# 2.7 Integrals, exponential functions, and logarithms  (Page 4/4)

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Let’s now apply this definition to calculate a differentiation formula for ${a}^{x}.$ We have

$\frac{d}{dx}{a}^{x}=\frac{d}{dx}{e}^{x\phantom{\rule{0.2em}{0ex}}\text{ln}\phantom{\rule{0.2em}{0ex}}a}={e}^{x\phantom{\rule{0.2em}{0ex}}\text{ln}\phantom{\rule{0.2em}{0ex}}a}\text{ln}\phantom{\rule{0.2em}{0ex}}a={a}^{x}\text{ln}\phantom{\rule{0.2em}{0ex}}a.$

The corresponding integration formula follows immediately.

## Derivatives and integrals involving general exponential functions

Let $a>0.$ Then,

$\frac{d}{dx}{a}^{x}={a}^{x}\text{ln}\phantom{\rule{0.2em}{0ex}}a$

and

$\int {a}^{x}dx=\frac{1}{\text{ln}\phantom{\rule{0.2em}{0ex}}a}{a}^{x}+C.$

If $a\ne 1,$ then the function ${a}^{x}$ is one-to-one and has a well-defined inverse. Its inverse is denoted by ${\text{log}}_{a}x.$ Then,

$y={\text{log}}_{a}x\phantom{\rule{0.2em}{0ex}}\text{if and only if}\phantom{\rule{0.2em}{0ex}}x={a}^{y}.$

Note that general logarithm functions can be written in terms of the natural logarithm. Let $y={\text{log}}_{a}x.$ Then, $x={a}^{y}.$ Taking the natural logarithm of both sides of this second equation, we get

$\begin{array}{ccc}\hfill \text{ln}\phantom{\rule{0.2em}{0ex}}x& =\hfill & \text{ln}\left({a}^{y}\right)\hfill \\ \hfill \text{ln}\phantom{\rule{0.2em}{0ex}}x& =\hfill & y\phantom{\rule{0.2em}{0ex}}\text{ln}\phantom{\rule{0.2em}{0ex}}a\hfill \\ \hfill y& =\hfill & \frac{\text{ln}\phantom{\rule{0.2em}{0ex}}x}{\text{ln}\phantom{\rule{0.2em}{0ex}}a}\hfill \\ \hfill {\text{log}}_{\phantom{\rule{0.2em}{0ex}}}x& =\hfill & \frac{\text{ln}\phantom{\rule{0.2em}{0ex}}x}{\text{ln}\phantom{\rule{0.2em}{0ex}}a}.\hfill \end{array}$

Thus, we see that all logarithmic functions are constant multiples of one another. Next, we use this formula to find a differentiation formula for a logarithm with base $a.$ Again, let $y={\text{log}}_{a}x.$ Then,

$\begin{array}{cc}\hfill \frac{dy}{dx}& =\frac{d}{dx}\left({\text{log}}_{a}x\right)\hfill \\ & =\frac{d}{dx}\left(\frac{\text{ln}\phantom{\rule{0.2em}{0ex}}x}{\text{ln}\phantom{\rule{0.2em}{0ex}}a}\right)\hfill \\ & =\left(\frac{1}{\text{ln}\phantom{\rule{0.2em}{0ex}}a}\right)\frac{d}{dx}\left(\text{ln}\phantom{\rule{0.2em}{0ex}}x\right)\hfill \\ & =\frac{1}{\text{ln}\phantom{\rule{0.2em}{0ex}}a}·\frac{1}{x}\hfill \\ & =\frac{1}{x\phantom{\rule{0.2em}{0ex}}\text{ln}\phantom{\rule{0.2em}{0ex}}a}.\hfill \end{array}$

## Derivatives of general logarithm functions

Let $a>0.$ Then,

$\frac{d}{dx}{\text{log}}_{a}x=\frac{1}{x\phantom{\rule{0.2em}{0ex}}\text{ln}\phantom{\rule{0.2em}{0ex}}a}.$

## Calculating derivatives of general exponential and logarithm functions

Evaluate the following derivatives:

1. $\frac{d}{dt}\left({4}^{t}·{2}^{{t}^{2}}\right)$
2. $\frac{d}{dx}{\text{log}}_{8}\left(7{x}^{2}+4\right)$

We need to apply the chain rule as necessary.

1. $\frac{d}{dt}\left({4}^{t}·{2}^{{t}^{2}}\right)=\frac{d}{dt}\left({2}^{2t}·{2}^{{t}^{2}}\right)=\frac{d}{dt}\left({2}^{2t+{t}^{2}}\right)={2}^{2t+{t}^{2}}\text{ln}\left(2\right)\left(2+2t\right)$
2. $\frac{d}{dx}{\text{log}}_{8}\left(7{x}^{2}+4\right)=\frac{1}{\left(7{x}^{2}+4\right)\left(\text{ln}\phantom{\rule{0.2em}{0ex}}8\right)}\left(14x\right)$

Evaluate the following derivatives:

1. $\frac{d}{dt}\phantom{\rule{0.2em}{0ex}}{4}^{{t}^{4}}$
2. $\frac{d}{dx}{\text{log}}_{3}\left(\sqrt{{x}^{2}+1}\right)$
1. $\frac{d}{dt}{4}^{{t}^{4}}={4}^{{t}^{4}}\left(\text{ln}\phantom{\rule{0.2em}{0ex}}4\right)\left(4{t}^{3}\right)$
2. $\frac{d}{dx}{\text{log}}_{3}\left(\sqrt{{x}^{2}+1}\right)=\frac{x}{\left(\text{ln}\phantom{\rule{0.2em}{0ex}}3\right)\left({x}^{2}+1\right)}$

## Integrating general exponential functions

Evaluate the following integral: $\int \frac{3}{{2}^{3x}}dx.$

Use $u\text{-substitution}$ and let $u=-3x.$ Then $du=-3dx$ and we have

$\int \frac{3}{{2}^{3x}}dx=\int 3·{2}^{-3x}dx=\text{−}\int {2}^{u}du=-\frac{1}{\text{ln}\phantom{\rule{0.2em}{0ex}}2}{2}^{u}+C=-\frac{1}{\text{ln}\phantom{\rule{0.2em}{0ex}}2}{2}^{-3x}+C.$

Evaluate the following integral: $\int {x}^{2}{2}^{{x}^{3}}dx.$

$\int {x}^{2}{2}^{{x}^{3}}dx=\frac{1}{3\phantom{\rule{0.2em}{0ex}}\text{ln}\phantom{\rule{0.2em}{0ex}}2}{2}^{{x}^{3}}+C$

## Key concepts

• The earlier treatment of logarithms and exponential functions did not define the functions precisely and formally. This section develops the concepts in a mathematically rigorous way.
• The cornerstone of the development is the definition of the natural logarithm in terms of an integral.
• The function ${e}^{x}$ is then defined as the inverse of the natural logarithm.
• General exponential functions are defined in terms of ${e}^{x},$ and the corresponding inverse functions are general logarithms.
• Familiar properties of logarithms and exponents still hold in this more rigorous context.

## Key equations

• Natural logarithm function
• $\text{ln}\phantom{\rule{0.2em}{0ex}}x={\int }_{1}^{x}\frac{1}{t}dt$ Z
• Exponential function $y={e}^{x}$
• $\text{ln}\phantom{\rule{0.2em}{0ex}}y=\text{ln}\left({e}^{x}\right)=x$ Z

For the following exercises, find the derivative $\frac{dy}{dx}.$

$y=\text{ln}\left(2x\right)$

$\frac{1}{x}$

$y=\text{ln}\left(2x+1\right)$

$y=\frac{1}{\text{ln}\phantom{\rule{0.2em}{0ex}}x}$

$-\frac{1}{x{\left(\text{ln}\phantom{\rule{0.2em}{0ex}}x\right)}^{2}}$

For the following exercises, find the indefinite integral.

$\int \frac{dt}{3t}$

$\int \frac{dx}{1+x}$

$\text{ln}\left(x+1\right)+C$

For the following exercises, find the derivative $dy\text{/}dx.$ (You can use a calculator to plot the function and the derivative to confirm that it is correct.)

[T] $y=\frac{\text{ln}\left(x\right)}{x}$

[T] $y=x\phantom{\rule{0.2em}{0ex}}\text{ln}\left(x\right)$

$\text{ln}\left(x\right)+1$

[T] $y={\text{log}}_{10}x$

[T] $y=\text{ln}\left(\text{sin}\phantom{\rule{0.2em}{0ex}}x\right)$

$\text{cot}\left(x\right)$

[T] $y=\text{ln}\left(\text{ln}\phantom{\rule{0.2em}{0ex}}x\right)$

[T] $y=7\phantom{\rule{0.2em}{0ex}}\text{ln}\left(4x\right)$

$\frac{7}{x}$

[T] $y=\text{ln}\left({\left(4x\right)}^{7}\right)$

[T] $y=\text{ln}\left(\text{tan}\phantom{\rule{0.2em}{0ex}}x\right)$

$\text{csc}\left(x\right)\text{sec}\phantom{\rule{0.2em}{0ex}}x$

[T] $y=\text{ln}\left(\text{tan}\left(3x\right)\right)$

[T] $y=\text{ln}\left({\text{cos}}^{2}x\right)$

$-2\phantom{\rule{0.2em}{0ex}}\text{tan}\phantom{\rule{0.2em}{0ex}}x$

For the following exercises, find the definite or indefinite integral.

${\int }_{0}^{1}\frac{dx}{3+x}$

${\int }_{0}^{1}\frac{dt}{3+2t}$

$\frac{1}{2}\text{ln}\left(\frac{5}{3}\right)$

${\int }_{0}^{2}\frac{x\phantom{\rule{0.2em}{0ex}}dx}{{x}^{2}+1}$

${\int }_{0}^{2}\frac{{x}^{3}dx}{{x}^{2}+1}$

$2-\frac{1}{2}\text{ln}\left(5\right)$

${\int }_{2}^{e}\frac{dx}{x\phantom{\rule{0.2em}{0ex}}\text{ln}\phantom{\rule{0.2em}{0ex}}x}$

${\int }_{2}^{e}\frac{dx}{{\left(x\phantom{\rule{0.2em}{0ex}}\text{ln}\left(x\right)\right)}^{2}}$

$\frac{1}{\text{ln}\left(2\right)}-1$

$\int \frac{\text{cos}\phantom{\rule{0.2em}{0ex}}x\phantom{\rule{0.2em}{0ex}}dx}{\text{sin}\phantom{\rule{0.2em}{0ex}}x}$

${\int }_{0}^{\pi \text{/}4}\text{tan}\phantom{\rule{0.2em}{0ex}}x\phantom{\rule{0.2em}{0ex}}dx$

$\frac{1}{2}\text{ln}\left(2\right)$

$\int \text{cot}\left(3x\right)dx$

$\int \frac{{\left(\text{ln}\phantom{\rule{0.2em}{0ex}}x\right)}^{2}dx}{x}$

$\frac{1}{3}{\left(\text{ln}\phantom{\rule{0.2em}{0ex}}x\right)}^{3}$

For the following exercises, compute $dy\text{/}dx$ by differentiating $\text{ln}\phantom{\rule{0.2em}{0ex}}y.$

$y=\sqrt{{x}^{2}+1}$

$y=\sqrt{{x}^{2}+1}\sqrt{{x}^{2}-1}$

$\frac{2{x}^{3}}{\sqrt{{x}^{2}+1}\sqrt{{x}^{2}-1}}$

$y={e}^{\text{sin}\phantom{\rule{0.2em}{0ex}}x}$

$y={x}^{-1\text{/}x}$

${x}^{-2-\left(1\text{/}x\right)}\left(\text{ln}\phantom{\rule{0.2em}{0ex}}x-1\right)$

$y={e}^{\left(ex\right)}$

$y={x}^{e}$

$e{x}^{e-1}$

$y={x}^{\left(ex\right)}$

$y=\sqrt{x}\phantom{\rule{0.2em}{0ex}}\sqrt[3]{x}\phantom{\rule{0.2em}{0ex}}\sqrt[6]{x}$

$1$

$y={x}^{-1\text{/}\text{ln}\phantom{\rule{0.2em}{0ex}}x}$

$y={e}^{\text{−}\text{ln}\phantom{\rule{0.2em}{0ex}}x}$

$-\frac{1}{{x}^{2}}$

For the following exercises, evaluate by any method.

${\int }_{5}^{10}\frac{dt}{t}-{\int }_{5x}^{10x}\frac{dt}{t}$

${\int }_{1}^{{e}^{\pi }}\frac{dx}{x}+{\int }_{-2}^{-1}\frac{dx}{x}$

$\pi -\text{ln}\left(2\right)$

$\frac{d}{dx}{\int }_{x}^{1}\frac{dt}{t}$

$\frac{d}{dx}{\int }_{x}^{{x}^{2}}\frac{dt}{t}$

$\frac{1}{x}$

$\frac{d}{dx}\text{ln}\left(\text{sec}\phantom{\rule{0.2em}{0ex}}x+\text{tan}\phantom{\rule{0.2em}{0ex}}x\right)$

For the following exercises, use the function $\text{ln}\phantom{\rule{0.2em}{0ex}}x.$ If you are unable to find intersection points analytically, use a calculator.

Find the area of the region enclosed by $x=1$ and $y=5$ above $y=\text{ln}\phantom{\rule{0.2em}{0ex}}x.$

${e}^{5}-6\phantom{\rule{0.2em}{0ex}}{\text{units}}^{2}$

[T] Find the arc length of $\text{ln}\phantom{\rule{0.2em}{0ex}}x$ from $x=1$ to $x=2.$

Find the area between $\text{ln}\phantom{\rule{0.2em}{0ex}}x$ and the x -axis from $x=1\phantom{\rule{0.2em}{0ex}}\text{to}\phantom{\rule{0.2em}{0ex}}x=2.$

$\text{ln}\left(4\right)-1\phantom{\rule{0.2em}{0ex}}{\text{units}}^{2}$

Find the volume of the shape created when rotating this curve from $x=1\phantom{\rule{0.2em}{0ex}}\text{to}\phantom{\rule{0.2em}{0ex}}x=2$ around the x -axis, as pictured here.

[T] Find the surface area of the shape created when rotating the curve in the previous exercise from $x=1$ to $x=2$ around the x -axis.

$2.8656$

If you are unable to find intersection points analytically in the following exercises, use a calculator.

Find the area of the hyperbolic quarter-circle enclosed by $x=2\phantom{\rule{0.2em}{0ex}}\text{and}\phantom{\rule{0.2em}{0ex}}y=2$ above $y=1\text{/}x.$

[T] Find the arc length of $y=1\text{/}x$ from $x=1\phantom{\rule{0.2em}{0ex}}\text{to}\phantom{\rule{0.2em}{0ex}}x=4.$

$3.1502$

Find the area under $y=1\text{/}x$ and above the x -axis from $x=1\phantom{\rule{0.2em}{0ex}}\text{to}\phantom{\rule{0.2em}{0ex}}x=4.$

For the following exercises, verify the derivatives and antiderivatives.

$\frac{d}{dx}\text{ln}\left(x+\sqrt{{x}^{2}+1}\right)=\frac{1}{\sqrt{1+{x}^{2}}}$

$\frac{d}{dx}\text{ln}\left(\frac{x-a}{x+a}\right)=\frac{2a}{\left({x}^{2}-{a}^{2}\right)}$

$\frac{d}{dx}\text{ln}\left(\frac{1+\sqrt{1-{x}^{2}}}{x}\right)=-\frac{1}{x\sqrt{1-{x}^{2}}}$

$\frac{d}{dx}\text{ln}\left(x+\sqrt{{x}^{2}-{a}^{2}}\right)=\frac{1}{\sqrt{{x}^{2}-{a}^{2}}}$

$\int \frac{dx}{x\phantom{\rule{0.2em}{0ex}}\text{ln}\left(x\right)\text{ln}\left(\text{ln}\phantom{\rule{0.2em}{0ex}}x\right)}=\text{ln}\left(\text{ln}\left(\text{ln}\phantom{\rule{0.2em}{0ex}}x\right)\right)+C$

#### Questions & Answers

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