Study Online Second Derivatives

Study online second derivatives involves the process differentiating the given polynomial function twice with respect to the given variables whereas all the process is clearly explained with the help of online. Generally the second derivative is discussed in calculus whereas it is mainly helps to find the rate of change of the given function with respect to the change in the input. The following are the solved example problems with detailed step by step solution for second derivatives study discussed in online.

Example 1:

Determine the second derivative from the polynomial.

f(b) = 5b 2 +5b 4  + 12

Solution:

The given function is

f(b) = 5b 2 +5b 4  + 12

The above function is differentiated with respect to b to find the first derivative

f '(b) = 5(2b  )+5(4 b 3 ) + 0

By solving above terms

f '(b) = 10b +20b3

The above function is again differentiated with respect to b to find second derivative

f ''(b) =  10(1 ) +20(3b2)

f ''(b) =  10 + 60b2 is the answer.

Example 2:

Determine the second derivative from the polynomial.

f(b) = 4b4 +5b 5 +6b 6  + 2b

Solution:

The given function is

f(b) = 4b4 +5b 5 +6b 6  + 2b

The above function is differentiated with respect to b to find the first derivative

f '(b) = 4(4b 3 )+5(5b 4 ) +6( 6b 5) +2

By solving above terms

f '(b) = 16b 3 +25b 4 +36 b 5 + 2

The above function is again differentiated with respect to b to find second derivative

f ''(b)= 16(3b 2) +25(4b 3) +36 (5b 4)

f ''(b)= 48b 2 +100b 3 +180b 4 is the answer.

Example 3:

Determine the second derivative from the polynomial.

f(b) = 2b6 + 2 b5 + 3 b4 + 3b

Solution:

The given equation is

f(b) = 2b6 + 2 b5 + 3 b4 + 3b

The above function is differentiated with respect to b to find the first derivative

f '(b) =  2(6b 5)  +2 (5 b4 ) +3(4 b3) + 3

By solving above terms

f '(b) =  12b 5  +  10b4  + 12 b3 – 3

The above function is again differentiated with respect to b to find second derivative

f ''(b) =  12(5b 4 ) – 10(4b3)  + 12(3b2)

f ''(b) =  60b 4 – 40b3 +36b2  is the answer.

Online second derivatives practice problems for study:

1) Determine the second derivative from the polynomial.

f(b) = b 3 + b 4 + b 5

Answer: f ''(b) = 6b +12b2+ 20b 3

2) Determine the second derivative from the polynomial.

f(b) = 2b 3+3b5 + 4b 6

Answer: f ''(b) = 12b + 60b3 + 120 b 4

Solving Negative Number

Definition:

Negative number is defined as the number which indicate by negative sign or minus ('-') . Negative number is less then zero and placed left to zero.

Ex: ...-5,-4,-3,-2,-1,0,1,2,3,4...

Comparison of Positive and Negative:

For each negative number , there is a positive number that is its opposite . Here we can write the opposite of negative number with a positive of same number or plus sign used In front of the number and call these numbers are positive numbers. Ex : 1,2,3 ,.....positive numbers are grater than zero. Similarly, the opposite of any positive number is a negative number .

Ex:    1,2,3 is -1,-2,-3.

Solving examples for negative numbers:

Zero cannot be taken as a negative number or positive number.
For every positive number x, there exists a negative number y such that x + y = 0
Positive number is denoted as plus ('+')sign and negative number is denoted as minus sign('-').
Example of negative number:-2,-43,-34 and example for positive number is 2,43,34.
negative and positive number may be written as mixed numbers or fraction numbers.

The equal fraction of negative numbers are given bellow:

(-3)/7,3/(-7),-(3/7) and -3/7 .

The equal mixed numbers are given bellow.

-2/5,-(2/5)       (-4)/9, 4/-(9) and -4/9

More about solving negative numbers:

Solving Addition of Negative Numbers:

To add the negative numbers which consist of minus sign. To provide the answer of addition of negative number.

Solving examples for addition negative numbers:

-12+(-6)=?

Solution:

-12+(-6)= -18.

Here the values of  -12 and -6are 12 and 6 adding the smaller from the larger gives -12+(-6)= -18, and since the larger  value was 12, so we can give result the same sign as -12, '-' so -12+(-6)= -18.

Example:

(-6) + (-6) = ?

Here the absolute values of -6 and -6 are 6 and 6.  Adding the smaller from the larger gives -6 - 6 = 12 ,  but here both has same value . In this case sign is the matter, here 12 and -12 are the not same and then -6 and -6 are same numbers. The property of all same number sum is 12. The addition of two number up to zero are called as additive inverses.

Multiplying Negative Numbers:

Solving example for multiplying negative numbers :

Product of negative number ,here we can take the product of their values.

(-3.3) × (-5) = ?


(-3.3) × (-5) = (-3.3) × (-5)

= (-3.3) × (-5)

=  +16.5.
Dividing Negative Numbers

Solving example for dividing negative number:

To divide two negative numbers, here we can divide the value of the first by the value of the second.

(-1.6) ÷ (-4) = (-1.6) ÷ (-4)

= (-1.6) ÷ (-4)

=  -0.4

Geometry Without Common Vertices

In geometry, some figures have common vertices. Mostly the geometric figures are without common vertices. If a triangle, quadrilateral and some geometric figures in a graph are lying without common vertices are refered same. In graph there are four quadrants in that some vertices are fall on common vertices, with out common vertices points of are plotted.

Geometric figure without common vertices

Without common vertices find the distance of two points

In geometry the triangle has three vertices; the vertices are not common vertices. The common vertices are formed only if two triangles are in same point without three common vertices. The distance between two un common vertices are find out by using the coordinates of the vertices (x1,x2) and (x2,x2) of the vertices.

Distance between two vertices = √(x2-x1)2 +(y2-y1)2

If the geometry figure having the common vertices in a graph



Examples for without common vertices

Distance of a vertices are find using distance formula:

Examples for distance between two vertices:

Ex 1:   Find the distance formed by without common vertices, Vertices A(4,5), B(7,4)

Sol :            Distance AB = √(x2-x1)2 +(y2-y1)2

X1=4    X2=7    Y1=5    Y2=4

AB = √(7-4)2+(4-5)2

= √(3)2+ (-1)2

= √9+1

= √10 units

Ex 2 :  Find the distance formed by without common vertices, Vertices A(3,2), B(5,4)

Sol :            Distance AB = √(x2-x1)2 +(y2-y1)2

X1=3    X2=5    Y1=2    Y2=4

AB = √(5-3)2+(4-2)2

= √(2)2+ (2)2

= √4+4

= √8 units

Ex 3 :      Find the distance formed by without common vertices, Vertices A(6,4), B(10,8)

Sol :            Distance AB = √(x2-x1)2 +(y2-y1)2

X1=6    X2=10    Y1=4    Y2=8

AB = √(10-6)2+(8-4)2

= √(4)2+ (4)2

= √16+16

= √32 units

Ex 4:    Find the distance formed by without common vertices, Vertices A(3,3), B(4,8)

Sol :            Distance AB = √(x2-x1)2 +(y2-y1)2

X1=3    X2=4    Y1=3    Y2=8

AB = √(4-3)2+(8-3)2

= √(1)2+ (5)2

= √1+25

= √26 units

Practice problems:

Q 1   Find the distance of two vertices A(1,1) B(1,2)   Answer: √1 units

Q 2   Find the distance of two vertices A(2,2) B(1,1)   Answer: √2 units

Straight Line Equation

The equation of straight line is generally written as    y = mx + b

Graphical Representation

where, m= slope or gradient of the straight line equation

b = the y-intercept

Suppose we want to find equation of a straight line that passes through a known point and has a known slope. Let (x ,y) represent the co-ordinates of any point on the line and let (x1 ,y1) represent the co-ordinates of other point. The slope of the straight line equation is given as,

m = (y-y_1)/(x-x_1)

After finding the slope m as we are given the co-ordinates of the point (x1 ,y1) in the equation    y= mx + b Then, the constant 'b' can also be found so that finally the straight line equation is obtained.

Other Forms of straight Line Equations

Different Forms of Straight Line Equations:

There are many other forms of Straight Line Equations.

1. Straight Line Equation through two points:

The line through two different points ( x1 ,y1) and ( x2 ,y2) is given by

y-y1  = [(y_2-y_1)/(x_2-x_1)]  . (x - x1)

2. Straight Line Equation in general form:

A straight line is defined by a linear equation as

Ax+By+C=0     where A , B are not both 0

3. Straight Line Equation in Intercept-intercept form:

Let us consider that a straight line intersects x-axis at (a , 0) and y-axis at (0 , b). Then it is defined by equation

(x/a) +(y/b) = 1

4. Straight Line Equation in Point-Slope Form:

The equation of straight line through the point (a , b) with slope m is

y = m ( x - a ) + b

Solved Examples on equations of straight line

Ex:1 Find the equation of a line passing through (2 , 3) and having a slope of 3?

sol: step 1: Compare the given point(2 , 3) with the general point (x1 , y1)

Now x1=2  and  y1=3

step 2: we have the straight line equation as  y-y1=m(x-x1)

now substitute the given point in the above equation

y-3=3(x-2)

y-3=3x-6

3x-y=6-3

3x-y-3=0

The straight line equation is 3x-y-3=0

Ex:2 Find out the straight line equation of the line passing through the points (1,2) and (2,4).

Sol: Given points are compared with (x1 , y1) and (x2 , y2) and substitute the points in the equation

y-y1  = [(y_2-y_1)/(x_2-x_1) ] . (x-x1)

y-2  =  [(4-2)/(2-1)] . (x-1)

y-2 = 2(x-1)

y-2 = 2x-2

2x - y = 0

The straight line equation is 2x-y=0.

Practice Problems on equations of straight line

Pro:1 Find the equation of the line passing through the points (-3,4) and (4,-2)

Ans: we have      

m = (y_2-y_1)/(x_2-x_1)

=(-2-4)/(4+3)

=(-6/7)

let (x,y) be compared to (-3,4)

y-4 = (-6/7) . (x-(-3))

7(y-4) = -6(x+3)

7y-28 = -6x-18

7y-28 = -6x-18

7y+6x = 10

Pro:2 Write equation of line having points and slopes as follows;

P(3,3) , m=-2

P(-2,-1) , m=1/3

P1(2,2) and P2(-4,-1)

y-intercept = 2 , m=3

Ans: The answers to the above given practice problems are

y+2x=11

3y-x=-1

2y-x=2

y-3x=2

Hyperbola Axis

The locus of a point whose distance from a point rest at center shows an unaltered ratio, higher than one to its length from the fixed line is known as hyperbola. In this article, you can learn about the axis of hyperbola, general equation of the hyperbola and definitions regarding hyperbola.

Axis of Hyperbola:

Transverse Axis:

Transverse axis can be defined as the line segment, which joins the vertices of the hyperbola. 2a is calculated as the difference between two vertices. The equation of the transverse axis is y = 0. This is showing that y-coordinate is zero.

Conjugate Axis:

Conjugate axis can be defined as the line segment, which joins the y-coordinates of the hyperbola. The distance between the two coordinates are 2a (i.e., the length of the Conjugate axis is 2a.). x = 0 is considered as the equation of conjugate axis. Therefore the x-coordinates of Conjugate axis is zero.

General Equation of the Hyperbola:

Some important points to be considered:

Fixed point is represented as F.
Fixed line as l.
Eccentricity as e, then it should be greater 1.
Moving point is represented as P(x,y).

Steps:

Fixed point F is plotted and also the fixed line ‘l’ is drawn.
Perpendicular (FZ) is dropped from F to l.
As next step drop PM which is perpendicular one from P to l.
Plot the points A, A’ which divides FZ internally and externally in the ratio e : 1            
respectively.
Take AA’ = 2a and treat it as x-axis.
Draw a perpendicular bisector of AA’ and treat it as y-axis.
Consider C as the origin, then the known points are C(0,0), A(a,0) and A’(-a,0).
The general equation of the hyperbola x^2/a^2 -y^2/b^2 = 1.

Definitions Regarding Hyperbola:

Focus:

The fixed point F is known as the focus of the hyperbola.

Directrix:

The directrix is nothing but the fixed line. Then the directrix equation is given by x = a/e.

Centre:

The centre of the hyperbola is a point, at which the transverse and conjugate axes intersect and they are represented as ‘C’.

Vertices:

The vertices of the hyperbola are the points, where the curve and its transverse axis meet. The vertices are A(a,0) and A’(-a,0).

Studying Standard Deviation Examples

The standard deviation of a statistical population, a data set, or a probability distribution is the square root of its variance. Standard deviation is a widely used measure of the variability or dispersion, being algebraically more tractable though practically less robust than the expected deviation or average absolute deviation. It shows how much variation there is from the "average" (mean) (or expected/budgeted value). It helps detect tampering of data. Examples for studying standard deviation is given below.

Formula for studying standard deviation examples:

Formula for studying standard deviation examples are defined below:

Measurement of standard deviation is prepared by taking square root for the addition of mean difference with the certain data divided by the total number of values subtracted by one. The following formula for standard deviation as shows given below.

s=v?(X-M)2/n-1

Here S = Sum of values

X = Individual value

M = Mean of total all value

N = Sample size (Total number of values)

Variance:

Variance = s2

Steps for calculating Standard Deviation examples:

• Step 1: calculate the average for given n numbers using the formula this is called mean of given numbers

• Step 2: Find distance between each given numbers in the Data set from the calculated average value. This is called  "deviation" from the mean value.

• Step 3: Take the Square of each deviation value found from mean. This is squared deviation from mean.

• Step 4: Calculate the sum for all the squared standard deviations.

• Step 5: Now apply the Standard Deviation formula and find standard deviation formula. It will be the square root of variance.

Examples for Studying Standard deviation:

Examples for Studying Standard deviation are as follows:

Pro 1:   Here are 4 measurements 4, 6, 7, 9 and 10Calculate the Standard Deviation

Sol :     Mean: Calculate the average for the given values. To find the mean.
4+6+7+9+ 10
x =   ------------------------
5-1

= 36/4

= 9

Standard Deviation,
v (4-9)2 + (6-9)2 + (7-9)2 + (9-9)2 + (10-9)2
S=  -------------------------------------------------------------
5-1

= v 81 /4

= v 20.25

=  4.5

Standard Deviation  S = 2.12132

Pro 2:  Here are 4 measurements 10, 20, 30, 40 and 50 Calculate the Standard Deviation

Sol :     Mean: Calculate the average for the given values. To find the mean.
10+20+30+40+60
x = ---------------------------
5-1

= 160/4

= 40

Standard Deviation,
v (10-40)2 + (20-40)2 + (30-40)2 + (40-40)2 + (60-40)2
S=  -----------------------------------------------------------------------------
5-1

= v 1600 /4

= v 400

S = 20

Standard Deviation S = 20

Absolutely Continuous Function

DEFINITION

Let [p,q] be the close bounded interval of C. Then a function f:[p,q]→ R will be an absolutely continuous function on [p,q], if for any δ>0 there will be a ε>0 such that the certain conditions which are mentioned below holds good

If (p1q1)..............(pnqn) is a collection which is finite with disjoint open intervals in [p,q] such that

Σni=1 (qi-pi) < ε

and

Σni=1 |f(qi)-f(pi)| < δ



EQUIVALENT DEFINITIONS

The condition on a real-valued function  "  f  "  on the compact interval [ p, q ] are equivalent if

1) f is an absolutely continuous function

2) ' f ' has a derivative f1 almost everywhere which is a Lebesgue integral and

f ( x ) = f ( p ) +  ∫x a  f1  ( c ) dt

for all x on [ p , q ]

3)There exists a Lebesgue integrable function such that g on [ p , q ] such that f ( x ) = f ( p ) + ∫x p  g ( c ) dt

for all x on [ p , q ]

If these conditions are satisfied by the function the definitely g = f' almost everywhere

Properties of the absolute continuous function

PROPERTIES OF ABSOLUTELY CONTINUOUS FUNCTION

1. The sum and difference of two absolute continuous function are also absolutely continuous. The products of two absolute continuous function defined on the bounded interval will also be a absolute continuous function.

2. If an absolutely absolute continuous function is defined on a bounded closed interval and is nowhere zero then its reciprocal is also an absolutely continuous function.

3. Every absolutely continuous function is an uniformly continuous function.

4. If  f: [ p , q ] → R is absolutely continuous, then it will be a function of the bounded variation on [ p,q ]