Function Derivative Calculator

Solving 2nd order derivatives of a function

1) Solve the derivative for the function  f(x) =  x^2 + 8x + 9

Solution :  The given function is  f(x) = x^2 + 8x + 9

Differentiate the above equation with respect to 'x' . It is represented as f'(x) .

f'(x)  =   `(d(x^2))/dx`   +  `(d(8x))/dx`   +  `(d(9))/dx` .

f'(x)   =     2x   +   8   +   0

f'(x)    =     2x    +   8   .

The answer is   f'(x)  =  2x  +  8 .

2) Solve the derivative for the function   f(y)  =  y^2  +  10y  + 3

Solution :   The given function is  f(y)  =  y^2  +  10y  +  3

Differentiate the above equation with respect to 'y' . It is represented as f'(y)  .

f'(y)  = `(d(y^2))/dy`   +  `(d(10y))/dy`   +   `(d(3))/dy`

f'(y)  =   2y   +  10   +   0

f'(y)  =   2y   +   10

The answer is  f'(y) = 2y  +  10

Solving third Order Derivative Functions

1) Find the derivative of the function  f(x)  =  x3 + 3x^2 + 18x  +  20

Solution : The given   function  f(x)  =  x3 + 3x^2 + 18x  +  20

Differentiate the above function with respect to 'x' .

f'(x)  =  3 x^2   +  3  ( 2 ) x   +  18   +  0

f'(x)   =  3x^2  +  6x  +  18  .

The answer is   f'(x)   =  3x^2  +  6x  +  18  .

2) Find the derivative for   f(x)  =  6x3  + 5x^2  +  3x  +  1

Solution : The given function is f(x)  =  6x3  + 5x^2  +  3x  +  1

Differentiate the above f(x) with respect to 'x'  .

f'(x)  =  6 (3)x^2  +  5 (2)x   +  3  + 0

f'(x)   =  18x^2   +  10x   +  3

The answer is  f'(x)   =  18x^2   +  10x   +  3

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Solving 4th Order Derivative Function

1) Solve the derivative for the function  f(y) = y^4  + 3y^3  +  5y^2  + 4y  +  9

Solution : The given function  f(y) = y^4  + 3y^3  +  5y^2  + 4y  +  9

DIfferentiate the above equation with respect to 'y' .

f'(y)  =  4y^3  +  3(3)y^2  +  5(2)y  +  4  +  0

f'(y)  =  4y^3   +  9y^2  + 10y   +  4

The answer is    f(y) = y^4  + 3y^3  +  5y^2  + 4y  +  9

2) Solve the derivative for the function f(y)  =  6y^4  +  y^3   +  y^2  + 10 y  + 3

Solution :  The given function  is    f(y)  =  6y^4  +  y^3   +  y^2  + 10 y  + 3

DIfferentiate the above equation with respect to 'y' .

f'(y)   =  6(4)y^3   +  3y^2    +  2y   +  10  +   0

f'(y)   =   24y^3   +  3y^2   +  2y   +  10

The answer is    f'(y)   =   24y^3   +  3y^2   +  2y   +  10

Algebra is widely used in day to day activities watch out for my forthcoming posts on Derivative Calculator and Derivative of Natural log. I am sure they will be helpful.

Free Algebra Classes

Algebra is a cluster of mathematics, which is used to create mathematical problems of valid-globe actions and control problems that we cannot explain using arithmetic.

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Free algebra mainly different kind of lesson handled here.

Algebra uses the cipher as calculation for addition, subtraction, multiplication and division and it includes constants, operating signs and variables.

Algebraic equations represent a collection, what is finished on one side of the range with a number to the other side of the range.

Below lesson topics explain the algebra classes.

Algebra Free Algebra Classes Topics:-

Exponents

Radicals

Polynomials

Factoring

Division of Polynomials

Solving Equations

Solving Inequalities

Lines

Solving Quadratic Equations

Complex Numbers

Graphing Quadratics (Parabolas)

Systems of Equations (2x^2's)

Systems of Equations (3x^3's)

Determinants and Cramer's Rule

Functions

Inverse Functions

Exponentials and Logarithms

Absolute Value Equations and Inequalities

Sequences and Series Combinatory

Advanced Graphing

Graphing Polynomials

Graphing Rational Functions

Matrices

Above the topics free algebra classes based process used here. And then free algebra mainly different kind of classes handle through now.

Free Algebra Classes - Problems:

Problem 1:

Solve for x: 3x+6 = 4x+8.

Solution:-

3x+6 = 4x+8

Subtract by 4x on both sides,

3x-4x+6 = 4x-4x+8

-x+6 = 8

Subtract by 6 on both sides,

-x+6-6 = 8-6

-x = 2

So, x = -2.

Order of Operations problem:-

Problem 2:

5 + 4 * 3y (x + 2)

Solution:

- 5 + 12xy + 6

Negative Exponents of Numbers:-

3 -3

Solution:- 1/27

Negative Exponents of Variables:-

K -3

Solution:- 1/ k^3

Negative Exponents in Fractions

X -2 y/x

Solution:- y/x 3

Substitution method:-

Problem:

Evaluate: w + a^2, where w = -12 and a = z + 3

-12 + (z + 3 )2 = z2 + 2(3(z)) + 3

Solution: Z2 + 6z + 9

Finding a Greatest Common Factor (GCF) of Two Terms:-

9 x^2y, 3 xy^3

3xy

Factoring a GCF from an Expression:-

4x^2 + 16x^3

4x^2(1 + 3x)

Factoring a Difference between Two Squares

x^2 – 9

(x + 3) (x – 3)

Factoring a Trinomial Completely:-

X^4 + 7x^3 + 12 x^2

x^2(x + 3) (x + 4)

Solve By Factoring

x^2 + 3x + 2

Solution:- -1, -2.


Practice Problems for Free Algebra Classes:

1. Find factors and root of the equation x^2 + 10x + 24

Answer: -6, -4

2.Find factors and root of the equation x^2 +11x + 28

Answer: -4, -7

3.Find factors and root of the equation x^2 – 2x + 8

Answer: 4, -2

4.Find factors and root of the equation 2x^2 – 15x + 8.

Answer: 1, 15

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Rate Histograms Tutoring

In rate histograms tutoring the word rate is a Greek term. In rate histograms the interval with each frequency value are equal.  In mathematical the rate histogram is used to counting the number of observations in the histogram. Tutor is a person who shares his knowledge with student which is called tutoring. Now the tutor will discuss about the study histogram in this tutorial.

Example for Rate Histograms Tutoring:

Example: 1

While doing a project for P.E. class, Alonzo researched the number of athletes competing in an international sporting event. By using this data complete the histogram below.

Solution:
In the above histogram the missing bar is for the range 41 to 50 athletes and count the number of values in this range. Two countries have between 41 and 50 athletes so set the height of the bar to 2.

Example: 2

Prepare a histogram  by using the following data in the frequency table

Solution:
From the above table take the class intervals on the X-axis and frequency on the Y-axis  and mark the Y-axis scales as 0,5,10,15,20 To avoid errors in our histogram we marking the class intervals so that  the histogram be more perfect  and draw the rectangle bars .

 Example: 3

Prepare a histogram  by using the following data in the frequency table

Solution:
From the above table take the class intervals on the X-axis and frequency on the Y-axis  and mark the Y-axis scales as 0,5,10,15,20,25,30,35. To avoid errors in our histogram we marking the class intervals so that  the histogram be more perfect  and draw the rectangle bars .

These are the examples of rate histograms tutoring.

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Mathematical Information of Rate Histogram:

In the mathematical terms, the rate histograms are commonly used for displaying and mapping of mi items. These items are generally used  for counting the number of observations that are present in the histogram. The mathematical formula for calculating the rate histogram is given below,

`sum_(P=i=1)^k`

where,

p is used to represent the sum of  number of observations in the given table.

mi is used to represent the certain conditions in the table,

k is used to represent the total number of bins present in the table.

This is the general information of rate histograms tutoring in mathematics. Tutoring is one of the process. Student can easily develop their knowledge through online tutoring . Tutoring is the correlation between student and tutor.

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Solve Equations with Exponent

Introduction to solving equations with exponents

Exponent equations are the equations in which variable appear as an exponent.
To solve these equations rules and laws of exponents are used. Exponent equations are of two types
(1) Exponent equations in which bases are same
(2) Exponent equations in which bases are different.

Steps to Solve Equations with Exponent

Solving Exponential Equations of the same base

1) Ignore the bases, and simply set the exponents equal to each other

2) Solve for the variable

 When the bases of the terms are different

1) Ignore the exponents; rewrite both of the bases as powers of same number. For    example if there are 2 and 4 in the bases, then convert base 4, in to base 2 by writing it again as (2)^2

2) once the bases are same , ignore them

3) Equalize the exponents

4) Solve for variable

Simple Problems of Equations with Exponents

Solve for variable Answer 1. 3m  =  35 Since the bases are the same, set the exponents equal to one another: m = 5 2. 5t   = 125 125can be expressed as a power of 5: 5t  = 53 t = 3  3.  493y=343 49 and 343 can be expressed as a power of 7: [(7)2]3y = 73 76y = 73 6y = 3 y = 1/2

More Problems of Equations with Exponents

Solve for x. Answer 1.  52x+1  =  53x-2 Since the bases are the same, set the exponents equal to one another: 2x + 1 = 3x - 2 3 = x 2.  32x-1  = 27x  27 can be expressed as a power of 3: 32x-1  = 33x 2x - 1 = 3x -1 = x 3.   43x-8  = 162x 16 can be expressed as a power of 4: 43x-8= [(4)2]2x 3x - 8 = 4x   -8 = x I like to share this multiplying rational expressions solver with you all through my article.

Ratios and Unit Rates

Introduction to Ratios and Unit rates:
                  In mathematics, there are more topics which perform arithmetic and binary operations. Ratios are the mathematical terms which are used to make a relationship between any two numbers or variables. They are used to compare two values of any quantities. A rate is a ratio which is used to compare any two terms. Unit rate is also a rate in which one quantity is corresponding to the unit of another quantity. Let us discuss about the ratios and unit rates with definitions.

Definitions for Ratios and Unit Rates:

           Here we see the definitions for rate and ratios.

Rate:

          The rate is defined as a ratio which is used for comparing any two different numbers. It is a ratio between two or more numbers.

Formula for rate:

              The term rate can be calculated or found as the distance travelled (d) to a time period (t) per hours or minutes in the form of proportion.

Unit rate:

          A unit rate is used to check the given equation or to compare the given quantity. It defines how many total units are there in one quantity to the other quantity.

Example for common unit rates:

         Salary per month, cost per item, kilometers per hour, dollars per pound.

Ratios:

        Ratios are defined as comparing of two numbers specified in a form by the use of a colon. It is a relationship between two or more things.

Example for ratio:

        The numbers 4 and 8 can be represented in ratio as 4: 8. It is also given as fraction or proportion form as `4/8` .

These are the definitions and explanation for ratios and unit rates.

Problems for Ratios and Unit Rates:

       We see some problems for ratios and unit rates.

Problem 1:

          Reena has bought 6 apples, 5 chocolates and 2 bananas.

 i) What is the ratio of apples to bananas?

ii) Mention the ratio of chocolates to bananas.

Solution:

      i) The ratio of apples to bananas can be written as one quantity of number to other.

So, the ratio is 6 : 2.

      ii) In this, the ratio of chocolates to bananas can be mentioned as number of chocolates to number of bananas.

That is, 5 : 2.

Problem 2:

     Find the unit rate for 52 kilometers per 3 hours.

Solution:

        The unit rate can be found as one quantity to other.

Therefore unit rate = `52/3`

= 14

Hence the unit rate = 14 kilometers per hour.

These are the problems for ratios and unit rates.

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Monomial Factor with Exponents

Introduction to monomial factor with exponents:

"Monomial factor with exponent" means the monomial term with exponents need to factorise. Here we need to understand each and every terms.

From the term mono we can understand it is talking about a single term, so "monomial" means it is an algebraic expression which is having only one single term. We can also say that monomial is a product of numbers and variables, where variables can be any letter or a power of letter( power or index is called as exponents).Monomial can be a single letter or a number also.So, Monomial factor with exponents can be a single term consist of a number with variable and also with exponents and which can be factorise also. Here the factor means when we multiply two elements together we can get the final product.
Now we come to the term "Exponent" here it means the number can be multiply how many times which depends upon the power. Now we have a small idea about the term "Monomial factor with exponent".

Examples on Monomial Factor with Exponents:

1. 12x²

This a monomial, here we have one single term which is 12x². Here 12 is a number with a variable x having the power as 2. x² is  called  x to the second power. Here x is the base and 2 is the exponent.

2. -24abc

This is also a monomial having a single term. The exponent here is 1 for each variables.

Above are the examples of monomial with exponents.

Let us take an example and learn to facrorise:

1)12 x⁴

first we will find the factors of 12 which is

1 x 12

2 x 6

3 x 4

Now we will factor x² .This can be written as

x.x³

x^2.x²

x⁰.x⁴

As per the exponent law (a)^m.(a)ⁿ=a^m+n

so if we add the power we get the final results as x⁴.

2) 6a⁴b¹⁰= (2a³b²) .(  ? )

Here we need to find the missing terms. As we have already one part, to find the other part let us find the factor.

6.............can be written as

1 x 6

2 x 3

As we have already 2 in the question so we have to consider the factor for 6 as 2 x 3.

Now factor a⁴b¹⁰

a⁴=a¹.a³                           and  b¹⁰=b¹.b⁹

a⁴=a².a²                                    b¹⁰=b².b⁸

a⁴=a⁰.a⁴                                    b¹⁰=b³.b⁷

                                                     b¹⁰=b⁴.b⁶

                                                     b⁵=b⁵.b⁵

In the question we have a³b². so from the above factorisation we can find as the final results as 3ab⁸

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Conclusion on Monomial Factor with Exponents:

In monomial factor with exponents always we have to remember that whenever we multiply the monomials we add the exponents also. Degree of monomials is the sum of exponents of all the letters.

Computing Fractions

Introduction:
          The whole part is divided in to different parts. Each parts is called fraction of the whole thing. The fraction is shown as a/b, where a is referred as numerator and b is referred as denominator. Those denominator and numerators are involed in fraction. Fractions are differentiated by their values of numerator and denominator. They are computed and described below.
The example of fraction is `2/6`

Common Factor for Computing Fractions:

This common factor is used for solving and reducing fractions:

Example:

16 =4*4

20 =4*5

       Take all common numbers.

       The product of common numbers and remaining numbers is called common factor.

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Description for Computing Fractions with Examples:

Computing fractions for addition:

Example: `2/5` + `3/5`

Here denominators are same.

So numerators are added  `(2 + 3)/5` .

Therefore the resultant fraction is `5/5`

The result for the addtion fraction is 1

Example: `2/5 ` + `3/4`

Here denominators are different. So we have to find least common divisor.

The lease common divisor for 4 and 5 is 20.

In `2/ 5` , the denominator 5 is 4 times in the least common divisor.

So we have to multiply the numeratory 2 by 4. 2 * 4 = 8.

In `3/4,` the donminator 4 is 5 times in the least common divisor. so we have to multiply the numerator 3 by 4. 3 * 5 = 15

Now, we can add the denominator. so we will get `(8 +15)/20` = `23/20` 

Computing fractions for subtraction: `8/5 - 3/5`

Here denominators are same. So we are doing like below.

Numerators are subtracted. `(8 - 3)/5` .

So we get `5/5` =1

Example: `2/5 - 3/4`

Here denominators are different. So we have to find least common divisor.

The lease common divisor for 4 and 5 is 20.

In `2/ 5` , the denominator 5 is 4 times in the least common divisor.

So we have to multiply the numeratory 2 by 4. 2 * 4 = 8.

In `3/4` , the donminator 4 is 5 times in the least common divisor.

So we have to multiply the numerator 3 by 4. 3 * 5 = 15

Now, we can subtract the denominator. so we will get `(8 -15)/20 = -7/20`

Computing fractions for multiplication:

`3/5 xx 5/7`

Here 3 and 5 are numerators and  5 and 7 are denominators

Multiplying the numerator by numerator:

3 * 5 = 15

Multiplying the denominator by deminator

5 * 7 = 35

Therefore the fraction is  `15/35`

therefore the result for the multiplication fractorion is `3/7` 

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Computing fractions for division:

`2/5-:3/5`

Firest we have to find  reciprocal for the second fraction is 5/3.

Then we have to multiply this reciprocal with dividend fraction like below

`2/5 xx 5/3`

So we will get `10/15` .

Therefore the result is `2/3`

Math Problems for Pre Algebra

Introduction :
Pre algebra is a branch of mathematics that is applied to make the mathematical model of the real-world situations and to handle a problems that we might not solve the problems using the simple arithmetic. Using words, algebra utilize a symbol to make statements. Algebra consist of real numbers, complex numbers, linear equations, vectors etc. In algebra, we are frequently using the letters that stands for the numbers in mathematics.  Algebra uses symbols of the arithmetic operations for adding, subtracting, dividing and multiplying. In this article we shall discuss about solve math problems for pre algebra.

Sample Problem for Solve Math Problems for Pre Algebra:

Solve math problems for pre algebra problem 1:

Solve the given math problem for pre algebra equation and find out x value of the equation 3(3x - 2) + 5 = 2(4x + 5) – 10.

Solution:

We are going to find the x value of the given pre algebra equation.

In the first we are going to multiply the term – 5 in both sides of the equation. We get

3(3x - 2) + 5 – 5 = 2(4x + 5) – 10 -5

3(3x - 2) = 2(4x + 5) - 15

In the next step grouping the same values in the above equation, we get

9x -8x = 6 + 10 – 15

x = 1

The value of the x in a given equation is 1.

Solve math problems for pre algebra problem 2:

Solve the given math problem for pre algebraic equation step by step and find the x value (5x – 6) + 8 = (6 x – 8) + 8.

Solution:

We are going to find the x value of the given pre algebra equation.

In the first we are going to multiply the value – 8 in both sides of the equation. We get

(5x – 6) + 8 – 8 = (6 x – 8) + 8 – 8

(5x – 6) = (6 x – 8)

In the next step grouping the same values in the above equation, we get

5x -6x = 6 – 8

-x = -2

x = 2

The solution of the x in a given equation is 2.

If we have any doubt in the solution means we check the answer of given problem. Substituting x = 2 in the given equation, we get

5(2) – 6 + 8 = 6(2) – 8 + 8

10 – 6 + 8 = 12 – 8 + 8

12 = 12

The left hand side and right hand side of the answers are same. So the solution is correct.

Practice Problem for Solve Math Problems for Pre Algebra:

Solve the given algebraic equation 1 + 2(1 + 4)² - 4

Answer: 47

Solve the given algebraic expression 2 + 3(1 + 4)² - 20

Answer: 57