Chain Rule

When a function is a composite of two or more functions then to find the derivative of that function, we use chain rule. Let us understand the proof of chain-rule:
Let y = f(u)
And z= g(x)
where y and z both are differentiable functions.
Then  y= f(g(x)) is a function of x
delta  y = f(u+delta  u ) – f(u)
And  delta  z = g(x+delta  x) – g(x)
Therefore delta  y/ delta  z = f(u+delta  u ) – f(u)/delta  t
And delta  z/ delta  x = g(x+ delta  x) – g(x)/delta x
We first suppose that for sufficiently small values of delta  x , delta  z not equal to 0. This will be the case if dz/dx not equal to 0. Then, g is differentiable, it is continuous and hence when delta  x ->0, x+delta x->x and g(x+delta x)->g(x).
Therefore z+delta z->z and delta z->0. Since delta z not equal to 0 as delta x->0, we may write
delta y/delta x= delta y/delta u . delta z/delta x
We can see that f and g, both are continuous functions as they are differentiable. Thus, delta u->0 when delta x->0 and delta y->0 when delta u->0.
Therefore, limit delta x->0 delta y/delta x = limit delta z->0 delta y/delta z multiply limit delta x->0 delta t/delta x
Hence dy/dx = dy/dz. dz/dx

For example, find derivative of sin(x^2)
Solution: Here we can clearly see that the function is function of a function.
Suppose y = sin(x^2) = sin z
Where z = x^2
Therefore dy/dz = cos z and dz/dx = 2x
Hence by this rule
Dy/dx = dy/dz. dz/dx
= cosz . 2x            [ substitute the values of dy/dz and dz/dx]
= 2xcos(x^2)         [ substitute the value of z]

Derivatives chain rule: Sometimes it is also known as chain-rule derivatives.  It means to find derivative of a function using this rule.

Reverse Chain Rule:
It is used for integration or say anti derivative of functions. It is equivalent to integration by parts.

The Chain Rule Multivariable:
It means many variables are included in this form. Suppose that y=f (a, b), where a’ and b themselves depend on one or more variables. This Rule gives us the liberty to differentiate y with respect to any of the variables (a, b) involved:
Let a=a (t) and b=b (t) be differentiable at t and let y=f (a, b) is differentiable at the point (a (t), b (t)). Then y=f (a (t), b (t)) is differentiable at t and
Dy/dt = dy/ da . da/dt + dy/ db . db/dt
Let us understand an example of the multivariate chain-rule:
For example: Solve it using the multivariate chain-rule:  Let y = a^2b-b^2 where a and b are parameterized as a=t^2 and b=2t
Then Dy/dt  = dy/ da . da/dt + dy/ db . db/dt
=2ab.2t +(a^2-2b)(2)
= (2t^2. 2t).2t+((t^2)^2 – 2(2t))(2)
= 8t^4 + 2t^4 – 8t
= 10t^4 -8t

Partial Fraction

Decomposing partial fractions: It is quite difficult to find out the integral of partial-fractions so to make our calculations easier we make use of decomposing partial fractions.
It means dividing a partial fraction into part of two or more simpler fractions. For decomposing purpose in calculus we have different types of methods.
Given below are some forms of decomposing-partial fractions:
Type 1:
Form of the rational function =(px+q)/(x-a)(x-b)
Form of the partial fraction = A/(x-a) + B/(x-b)
Given below is the integration by partial fractions examples of the above type:
Partial fractions problems 1:  Convert 1/(x+1)(x+2) into integral partial fractions
Solution:
1/(x+1)(x+2) = A/(x+1) +B/(x+2)
1 = A(x+2) + B (x+1)
Equating the coefficients of x and constant term, we get
A + B = 0…….(1)
2A + B =1……(2)
Solving these two equations we get
A = 1
B = -1
1/(x+1)(x+2) = 1/(x+1) -1/(x+2)
problem 2:  Convert x²+1/x²-5x+6 into integral partial fractional forms
Solution: Here the integrand x²+1/x²-5x+6 is not proper rational function, so we divide x²+1 by x²-5x+6 and find that
x²+1/x²-5x+6 = 1 + (5x-5)/ x²-5x+6
=1 + (5x-5)/ (x-2)(x-3)
(5x-5)/ (x-2)(x-3) =A/(x-2) + B/(x-3)
A(x-3) +B(x-2) =5x -5
Equating the coefficients of x and constant term, we get
A+B = 5……………(1)
-3A -2B = -5
3A +2B =5…………..(2)
Solving these two equations we get
A =-5 and B = 10
Therefore x²+1/x²-5x+6 = 1 + 5/(x-2) + 10/(x-3)
Type 2:
Form of the rational function =(px+q)/(x-a)²
partial form of fraction = A/(x-a) + B/(x-a)²
Given below is the integration by partial-fractions examples of the above type:
Problem 3:  Convert (3x-2)/(x+1)² into integral partial form of fractions
Solution: (3x-2)/(x+1)² = A/(x+1) +B/(x+1)²
A(x+1) + B = 3x-2
Equating the coefficients of x and constant term, we get
A =3
A + B = -2
B= -5
Therefore  (3x-2)/(x+1)² = 3/(x+1) -5/(x+1)²
Type 3:
Form of the rational function =(px²+qx +c)/(x-a)(x-b)(x-c)
Form of the partial fraction = A/(x-a) + B/(x-b) +C/(x-c)
Given below is the integration by partial-fractions examples of the above type:

Problem 4:  Convert (3x²+2x +5)/(x-1)(x-2)(x-3) into integral partial form of fractions
Solution: (3x²+2x +5)/(x-1)(x-2)(x-3) =A/(x-1) +B/(x-2) +C/(x-3)
A(x-2)(x-3) +B(x-1)(x-3) + C(x-1)(x-2) =3x²+2x+5
A(x²-5x+6) +B(x²-4x+3) +C(x²-3x+2) =(3x²+2x +5)
Equating the coefficients of x and constant term, we get
A +B +C = 3…………..(1)
-5A -4B -3C =2………….(2)
6A + 3B+2C =5………….(3)
Solving these equations we get
A =5,B=-21, C=19
Therefore
(3x²+2x +5)/(x-1)(x-2)(x-3) =5/(x-1) -21/(x-2) +19/(x-3)
Type 4:
Form of the rational function =(px²+qx +c)/(x-a)²(x-b)
Form of the partial fraction = A/(x-a) + B/(x-a) ²+C/(x-b)
Given below is the integration by partial-fractions examples of the above type:
Problem 4:  Convert (x²+2x +1)/(x-1)²(x-2) into integral partial form of fractions
Solution: (x²+2x +1)/(x-1)²(x-2) = A/(x-1) + B/(x-1) ²+C/(x-2)
(x²+2x +1) = A(x-1)(x-2)+B(x-2)+C(x-1)²
(x²+2x +1) = A(x²-3x+2) +B(x-2)+C(x²-2x+1)
Equating the coefficients of x and constant term, we get
A + C = 1……………(1)
-3A + B -2C =2……….(2)
2A -2B +C = 1…………(3)
Solving these equations we get
A =-8
B =-4
And C=9
Therefore,( x²+2x +1)/(x-1)²(x-2) =-8/(x-1) -4/(x-1)²+9/(x-2)

Type 5:
Form of the rational function =(px²+qx +c)/(x-a)(x²+bx+c)
Form of the partial fraction = A/(x-a) + Bx +C/(x²+bx +c)
Given below is the integration by partial-fractions examples of the above type:
Problem 5:  Convert (x²+x +1)/(x²+1)(x+2) into integral partial form of fractions
Solution: (x²+x +1)/(x²+1)(x+2) = A/(x+2) + Bx +C/(x²+1)
A(x²+1) + (Bx +C)(x+2) =(x²+x +1)
Ax²+A + (Bx² +Cx+2Bx+2C) =(x²+x +1)
Equating the coefficients of x and constant term, we get
A +B = 1…………………….(1)
2B +C =1…………………….(2)
A +2C  =1…………………….(3)
Solving these equations we get
A =3/5
B =2/5
C =1/5
Therefore, (x²+x +1)/(x²+1)(x+2) = 3/5(x+2) + (2x/5 +1/5)/(x²+1)
(x²+x +1)/(x²+1)(x+2) = 3/5(x+2) + (2x +1)/5(x²+1)

Evaluate the algebraic expressions

If the phrase is given as the sum of one and eight then expression is like that 1+8. It is known to be conversion of the word into algebraic expressions. Qualitative analysis is not measurable but if you convert it into the quantitative analysis or which is to be easily quantifiable then you are completed in writing an algebraic expressions. Numerical expression and algebraic equation is one and the same thing as which is quantify in terms of the numbers then it is known to be as algebraic expressions.

Sometimes it is much complex word problem is given and you have to write algebraic expressions then you have to go step by step procedure as first you have to analyze what the questions wants in the solution and what are the expressions that are easily frame into the quantitative form. Then you have to use your brain and your logical reasoning to frame the equations.

Let us take an example like there are 5 people and each one of them is using the pens in the class. In the sequence individually use the no of pens as 4, 5, 3, 1 and 2 respectively. So the formation of the word problem is very easy. You have to see the need of the problem. Writing algebraic expression is easy if you understand the problem well. Now select the first variable and assign the variable as x for first person and multiply it by the number given as no of pens as 4 as 4x.

After solving the equation initially now you have to evaluate algebraic expressions. Here the expression is very easy as something in the form of summation of 5 persons who are using pens in the class. It is then easily solved further if another question is there. You have to check the linking between the questions.

We have to frame and grab every single digit and consider it which is beneficial for our calculation. Sometimes the numerals in the question given are not relevant just to confuse so that you will consider it and loose the marks.

We should have a deep thinking during solving the problem. The good thing is that as much as you do practice over these kinds of problems you are more confident enough to solve similar problems as maximum times the problem pattern is the same but solution is different. It is very rare that complex equation will come in your exam.

Boolean algebra

Boolean Algebra Help:   It is a mathematical discipline that is used for designing digital circuits in a digital computer. It describes the relation between the inputs and outputs of a digital circuit.
A Boolean algebra is a variable having only two possible values such as TRUE or FALSE, or  as  1 or 0.
A Boolean expression is a combination of Boolean variables, Boolean constants and logical operators which are AND, OR and NOT.

Boolean algebra proofs :

1.   a + a • b = a + b

Proof:
a + a • b
= (a + a) • (a +b)
= 1 • (a + b)
= a+ b

2.    a • (a + b) = a • b

Proof:
a • (a + b)
= a • a + a • b
= 0 + a • b
= a • b

3.       a • b + a • b = a

Proof:
a • b + a • b
= a • (b + b)
= a • 1
= a

4.    (a + b) • (a + b) = a

Proof:
(a +b) • (a + b)
= a + (b • b)
= a + 0
= a

Simplify Boolean algebra
Here is the list of rules used for the boolean expression simplifications.
The Idempotent Laws  AA = A   and  A+A = A
The Associative Laws states that  (AB)C = A(BC) and (A+B)+C = A+(B+C)
The Commutative Law states that    AB = BA   and  A+B = B+A
The Distributive Laws  A(B+C) = AB+AC  and  A+BC = (A+B)(A+C)
B

Boolean algebra truth tables

Digital systems are said to be constructed by using logic gates. These gates are the AND, OR, NOT, NAND, NOR, EXOR and EXNOR gates.

AND gate

The AND gate gives a high output (1) only if all its inputs are high.
Y= A.B.  That is  if A=0, B=0 then Y =0,  If A=0, B=1 then Y=0,  If A =1, B=0 then Y=0 , If A=1, B=1 then Y = 1

OR gate

The OR gate gives a high output (1) if one or more of its inputs are high.  A plus (+) is used to show the OR operation.
Y= A+B.  That is  if A=0, B=0 then Y =0,  If A=0, B=1 then Y=1,  If A =1, B=0 then Y=1 ,
If A=1, B=1 then Y = 1

NOT gate

The NOT gate produces an inverted version of the input at its output.
Y = A.  If  A=1 , Y=0 and A=0, Y=1

The Ac-method of Factoring Polynomials

The concept of polynomial has to be clear before the concept of this factoring method.  The polynomial can be of different degrees. A polynomial can also of degree one. This is the simplest of the polynomials. It contains variables and constants in it.

The value of constants remains the same throughout and the value of the variables will change depending on the values given to them. So, one must be very careful while dealing with the constants and the variables. There is this basic difference between the two. A number which divides another number completely can be taken as the factor of the number.

An example will help in explaining the concept better. The number ‘10’ can be obtained by multiplying ‘2’ and ‘5’. These numbers ‘2’ and ‘5’ can be taken as the factors of the number ‘10’. This is the process of factorization  The ac method is used in the factorization of a given polynomial.

The term ‘c’ represents the constant present in the polynomial. This constant can take various values. It can take values which can be negative and sometimes can take values which will be positive. So, one must be very careful. The next term in consideration is ‘a’.

This is the coefficient of the term which has the power two. In this method both these terms are multiplied to arrive at the final solution and that is why the name ac factoring method is given to it. This method is basically used to find the solution of a polynomial, which has a degree ‘2’. So, this method is very helpful in finding the solution of the polynomial with degree ‘2’.

The polynomial will have three terms in it. One is the term with degree ‘2’ and then the term with degree ‘1’ and then the constant which can take both the positive and negative values available. The Ac-method can be very helpful in solving these kinds of polynomials.

The solutions will be in such a way that that their product must be equal to the product of the terms ‘a’ and ‘c’ and their sum must be equal to the middle term. The middle term in the polynomial can be represented by the term ‘b’. This is the basic principle behind this method. So, the numbers which satisfy this principle will be the solutions of the given polynomial. The final solutions will satisfy the polynomial.