# 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,

questions solve y=sin x
Solve it for what?
Tim
you have to apply the function arcsin in both sides and you get arcsin y = acrsin (sin x) the the function arcsin and function sin cancel each other so the ecuation becomes arcsin y = x you can also write x= arcsin y
Ioana
what is the question ? what is the answer?
Suman
there is an equation that should be solve for x
Ioana
ok solve it
Suman
are you saying y is of sin(x) y=sin(x)/sin of both sides to solve for x... therefore y/sin =x
Tyron
or solve for sin(x) via the unit circle
Tyron
what is unit circle
Suman
a circle whose radius is 1.
Darnell
the unit circle is covered in pre cal...and or trigonometry. it is the multipcation table of upper level mathematics.
Tyron
what is function?
A set of points in which every x value (domain) corresponds to exactly one y value (range)
Tim
what is lim (x,y)~(0,0) (x/y)
limited of x,y at 0,0 is nt defined
Alswell
But using L'Hopitals rule is x=1 is defined
Alswell
Could U explain better boss?
emmanuel
value of (x/y) as (x,y) tends to (0,0) also whats the value of (x+y)/(x^2+y^2) as (x,y) tends to (0,0)
NIKI
can we apply l hospitals rule for function of two variables
NIKI
why n does not equal -1
Andrew
I agree with Andrew
Bg
f (x) = a is a function. It's a constant function.
proof the formula integration of udv=uv-integration of vdu.?
Find derivative (2x^3+6xy-4y^2)^2
no x=2 is not a function, as there is nothing that's changing.
are you sure sir? please make it sure and reply please. thanks a lot sir I'm grateful.
The
i mean can we replace the roles of x and y and call x=2 as function
The
if x =y and x = 800 what is y
y=800
800
Bg
how do u factor the numerator?
Nonsense, you factor numbers
Antonio
You can factorize the numerator of an expression. What's the problem there? here's an example. f(x)=((x^2)-(y^2))/2 Then numerator is x squared minus y squared. It's factorized as (x+y)(x-y). so the overall function becomes : ((x+y)(x-y))/2
The
The problem is the question, is not a problem where it is, but what it is
Antonio
I think you should first know the basics man: PS
Vishal
Yes, what factorization is
Antonio
Antonio bro is x=2 a function?
The
Yes, and no.... Its a function if for every x, y=2.... If not is a single value constant
Antonio
you could define it as a constant function if you wanted where a function of "y" defines x f(y) = 2 no real use to doing that though
zach
Why y, if domain its usually defined as x, bro, so you creates confusion
Antonio
Its f(x) =y=2 for every x
Antonio
Yes but he said could you put x = 2 as a function you put y = 2 as a function
zach
F(y) in this case is not a function since for every value of y you have not a single point but many ones, so there is not f(y)
Antonio
x = 2 defined as a function of f(y) = 2 says for every y x will equal 2 this silly creates a vertical line and is equivalent to saying x = 2 just in a function notation as the user above asked. you put f(x) = 2 this means for every x y is 2 this creates a horizontal line and is not equivalent
zach
The said x=2 and that 2 is y
Antonio
that 2 is not y, y is a variable 2 is a constant
zach
So 2 is defined as f(x) =2
Antonio
No y its constant =2
Antonio
what variable does that function define
zach
the function f(x) =2 takes every input of x within it's domain and gives 2 if for instance f:x -> y then for every x, y =2 giving a horizontal line this is NOT equivalent to the expression x = 2
zach
Yes true, y=2 its a constant, so a line parallel to y axix as function of y
Antonio
Sorry x=2
Antonio
And you are right, but os not a function of x, its a function of y
Antonio
As function of x is meaningless, is not a finction
Antonio
yeah you mean what I said in my first post, smh
zach
I mean (0xY) +x = 2 so y can be as you want, the result its 2 every time
Antonio
OK you can call this "function" on a set {2}, but its a single value function, a constant
Antonio
well as long as you got there eventually
zach
2x^3+6xy-4y^2)^2 solve this
femi
moe
volume between cone z=√(x^2+y^2) and plane z=2
Fatima
It's an integral easy
Antonio
V=1/3 h π (R^2+r2+ r*R(
Antonio
How do we find the horizontal asymptote of a function using limits?
Easy lim f(x) x-->~ =c
Antonio
solutions for combining functions
what is a function? f(x)
one that is one to one, one that passes the vertical line test
Andrew
It's a law f() that to every point (x) on the Domain gives a single point in the codomain f(x)=y
Antonio
is x=2 a function?
The
restate the problem. and I will look. ty
is x=2 a function?
The