Antiderivative of Log X

Introduction to anti-derivative of log x:

The natural logarithm is the logarithm to the base e, where e is an irrational constant approximately equal to 2.718. The natural logarithm is generally written as ln(x), loge(x) or sometimes, if the base of e is implicit, as simply log(x). Formally, ln(a) may be defined as the area under the graph of `1/x ` from 1 to a, that is as the anti-derivatives or integral,

ln a  = `int_1^a(1/x)dx`

This defines a logarithm because it satisfies the fundamental property of a logarithm:

ln (ab) = ln a + ln b

Source Wikipedia.

Anti-derivative Logarithmic Formulas:

1. `int` `(1/x)` dx = log x + c

2.  `int` e x dx = e x + c

3. `int (dx) / (a^2 - x^2) ` = `(1/(2a)) log [(a + x) / (a - x)] + c`

4. `int (dx) / (x^2 - a^2)` = `(1/(2a)) log [(x - a) / (x + a)] + c`

5. `int (dx) / sqrt(a^2 - x^2)` = `sin^-1(x / a) + c`

6. `int (dx) / sqrt(x^2 - a^2) ` = `log [(x + sqrt(x^2 - a^2)] + c`

Anti-derivative Logarithmic Problems:

Anti-derivative logarithmic problem 1:

Find the anti-derivative of given logarithmic function, log x  with respect to x

Solution:

Given logarithmic function, ` int ` log x. dx

Let,    u = log x                            dv = dx.

`(du)/(dx) ` = `1/x`                                    v = x

du = `1/x` dx

We know anti-derivative parts formula,  `int ` u dv = uv - `int ` v du

`int ` log x. dx  = log x . x - `int`` x ((dx) /x) `

= x. log x - `int` dx

= x. log x - x + c

= x( log x - 1) + c

Answer: Anti-derivative of log x is    x( log x - 1) + c

Anti-derivative logarithmic problem 2:

Find the anti-derivative of given logarithmic function, `(1 + 25x)/x^2`   with respect to x

Solution:

Given function, ` int` `(1 + 25x)/x^2` . dx

`int``(1 + 25x)/x^2`. dx  =` int` `dx/x^2` + `int` ` (25x)/x^2 ` dx            

= `int` `x^(-2) ` dx + ` int `` 25x^(-1)` dx               

= `x^(-1)` +  25 log x + c

Answer:  Anti-derivative of  `(1 + 25x)/x^2`  is x-1 + 25 log x + c

Anti-derivative logarithmic problem 3:

Find the anti-derivative of given logarithmic function, `e^(3x)/(1- e^(3x))` with respect to x

Solution:

Let u = 1- e3x           du = - `3` `e^(3x)` dx      

So, substitute the u and du

`int `  `e^(3x)/(1- e^(3x))` dx = `int`` (-1/3)(du)/u`                                 

= `(-1/3)` ` int` `1/u` du

= `(-1)/3` ln u + c                                                      we know u = 1- e3x

=` (- ln (1-e^(2x)))/3` +c

= `((-1)/3)` ln(e3x -1) + c

Answer:  Anti-derivative of `e^(3x)/(1- e^(3x))` is  `((-1)/3)` ln(e3x -1) + c

Subtracting Mixed Fractions

Introduction subtracting mixed numbers:

Natural number and a fraction is called mixed number. If numerator is less than denominator then it is said to be proper fraction. Improper fraction is formed when numerator is greater than denominator. For subtracting mixed numbers we have to first convert it in to proper fractions and then subtract the numbers so obtained.

Problem Subtracting Mixed Numbers with Different Denominators:

Pro 1 : 2 `3/5` - 3` 5/7`

Sol : Here the both are mixed fraction,

Step 1:  convert into the proper fraction and subtracting mixed numbers

2 `3/5` first we have to change into the proper numbers `(5xx2+3)/5` , we get

`13/5` is the proper numbers

3 `5/7=` `[(7xx3)+5]/7` = `26/7 ` is the proper number

Here we round mixed numbers and adding both functions

`13/5` -` 26/7`

In this step subtracting mixed numbers taking the Least Common Multiplier of 5 and 7 is 35. `[(13 xx 7) - (26 xx 5)]/(35)`  = `(91-130)/35 = (-39)/(35) `= -1.1142

Problem Subtracting Mixed Numbers with same Denominator:

Ex 1:      2 `3/4` - 3 `5/4`

Sol:  Here the fraction 4 is the equivalent fraction mixed number, subtracting both mixed numbers

2 `3/4` first we have to change into the proper numbers `(4xx2+3)/4` we get

`11/4 ` is the proper numbers

3 `5/4 = [(4xx3)+5]/(4) = (17)/(4)`

Here we round mixed numbers and adding both functions

`11/4 ` - `17/4 = (11-17)/(4) = (-6)/(4) = (-3/2) = -1.5`

Tangent Function Graph

Introduction to tangent function graph:
In geometry, the tangent line (or simply the tangent) to a curve at a given point is the straight line that "just touches" the curve at that point (in the sense explained more precisely below). As it passes through the point where the tangent line and the curve meet, or the point of tangency, the tangent line is "going in the same direction" as the curve, and in this sense it is the best straight-line approximation to the curve at that point. The same definition applies to space curves and curves in n-dimensional Euclidean space.

Source: Wikipedia.


Definitions of a Trigonometry Tangent Function:

Definition of tangent function is defined by using the right angle triangle

Tangent of an angle is the ratio of length of the adjacent and the opposite side.
tan x = opposite / adjacent

The tangent is the inverse of the cotangent and is given by
tan x = 1 / cot x

Quotient of sine and cosine functions is called as tan or tangent function.
tan x = sin x / cos x

Properties of a Tangent Function Using the Graph



Example Problem for Tangent of a Graph:

Example 1:

Find the function value of tan 35 o.

Solution:       

Use the tangent's co function identity to solve the problem.

tan x = cot (90o – x)
tan 35o = cot (90o – 35o)
= cot (55o)
tan 35o = 0.70021
The tangent of 60o is 0.70021.

Example 2:

Find the hypotenuse of the right angle triangle using the Pythagorean Theorem and find the tangent of an angle value.

Solution:

Use the Pythagorean Theorem, for finding the hypotenuse

In the given right angle triangle

AC2 = AB2 + BC2

Here,

AB = Opposite side

BC = Adjacent side

AC = Hypotenuse

AC2 = AB2 + BC2

= 42 + 92

= 16 + 81

AC2 = 97

AC = 9.84

Hypotenuse for the given right angle triangle is 9.84

Tangent function:

Tan ? = Sin ?/ Cos ?

= adj / opp

= 4 / 9

Tan ? = 4 / 9

? = tan -1(4/9)

? = 23.96 °

Solving Distance Formula

Introduction to solving distance formula:
Distance formula is a formula which is used to find the distance between two points (x1, y1) and (x2, y2). This formula is based on the Pythagorean Theorem.Solving problems using distance formula is one of the easier method to find the distance between two given points.

The distance d between the points (x1, y1) and (x 2, y2) is given as,

d = v ((x2 - x1)2 + (y2 - y1)2)

Example Problems on Solving Distances Using Distance Formula:


1) Find the distance between the points (-2, -3) and (-4, 4).

Solution:

Let (x1, y1) = (-2, -3)

(x2, y2) = (-4, 4)

By distance formula, d = v ((x2 - x1)2 + (y2 - y1)2)

=  v (((-4 - (-2))2 + (4 - (-3))2)

=  v ((-2)2  + (7)2)

=  v (4 + 49)

= v53

Solving for d, we get d  ˜ 7.28

2) Find the distance between the points (2, 4) and (5, -1).

Solution:

Let (x1, y1) = (2, 4)

(x2, y2) = (5, -1)

By distance formula, d = v ((x2 - x1)2 + (y2 - y1)2)

=  v ((5 - 2)2 + (-1-4)2)

=  v (32 + (-5)2)

=  v (9 + 25)

= v 34

Solving for d, we get d ˜ 5.83

3) Find the distance between the points (-2, 1) and (1, 5).

Solution:

Let (x1, y1) = (-2, 1)

(x2, y2) = (1, 5)

By distance formula, d = v ((x2 - x1)2 + (y2 - y1)2)

=  v ((1 - (-2))2 + (5 - 1)2

=  v (32 + 42)

=  v (9 + 16)

d  = v 25

Solving for d, we get   d = 5

4) If the distance from the point is (1, 2) to the point (3,y) is v8.  Find the value of  y.

Solution:

Let (x1, y1) = (1, 2)

(x2, y2) = (3, y)

d = v8

By distance formula, d = v ((x2 - x1)2 + (y2 - y1)2)

v8 = v ((3 - 1)2 + (y - 2)2

8 = 22 + (y - 2)2

8 = 4 + (y - 2)2

8 - 4 = (y 2 - 4y + 4)

4 = y 2 - 4y + 4

0 =  y 2 - 4y

0 = y (y - 4)

Solving for y. we get

y = 0 ;   y = 4
   


Solve the Problems Using Distance Formula:

Find the distance between the following points,

(1, 4) and (4, 0)

(2, 8) and  (16, 4)

(8, 5) and (9, 6)

(5, 6) and (-12, 40)

Solutions:

d = 5
d = 14.56
d = 1.41
d = 38.01

Tangent and Secant of a Circle

Introduction to tangent and secant of a circle:

A tangent of a circle is a line drawn from a point passing through the circle just at one point and secant is a line passing through two points of a circle. In case, a tangent and a secant are drawn from the same point outside the circle, we can find an interesting relation. Similar is the case when two secants are drawn from the same point outside the circle.

Let us study these two situations.

Tangent and Secant of a Circle – a Tangent and a Secant

Look at the above diagram.

A tangent OT is drawn from O touching the circle at P. From the same point O, a secant OS is drawn passing through the points Q and R on the circle. Join PR and PQ.

The angle OPQ and ORP are congruent as both of them are subtended by the intercepted arc PQ.

Angle POR is subtended both by the chords PQ and PR at O. Therefore the triangles OPQ and ORP are similar.

Applying the rule of similarity,

OP/OR = OQ/OP

or,  OP2 = OQ*OR



Tangent and Secant of a Circle – Two Secants

Look at the above diagram.

Two secants are drawn from the same point O to the circle. One secant passes through the points A and B on the circle and the other passes through the points C and D on the circle. Join BC and DA.

The angles BAD and BCD are subtended by the same arc BD on the circumference on the circle. Hence these two angles are congruent. The angle at O is common to the triangle AOD and BOC. Therefore, the triangles AOD and BOC are similar.

Applying the rule of similarity,

OC/OA = OB/OD

or,   OA*OB = OC*OD 

The two properties derived are very useful in solving problems related circles with tangents and secants.

Adding Negative Exponents

Introduction :

The term exponent in math is used to find the exponential value of the particular value it may be integer  or fraction . We can easy to get  the power of a  number using online calculator ,Consider  the unknown number , Here x is base and y is the power  of x

Adding Exponents means how many times to divide the number .

that is , `a ^(-n) = 1/a^n`

Steps for Adding Negative Exponents :

Negative exponents are added in the same way as the exponents are added with just a negative sign.

The given terms exponents are combined in a way such that the negative exponents are added or combined in case of same base.
In case of different bases , we have to simplify by convert it to positive exponent ,
Then we have to simplify the exponent value . Make the negative exponent to the positive one .
Now simplify further to get the results.
Ex  :   `5^-2 + 5^-4 = 5^((-2-4))`

`=5^-6`

=`1/5^6`

=`1/15625`

= 0.000064



Examples to Add Negative Exponents:

Ex 1:  A number  with negative exponents `8^(-3)`

Sol :     `8^(-3)` = `1/ (8^3)`

=`1/ (8 *8*8)`

= `1/ 512`

=0.00195

Ex 2:    Add the neagtive exponents` 7^-3`    + `5^-2`

Sol :    `=1/7^3 + 1/5^2`

`=1/343 +1/25`

=0.0029 +0.04

=0.0429

Ex 3:   Add the negative exponents `6^-2 + 6^-3`

Sol :  Using the rule , we have to add the exponents as 6 is the common base

= `6^(-2 + -3)`

` 6^(-2-3)`

=`6^-5`



= `1/(6^5)`

= 0.00012

Example 4:

Add the negative exponents  of `5^-3 +5^-5 +5^-2`

Solution:

Use the property to add the negative exponents , we have

=`5^-3 +5^-5 +5^-2`

=`5^(-3 + -5 + -2)`

=`5^(-3 -5 -2)`

= `5^-10`

= `1/(5^10)`

=0.04832

Example 5:

Add the negative exponents  of` 3^-3 + 3^-6`

Solution:

Use `a^-n = 1/a^n`   rule

We  get ,

=`1/3^3 +1/3^6`

=0.03703 +0.0013

=0.03833

Example 6:

Add the negative exponents  of `5^-2 + 5^-3`

Solution:

Use a ^-n = 1/a^n

Then we  get ,

=`1/5^2 +1/5^3` 

=0.04+0.008

=0.048

Practice Problem to Add Negative Exponents:

Pro1 : Add the negative exponents  of` 2^-2 +2^-4`

Ans : 0.3125

Pro2 :Add the negative exponents  of` 3^-3 +3^-2`

Ans : 0.1481

Selecting on the Dependent Variable

Introduction about selecting on the dependent variable:

Independent variable is a variable which does not depends on any other variable. But the dependent variable should depend only on independent variable. The dependent variable’s value depends on direct or indirect variation. In direct variation if the value of independent variable increases then the value of dependent variable also increases and if independent variable decreases then the value of dependent variable also decreases. Here we are going to learn about some example problems of selecting on the dependent variable.

Looking out for more help on Confounding Variable in algebra by visiting listed websites.

Simple Example Problems of Selecting on the Dependent Variable.

Example 1:

What is the dependent variable in the function f(x) =4-x?

Solution:

In the function f(x) = 4 - x, the value of f(x) depends on the value of x. So, f(x) is dependent variable.

Example 2:

What is the dependent variable in the function f(x) =10-x?

Solution:

In the function f(x) = 10- x, the value of f(x) depends on the value of x. So, f(x) is dependent variable.

These are the examples of selecting on the dependent variable.



Some more Examples of Selecting on the Dependent Variable:

Example 3:

Check whether the d variable dependent or not in the equation d=8b

Solution:

Step 1: The given equation is d=8b.Here =d is dependent variable and b is independent variable.

Step 2: Now plug different values for b and check the value of d for each value of b.

Step 3: When b = 1 =the value of d is 8.

Step 4: When b = 2 =the value of d is 16.

Step 5: When b = 3 =the value of d is 24.

Step 6: When b = 4 =the value of d is 32.

Step 7: When b = 5 =the value of d is 40.

Step 8: If the value of b increases then the value of d is also an increase.

Step 9: So, d is a dependent variable.

These are the example problems selecting on the dependent variable.