# 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 & Answers

Find the arc length of the graph of f(x) = In (sinx) on the interval [Π/4, Π/2].
mukul Reply
Sand falling freely from a lorry form a conical shape whose height is always equal to one-third the radius of the base. a. How fast is the volume increasing when the radius of the base is (1m) and increasing at the rate of 1/4cm/sec Pls help me solve
ade
show that lim f(x) + lim g(x)=m+l
BARNABAS Reply
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Differentiation and integration
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terms of an AP is 1/v and the vth term is 1/u show that the sum of uv terms is 1/2(uv+1)
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the function at x
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also known as the y value so I could say y=2x or f(x)= 2x same thing just using functional notation your next question is what is dependent and independent variables. I am Dyslexic but know math and which is which confuses me. but one can vary the x value while y depends on which x you use. also
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Marc
I actually wanted to ask another questions on sets if u dont mind please?
Inembo
I have so many questions on set and I really love dis app I never believed u would reply
Inembo
Hmm go ahead and ask you got me curious too much conversation here
Adri
am sorry for disturbing I really want to know math that's why *I want to know the meaning of those symbols in sets* e.g n,U,A', etc pls I want to know it and how to solve its problems
Inembo
and how can i solve a question like dis *in a group of 40 students, 32 offer maths and 24 offer physics and 4 offer neither maths nor physics , how many offer both maths and physics*
Inembo
next questions what do dy mean by (A' n B^c)^c'
Inembo
The sets help you to define the function. The function is like a magic box where you put inside stuff(numbers or sets) and you get out the stuff but in different shapes (forms).
Adri
I dont understand what you wanna say by (A' n B^c)^c'
Adri
(A' n B (rise to the power of c)) all rise to the power of c
Inembo
Aaaahh
Adri
Ok so the set is formed by vectors and not numbers
Adri
A vector of length n
Adri
But you can make a set out of matrixes as well
Adri
I I don't even understand sets I wat to know d meaning of all d symbolsnon sets
Inembo
Wait what's your math level?
Adri
High-school?
Adri
yes
Inembo
am having big problem understanding sets more than other math topics
Inembo
So f:R->R means that the function takes real numbers and provides real numer. For ex. If f(x) =2x this means if you give to your function a real number like 2,it gives you also a real number 2times2=4
Adri
pls answer this question *in a group of 40 students, 32 offer maths and 24 offer physics and 4 offer neither maths nor physics , how many offer both maths and physics*
Inembo
If you have f:R^n->R^n you give to your function a vector of length n like (a1,a2,...an) where all a1,.. an are reals and gives you also a vector of length n... I don't know if i answering your question. Otherwise on YouTube you havr many videos where they explain it in a simple way
Adri
I would say 24
Adri
Offer both
Adri
Sorry 20
Adri
Actually you have 40 - 4 =36 who offer maths or physics or both.
Adri
I know its 20 but how to prove it
Inembo
You have 32+24=56who offer courses
Adri
56-36=20 who give both courses... I would say that
Adri
solution: In a question involving sets and Venn diagram, the sum of the members of set A + set B - the joint members of both set A and B + the members that are not in sets A or B = the total members of the set. In symbolic form n(A U B) = n(A) + n (B) - n (A and B) + n (A U B)'.
Mckenzie
In the case of sets A and B use the letters m and p to represent the sets and we have: n (M U P) = 40; n (M) = 24; n (P) = 32; n (M and P) = unknown; n (M U P)' = 4
Mckenzie
Now substitute the numerical values for the symbolic representation 40 = 24 + 32 - n(M and P) + 4 Now solve for the unknown using algebra: 40 = 24 + 32+ 4 - n(M and P) 40 = 60 - n(M and P) Add n(M and P), as well, subtract 40 from both sides of the equation to find the answer.
Mckenzie
40 - 40 + n(M and P) = 60 - 40 - n(M and P) + n(M and P) Solution: n(M and P) = 20
Mckenzie
thanks
Inembo
Simpler form: Add the sums of set M, set P and the complement of the union of sets M and P then subtract the number of students from the total.
Mckenzie
n(M and P) = (32 + 24 + 4) - 40 = 60 - 40 = 20
Mckenzie
how do i evaluate integral of x^1/2 In x
ayo Reply
first you simplify the given expression, which gives (x^2/2). Then you now integrate the above simplified expression which finally gives( lnx^2).
Ahmad
by using integration product formula
Roha
find derivative f(x)=1/x
Mul Reply
-1/x^2, use the chain rule
Andrew
f(x)=x^3-2x
Mul
what is domin in this question
noman
all real numbers . except zero
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Roha
?
Roha
the domain of the function is all real number excluding zero, because the rational function 1/x is a representation of a fractional equation (precisely inverse function). As in elementary mathematics the concept of dividing by zero is nonexistence, so zero will not make the fractional statement
Mckenzie
a function's answer/range should not be in the form of 1/0 and there should be no imaginary no. say square root of any negative no. (-1)^1/2
Roha
domain means everywhere along the x axis. since this function is not discontinuous anywhere along the x axis, then the domain is said to be all values of x.
Andrew
Derivative of a function
Waqar
right andrew ... this function is only discontinuous at 0
Roha
of sorry, I didn't realize he was taking about the function 1/x ...I thought he was referring to the function x^3-2x.
Andrew
yep...it's 1/x...!!!
Roha
true and cannot be apart of the domain that makes up the relation of the graph y = 1/x. The value of the denominator of the rational function can never be zero, because the result of the output value (range value of the graph when x =0) is undefined.
Mckenzie
👍
Roha
Therefore, when x = 0 the image of the rational function does not exist at this domain value, but exist at all other x values (domain) that makes the equation functional, and the graph drawable.
Mckenzie
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Roha
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