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

 Page 2 / 4

## 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,

#### Questions & Answers

What is a independent variable
a variable that does not depend on another.
Andrew
solve number one step by step
x-xcosx/sinsq.3x
Hasnain
x-xcosx/sin^23x
Hasnain
how to prove 1-sinx/cos x= cos x/-1+sin x?
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
take sin x common. you are left with 1-cos^2x which is sin^2x. multiply back sinx and you get sin^3x.
navin
Left side=sinx-sinx cos^2x =sinx-sinx(1+sin^2x) =sinx-sinx+sin^3x =sin^3x thats proved.
Alif
how to prove tan^2 x/tan^2 x+1= sin^2 x
Rochel
not a bad question
Salim
what is function.
what is polynomial
Nawaz
an expression of more than two algebraic terms, especially the sum of several terms that contain different powers of the same variable(s).
Alif
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
Nawaz
what is hyperbolic function
find volume of solid about y axis and y=x^3, x=0,y=1
3 pi/5
vector
what is the power rule
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.
Timothy
how do i deal with infinity in limits?
Add the functions f(x)=7x-x g(x)=5-x
f(x)=7x-x g(x)=5-x
Awon
5x-5
Verna
what is domain
difference btwn domain co- domain and range
Cabdalla
x
Verna
The set of inputs of a function. x goes in the function, y comes out.
Verna
where u from verna
Arfan
If you differentiate then answer is not x
Raymond
domain is the set of values of independent variable and the range is the corresponding set of values of dependent variable
Champro
what is functions
give different types of functions.
Paul
how would u find slope of tangent line to its inverse function, if the equation is x^5+3x^3-4x-8 at the point(-8,1)
pls solve it i Want to see the answer
Sodiq
ok
Friendz
differentiate each term
Friendz
why do we need to study functions?
to understand how to model one variable as a direct relationship to another variable
Andrew