# 6.7 Integrals, exponential functions, and logarithms  (Page 2/4)

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## Calculating derivatives of natural logarithms

Calculate the following derivatives:

1. $\frac{d}{dx}\text{ln}\left(5{x}^{3}-2\right)$
2. $\frac{d}{dx}{\left(\text{ln}\left(3x\right)\right)}^{2}$

We need to apply the chain rule in both cases.

1. $\frac{d}{dx}\text{ln}\left(5{x}^{3}-2\right)=\frac{15{x}^{2}}{5{x}^{3}-2}$
2. $\frac{d}{dx}{\left(\text{ln}\left(3x\right)\right)}^{2}=\frac{2\left(\text{ln}\left(3x\right)\right)·3}{3x}=\frac{2\left(\text{ln}\left(3x\right)\right)}{x}$

Calculate the following derivatives:

1. $\frac{d}{dx}\text{ln}\left(2{x}^{2}+x\right)$
2. $\frac{d}{dx}{\left(\text{ln}\left({x}^{3}\right)\right)}^{2}$
1. $\frac{d}{dx}\text{ln}\left(2{x}^{2}+x\right)=\frac{4x+1}{2{x}^{2}+x}$
2. $\frac{d}{dx}{\left(\text{ln}\left({x}^{3}\right)\right)}^{2}=\frac{6\phantom{\rule{0.2em}{0ex}}\text{ln}\left({x}^{3}\right)}{x}$

Note that if we use the absolute value function and create a new function $\text{ln}\phantom{\rule{0.2em}{0ex}}|x|,$ we can extend the domain of the natural logarithm to include $x<0.$ Then $\left(d\text{/}\left(dx\right)\right)\text{ln}\phantom{\rule{0.2em}{0ex}}|x|=1\text{/}x.$ This gives rise to the familiar integration formula.

## Integral of (1/ u ) du

The natural logarithm is the antiderivative of the function $f\left(u\right)=1\text{/}u\text{:}$

$\int \frac{1}{u}du=\text{ln}\phantom{\rule{0.2em}{0ex}}|u|+C.$

## Calculating integrals involving natural logarithms

Calculate the integral $\int \frac{x}{{x}^{2}+4}dx.$

Using $u$ -substitution, let $u={x}^{2}+4.$ Then $du=2x\phantom{\rule{0.2em}{0ex}}dx$ and we have

$\int \frac{x}{{x}^{2}+4}dx=\frac{1}{2}\int \frac{1}{u}du\frac{1}{2}\text{ln}\phantom{\rule{0.2em}{0ex}}|u|+C=\frac{1}{2}\text{ln}\phantom{\rule{0.2em}{0ex}}|{x}^{2}+4|+C=\frac{1}{2}\text{ln}\left({x}^{2}+4\right)+C.$

Calculate the integral $\int \frac{{x}^{2}}{{x}^{3}+6}dx.$

$\int \frac{{x}^{2}}{{x}^{3}+6}dx=\frac{1}{3}\text{ln}\phantom{\rule{0.2em}{0ex}}|{x}^{3}+6|+C$

Although we have called our function a “logarithm,” we have not actually proved that any of the properties of logarithms hold for this function. We do so here.

## Properties of the natural logarithm

If $a,b>0$ and $r$ is a rational number, then

1. $\text{ln}\phantom{\rule{0.2em}{0ex}}1=0$
2. $\text{ln}\left(ab\right)=\text{ln}\phantom{\rule{0.2em}{0ex}}a+\text{ln}\phantom{\rule{0.2em}{0ex}}b$
3. $\text{ln}\left(\frac{a}{b}\right)=\text{ln}\phantom{\rule{0.2em}{0ex}}a-\text{ln}\phantom{\rule{0.2em}{0ex}}b$
4. $\text{ln}\left({a}^{r}\right)=r\phantom{\rule{0.2em}{0ex}}\text{ln}\phantom{\rule{0.2em}{0ex}}a$

## Proof

1. By definition, $\text{ln}\phantom{\rule{0.2em}{0ex}}1={\int }_{1}^{1}\frac{1}{t}dt=0.$
2. We have
$\text{ln}\left(ab\right)={\int }_{1}^{ab}\frac{1}{t}dt={\int }_{1}^{a}\frac{1}{t}dt+{\int }_{a}^{ab}\frac{1}{t}dt.$

Use $u\text{-substitution}$ on the last integral in this expression. Let $u=t\text{/}a.$ Then $du=\left(1\text{/}a\right)dt.$ Furthermore, when $t=a,u=1,$ and when $t=ab,u=b.$ So we get
$\text{ln}\left(ab\right)={\int }_{1}^{a}\frac{1}{t}dt+{\int }_{a}^{ab}\frac{1}{t}dt={\int }_{1}^{a}\frac{1}{t}dt+{\int }_{1}^{ab}\frac{a}{t}·\frac{1}{a}dt={\int }_{1}^{a}\frac{1}{t}dt+{\int }_{1}^{b}\frac{1}{u}du=\text{ln}\phantom{\rule{0.2em}{0ex}}a+\text{ln}\phantom{\rule{0.2em}{0ex}}b.$
3. Note that
$\frac{d}{dx}\text{ln}\left({x}^{r}\right)=\frac{r{x}^{r-1}}{{x}^{r}}=\frac{r}{x}.$

Furthermore,
$\frac{d}{dx}\left(r\phantom{\rule{0.2em}{0ex}}\text{ln}\phantom{\rule{0.2em}{0ex}}x\right)=\frac{r}{x}.$

Since the derivatives of these two functions are the same, by the Fundamental Theorem of Calculus, they must differ by a constant. So we have
$\text{ln}\left({x}^{r}\right)=r\phantom{\rule{0.2em}{0ex}}\text{ln}\phantom{\rule{0.2em}{0ex}}x+C$

for some constant $C.$ Taking $x=1,$ we get
$\begin{array}{ccc}\hfill \text{ln}\left({1}^{r}\right)& =\hfill & r\phantom{\rule{0.2em}{0ex}}\text{ln}\left(1\right)+C\hfill \\ \hfill 0& =\hfill & r\left(0\right)+C\hfill \\ \hfill C& =\hfill & 0.\hfill \end{array}$

Thus $\text{ln}\left({x}^{r}\right)=r\phantom{\rule{0.2em}{0ex}}\text{ln}\phantom{\rule{0.2em}{0ex}}x$ and the proof is complete. Note that we can extend this property to irrational values of $r$ later in this section.
Part iii. follows from parts ii. and iv. and the proof is left to you.

## Using properties of logarithms

Use properties of logarithms to simplify the following expression into a single logarithm:

$\text{ln}\phantom{\rule{0.2em}{0ex}}9-2\phantom{\rule{0.2em}{0ex}}\text{ln}\phantom{\rule{0.2em}{0ex}}3+\text{ln}\left(\frac{1}{3}\right).$

We have

$\text{ln}\phantom{\rule{0.2em}{0ex}}9-2\phantom{\rule{0.2em}{0ex}}\text{ln}\phantom{\rule{0.2em}{0ex}}3+\text{ln}\left(\frac{1}{3}\right)=\text{ln}\left({3}^{2}\right)-2\phantom{\rule{0.2em}{0ex}}\text{ln}\phantom{\rule{0.2em}{0ex}}3+\text{ln}\left({3}^{-1}\right)=2\phantom{\rule{0.2em}{0ex}}\text{ln}\phantom{\rule{0.2em}{0ex}}3-2\phantom{\rule{0.2em}{0ex}}\text{ln}\phantom{\rule{0.2em}{0ex}}3-\text{ln}\phantom{\rule{0.2em}{0ex}}3=\text{−}\text{ln}\phantom{\rule{0.2em}{0ex}}3.$

Use properties of logarithms to simplify the following expression into a single logarithm:

$\text{ln}\phantom{\rule{0.2em}{0ex}}8-\text{ln}\phantom{\rule{0.2em}{0ex}}2-\text{ln}\left(\frac{1}{4}\right).$

$4\phantom{\rule{0.2em}{0ex}}\text{ln}\phantom{\rule{0.2em}{0ex}}2$

## Defining the number e

Now that we have the natural logarithm defined, we can use that function to define the number $e.$

## Definition

The number $e$ is defined to be the real number such that

$\text{ln}\phantom{\rule{0.2em}{0ex}}e=1.$

To put it another way, the area under the curve $y=1\text{/}t$ between $t=1$ and $t=e$ is $1$ ( [link] ). The proof that such a number exists and is unique is left to you. ( Hint : Use the Intermediate Value Theorem to prove existence and the fact that $\text{ln}\phantom{\rule{0.2em}{0ex}}x$ is increasing to prove uniqueness.)

The number $e$ can be shown to be irrational, although we won’t do so here (see the Student Project in Taylor and Maclaurin Series ). Its approximate value is given by

$e\approx 2.71828182846.$

## The exponential function

We now turn our attention to the function ${e}^{x}.$ Note that the natural logarithm is one-to-one and therefore has an inverse function. For now, we denote this inverse function by $\text{exp}\phantom{\rule{0.2em}{0ex}}x.$ Then,

What is a independent variable
a variable that does not depend on another.
Andrew
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x-xcosx/sinsq.3x
Hasnain
x-xcosx/sin^23x
Hasnain
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1-sin x/cos x= cos x/-1+sin x
Rochel
how to prove 1-sun x/cos x= cos x / -1+sin x?
Rochel
how to prove tan^2 x=csc^2 x tan^2 x-1?
divide by tan^2 x giving 1=csc^2 x -1/tan^2 x, rewrite as: 1=1/sin^2 x -cos^2 x/sin^2 x, multiply by sin^2 x giving: sin^2 x=1-cos^2x. rewrite as the familiar sin^2 x + cos^2x=1 QED
Barnabas
how to prove sin x - sin x cos^2 x=sin^3x?
sin x - sin x cos^2 x sin x (1-cos^2 x) note the identity:sin^2 x + cos^2 x = 1 thus, sin^2 x = 1 - cos^2 x now substitute this into the above: sin x (sin^2 x), now multiply, yielding: sin^3 x Q.E.D.
Andrew
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navin
Left side=sinx-sinx cos^2x =sinx-sinx(1+sin^2x) =sinx-sinx+sin^3x =sin^3x thats proved.
Alif
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what is polynomial
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an expression of more than two algebraic terms, especially the sum of several terms that contain different powers of the same variable(s).
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a term/algebraic expression raised to a non-negative integer power and a multiple of co-efficient,,,,,, T^n where n is a non-negative,,,,, 4x^2
joe
An expression in which power of all the variables are whole number . such as 2x+3 5 is also a polynomial of degree 0 and can be written as 5x^0
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find volume of solid about y axis and y=x^3, x=0,y=1
3 pi/5
vector
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Is a rule used to find a derivative. For example the derivative of y(x)= a(x)^n is y'(x)= a*n*x^n-1.
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f(x)=7x-x g(x)=5-x
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5x-5
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difference btwn domain co- domain and range
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x
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The set of inputs of a function. x goes in the function, y comes out.
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If you differentiate then answer is not x
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domain is the set of values of independent variable and the range is the corresponding set of values of dependent variable
Champro
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give different types of functions.
Paul
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ok
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differentiate each term
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to understand how to model one variable as a direct relationship to another variable
Andrew