convenient to rewrite the function:
\[y = \ln^2 \frac{1}{\sqrt[4]{x^3}} = \left (\ln x^{-3/4} \right )^2 = \left (-\frac{3}{4}\ln x \right )^2 = \frac{9}{16}\ln^2x.\]
\[y’=
\frac{d}{dx}
\frac{9}{16}\ln^2x = \frac{9}{16}\cdot 2 \ln x\cdot \frac{1}{x} = \frac{9}{8}\frac{\ln x}{x} .\]

(d) \(x\in D_y \Longleftrightarrow x \neq 0 \Longleftrightarrow D_y = (-\infty, 0) \cup (0, \infty)\).
Note that
\[y =\sin \left [ \exp \left (\ln^3 x^2\right )\right ] \underset{?}{=} \sin \left [\exp \left ( \left (2\ln| x|\right )\right )^3\right ] =
\sin \left [\exp \left (8\ln^3| x|\right ) \right ]. \]
Then
\begin{align*}
y’ &= \cos \left [\exp \left (8\ln^3| x|\right ) \right ] \cdot \frac{d}{dx} \exp \left (8\ln^3| x|\right) =
\cos \left [\exp \left (8\ln^3| x|\right ) \right ] \cdot \exp \left (8\ln^3| x|\right)\cdot \frac{d}{dx} 8\ln^3| x| = \\
&=
\cos \left [\exp \left (8\ln^3| x|\right ) \right ] \cdot \exp \left (8\ln^3| x|\right)\cdot 24\ln^2| x|\cdot\frac{1}{x}.
\end{align*}

4.8.6 Derivatives of Generic Exponential and Logarithmic Functions

Proposition 3. Let \(a \in (0,1)\cup (1,\infty)\). Then
\begin{eqnarray}
\frac{d}{dx} a^x=a^x \ln a &\forall \, x \in (-\infty, \infty). \label{ch04-08-eq04}\tag{4.12} \\
\frac{d}{dx} \log_a |x|= \frac{1}{x\ln a} & \forall \, x \in (-\infty, 0)\cup (0, \infty). \label{ch04-08-eq05}\tag{4.13}
\end{eqnarray}

Proof. (\ref{ch04-08-eq04}).
\[y = a^x\;\; \Longleftrightarrow \;\; \ln y = \ln a^x \;\; \Longleftrightarrow \;\; \ln y= x\ln a.\]
Differentiating both side of the last equation with respect to \(x\), and applying the chain rule
\[\frac{1}{y}y’ = \ln a \;\; \Longleftrightarrow \;\; y’ = y \ln a \;\; \Longleftrightarrow \;\; y’ = a^x\ln a.\]
(\ref{ch04-08-eq05}).
\[y = \log_a |x| \;\; \Longleftrightarrow \;\; a^{ y} = |x| \;\; \Longleftrightarrow \;\; \ln a^{y} = \ln |x| \;\;\Longleftrightarrow \;\; y \ln a = \ln |x| \;\; \Longrightarrow\]
\[\Longrightarrow \frac{d}{dx} \left ( y \ln a \right ) = \frac{d}{dx} \ln |x| \;\; \Longrightarrow \;\; y’\ln a = \frac{1}{x} \;\; \Longrightarrow \;\; y’= \frac{1}{x\ln a}.\]

Example 1. Find \(D_f\) and \(f’\) for each function.     (a) \(\displaystyle f(x) = \log\left ( 4^{x}-9\right )\);     (b) \(\displaystyle f(x) = \log^2 \frac{1}{\sqrt[4]{2^{x^3}}}\);

(c) \(f(x) =\log_8 [\log_4(\log_2 x) ]\);     (d) \(f(x) =\arcsin \left (\log_2^3 x^2\right )\).

Solution.
(a)
\[x\in D_f \,\Longleftrightarrow \, 4^{x}-9 > 0 \, \underset{?}{\Longleftrightarrow } \, \log_4 4^{x} > \log_4 9 \, \Longleftrightarrow \, x > \log_4 9 \, \Longleftrightarrow D_f=\left ( \log_4 9 , \infty \right ).\]
From (\ref{ch04-08-eq05}) and the chain rule, with \(g(x) = 4^{ x}-9\),
\[f'(x) = \frac{1}{g(x) \ln 10}\cdot g'(x)
\underset{\text{(\ref{ch04-08-eq04})}}{=} \frac{1}{\left (4^{ x}-9\right )\ln 10}\cdot 4^{ x} \ln 4 .\]
(b) Since \(2^{x^3}>0\) for all \(x\in \mathbb{R}\), \(D_f=\mathbb{R}\). Note that
\[ f(x) = \log^2 \frac{1}{\sqrt[4]{2^{x^3}}} = \left (\log 2^{-x^3/4}\right )^2 =
\left (-\frac{x^3}{4} \log 2 \right )^2 = \frac{\log^2 2}{16}x^6 \Longrightarrow f'(x) = \frac{3\log^2 2}{8}x^5.\]
(c)
\[x\in D_f \;\Longleftrightarrow \;
\left \{ \renewcommand{\arraystretch}{1.2}
\begin{array}{l} x > 0 \\ \log_2x > 0 \\ \log_4 (\log_2 x) > 0 \end{array} \right .
\; \underset{?}{\Longleftrightarrow} \;
\left \{ \renewcommand{\arraystretch}{1.2}
\begin{array}{l} x > 1 \\ \log_2 x > 1 \end{array} \right .
\;\Longleftrightarrow D_f = (2, \infty).\]
To compute the derivative of \(f\), we apply (\ref{ch04-08-eq05}) thrice, with \(g(x) = \log_4(\log_2 x) \), and \(h(x) = \log_2 x\),
\begin{align*}
f'(x) &= \frac{1}{g(x)\ln 8}\cdot g'(x) =
\frac{1}{\log_4(\log_2 x) \cdot \ln 8}\cdot \frac{1}{h(x) \ln 4} \cdot h'(x) = \\
& \\
&=
\frac{1}{\log_4(\log_2 x) \cdot \ln 8}\cdot \frac{1}{\log_2 x \cdot \ln 4} \cdot \frac{1}{x\ln 2}
\underset{?}{=}
\frac{1}{\log_4(\log_2 x) }\cdot \frac{1}{\log_2 x }\cdot \frac{1}{3\cdot2 \ln^3 2}.
\end{align*}
(d)
\begin{align*}
x\in D_f &\; \Longleftrightarrow \; -1\leq \log_2^3 x^2 \leq 1
\;\Longleftrightarrow \; -1 \leq \left (2\log_2|x|\right )^3 \leq 1
\;\Longleftrightarrow \; -1 \leq 2\log_2|x| \leq 1 \; \Longleftrightarrow \\
& \: \Longleftrightarrow D_f= \left [- \sqrt{2}, \sqrt 2\right ].
\end{align*}
Let \(g(x) = \log_2^3 x^2 = 8 \log_2^3|x|\). Then, by the chain rule and (\ref{ch04-08-eq05}),
\[f'(x) = \frac{1}{\sqrt{\displaystyle 1-(g(x))^2}}\cdot g'(x) = \frac{1}{\sqrt{\displaystyle 1-\left (8 \log_2^3|x|\right )^2}}\cdot 24 \log_2^2|x|\cdot \frac{1}{x\ln 2}=
\frac{24 \log_2^2|x|}{x\ln 2\,\sqrt{\displaystyle 1-64\log_2^6|x|}}
.\]

4.8.7 Logarithmic Differentiation

Logarithmic differentiation is a very useful technique to compute derivatives of functions that involve products and exponential expressions. In essence, it amounts to a direct application of the chain rule and properties of logarithms.

Example 1. Compute the derivative of the function.     (a) \(\displaystyle f(x) = x^x\);     (b) \(\displaystyle g(x) = (4x^2+ 1)^{\sin \pi x}\);

(c) \(\displaystyle f(x) =\frac{x^2e^{-x^2}\sqrt{x^2+1}}{\cos^3x}\);     (d) \(f(x) =x^{\arccos x}+ \left (\arccos x\right )^x\).

Solution.
(a) Note that \( f(x) = x^x\) has domain \(D_f=(0, \infty)\), and hence, \(f(x) > 0\). Thus, \(\ln f(x)\) is well defined, and we have
\[\ln f(x) = \ln x^x \;\; \Longrightarrow \;\; \ln f(x) = x\ln x.\]
Differentiation both sides of the last equation
\[\frac{d}{dx} \ln f(x) = \frac{d}{dx} ( x\ln x) \;\; \Longrightarrow \;\; \frac{f'(x)}{f(x)} = \ln x + 1 \;\; \Longrightarrow\;\; f'(x) = f(x) (\ln x + 1) = x^x(\ln x + 1).\]
(b) Since \(g(x) = (4x^2+ 1)^{\sin \pi x} > 0\), \(\ln g(x)\) is well defined, and
\[\ln g(x) = \ln (4x^2+ 1)^{\sin \pi x} = \sin \pi x \ln (4x^2+ 1).\]
Differentiating
\[\frac{d}{dx}\ln g(x) = \frac{g'(x)}{g(x)} =\pi \cos \pi x \ln(4x^2 +1) + \sin \pi x \frac{8x}{4x^2+1} \]
Thus
\[ g'(x) = g(x) \left (\pi \cos \pi x \ln(4x^2 +1) + \sin \pi x \frac{8x}{4x^2+1}\right ),\]
that is,
\[g'(x) = (4x^2+ 1)^{\sin \pi x}\left (\pi \cos \pi x \ln(4x^2 +1) + \sin \pi x \frac{8x}{4x^2+1}\right ).\]
(c) Since, \(f(x)\) may be negative, we are going to compute the derivative of \(|f(x)|\):
\begin{align*}
\ln |f(x) | =
\ln\left | \frac{x^2e^{-x^2}\sqrt{x^2+1}}{\cos^3x }\right | &= \ln x^2 + \ln e^{-x^2}+ \ln \sqrt{x^2+1} – \ln |\cos^3x |
=\\
&=
2\ln |x| -x^2 + \frac{1}{2}\ln (x^2+1) – 3\ln |\cos x |.
\end{align*}
Differentiating, we have
\[\frac{f'(x)}{f(x)} = \frac{2}{x}- 2x + \frac{2x}{x^2+1} – 3\frac{-\sin x}{\cos x}.\]
Then
\[f'(x) = \frac{x^2e^{-x^2}\sqrt{x^2+1}}{\cos^3x }\left (\frac{2}{x}- 2x + \frac{2x}{x^2+1} + 3\tan x\right ).\]
(d) In the case of \(f(x) =x^{\arccos x}+ \left (\arccos x\right )^x\),taking the logarithm on the right-hand-side does not lead to a simplification because it is a sum and not a product. Thus, what we need to do is to compute first the derivative of each term separately.

Let \(g(x) = x^{\arccos x} \), then
\[\ln g(x) = x^{\arccos x} = \arccos x\ln x \]
Differentiating
\[\frac{ g'(x)}{g(x)} = – \frac{1}{\sqrt{1-x^2}}\cdot \ln x+ \arccos x \cdot \frac{1}{x} \Longrightarrow \frac{d}{dx} x^{\arccos x} = x^{\arccos x}\left ( – \frac{\ln x}{\sqrt{1-x^2}} + \frac{\arccos x }{x} \right ) .\]
Similarly (verify this),
\[\frac{d}{dx} \left (\arccos x\right )^x = \left (\arccos x\right )^x\left (\ln (\arccos x ) -\frac{x}{\sqrt{1-x^2}\, \arccos x} \right ).\]
Thus,
\[
\frac{d}{dx}\left (x^{\arccos x}+ \left (\arccos x\right )^x\right ) =
x^{\arccos x}\left ( – \frac{\ln x}{\sqrt{1-x^2}} + \frac{\arccos x }{x} \right ) +\;\;\;\;\; \]
\[ \hspace{2.9in}
+ \left (\arccos x\right )^x\left (\ln (\arccos x ) -\frac{x}{\sqrt{1-x^2}\, \arccos x} \right ).\]