Cartesian Product

In the previous post we have discussed about Differential Calculus and In today's session we are going to discuss about Cartesian Product. An ordered pair, usually denoted by (x , y) is pair of elements x and y of some sets. Like the Cartesian product of two sets A and B is set of those pairs whose first co coordinate is an element of A and second co-ordinate is an element B. The set is denoted by A x B and is read as ‘A cross B’ or ‘product set of A and B’. Basically these sets are work as ordered pair usually denoted by (x, y) is a pair of element x and y of some sets, which is ordered in the sense that (x, y) ≠ (y, x) whenever x≠y. Here x is called first co-ordinate and y is called second co-ordinate of the ordered pair (x, y), for example ordered pair (1, 2) and (2, 1) consist of same element 1 and 2 but they are different because they represent different points in the co-ordinate plane. (know more about Cartesian Product, here)

Now we use this concept in Cartesian product as following manner:
A x B = (x, y): xϵ A  É… yϵ B, A and B are different sets which are multiplying with each other.
Let the value of set  A = 1, 2, 3 and the value set  B=3, 5;
A x B=1, 2, 3 x 3, 5;
=(1, 3), (1, 5), (2, 3), (2, 5), (3, 3), (3, 5) (Here we get many ordered pair of A x B  which are called Cartesian product value of A x B),
And B x A=(3, 1), (3, 2), (3, 3), (5, 1), (5, 2), (5, 3) (Here we get many ordered pair of B x A which are called Cartesian product value of B x A ),
A x B ≠ B x A (it means both product are not same).
Similarly, we can define Cartesian product for n sets A1, A2, A3 …….An
A1, A2, A3 …….An  = ( x1, x2, x3 …….xn  )  x1 Ïµ A1 É…  x2ϵ A2 É…  x3ϵ A3 ……………. É… xnϵ An
The element (x1, x2, x3 …….xn  ) is called as n-type elements.
In this way we can solve the all Cartesian product problem.
Now we discuss what is Logarithmic Differentiation? We know exponent expression is defined as in the power expression like if y = ax, where x is logarithmic of y to the base a, in this case we will use the logarithmic differentiation. ISEET Physics syllabus helps students for their better study.

Differential Calculus

In the previous post we have discussed about limit laws and In today's session we are going to discuss about Differential Calculus. Dear students in today’s class we will study a very important and very interesting topic that is calculus. It is a branch of mathematics. We will take a brief idea about this topic.“Calculus” is a Latin word which means a tiny stone that is used for counting. It is a branch of mathematics which is basically based on functions, limit, differentiation and  integration. It constitutes of two major branches. One of which is differential calculus and another is integral calculus. Differential calculus is the study of change of limit, function and derivative . Differential calculus has vast  applications in the field of science, engineering and  sometime in economics also.  (know more about Differential Calculus, here)
It is also called that differential calculus is a very important method or system of calculation which is explained by the representative manipulation of expressions. There are many examples of differential calculus. Differential Calculus was the first success of modern mathematics and it is very tough to avoid its importance.  It only clears more efficiently the system of mathematical analysis.
Calculus is generally developed by manipulating very small quantities. Calculus has been used in every branch of the computer science, physical science, statistic, engineering, economics, business,etc. Calculus can also be used in combination with other mathematical branches.
Lets take a example: it can be used in theory of probability to settle on the probability of a nonstop arbitrary variable from an assumed density function. it can also be used with linear algebra to determine  the "best fit" linear approximation.
Calculus is also applied to evaluate estimated solutions to equations.  It is the very important way to solve differential equations in most applications.
So, we can say that calculus has very wide applications in the field of mathematics and even imagine of mathematics is not possible without “Calculus”.
There are different types of carbohydrates. It can be seen in icse syllabus 2013. 

limit laws

Limit is used to define the value in which the function is approaches as an input. Limit is also used in calculus and also used to define the continuity function. Let’s talk about the limit laws. In mathematics, there are different types of laws which are given below. Now see all the limit laws one by one.
1.      Addition law: - The additive law is given by: Let the lim _{y ⇥b} f (y) and lim _{y ⇥b} g (y) both are exist then
lim _{y ⇥b} f (y) + lim g (y) = lim _{y ⇥b} f (y) + lim _{y ⇥b} g (y).
2.      Subtraction law: - The subtraction law of limit is given by: Let the lim _{y ⇥b} f (y) and
lim _{y ⇥b} g (y) both are exist then
lim _{y ⇥b} f (y) - lim g (y) = lim _{y ⇥b} f (y) - lim _{y ⇥b} g (y).
3.    Constant law: - The constant law of limit is given as: Let ‘c’ is a constant, and the limit lim _{y ⇥b} f (y) exist, the constant law can be written as: lim _{y ⇥b} c. f (y) = c. lim _{y ⇥b} f (y).
4.      Multiplication law: - The multiplication law is given by: Let the lim _{y ⇥b} f (y) and
lim _{y ⇥b} g (y) both are exist then
lim _{y ⇥b} f (y) . lim g (y) = lim _{y ⇥b} f (y) . lim _{y ⇥b} g (y). (know more about limit laws, here)
5.      Division laws: - The division law of limit is given by: Let the lim _{y ⇥b} f (y) and
lim _{y ⇥b} g (y) both are exist and lim _{y ⇥b} g (y) ≠ 0, then the division law can be written as: lim _{y ⇥b} f (y) / g (y) = lim _{y ⇥b} f (y) / lim _{y ⇥b} g (y). This is all about the limit laws. Now we will see the Triangle Inequality Theorem. It says that the sum of two sides of a triangle is always larger than the third side. To get more information then prefer online tutorial of cbse syllabus for class 9 and In the next session we will discuss about Differential Calculus. 

arctan

In the previous post we have discussed about arcsin and In today's session we are going to discuss about arctan. Hi friends in mathematics we will study different types of trigonometric functions. Here we will discuss one of the trigonometric function that is arctan. Before learn the concept of arc tan first we will discuss about trigonometry. Trigonometry can be defined as a branch of mathematics which is used to show the relationship among the angles and sides of triangle. 

Now we will discuss inverse tangent function of arc tan. The inverse of arc tan is given as:

=> arc tan p = 2 arc tan p / 1 + √ (1 + p²);

The derivative of tan inverse ‘p’ is 1 / 1 + p²;

Let’s see the proof of derivative tan inverse ‘p’.

First write the tan^-1 p in the derivative form:

=> d / dp tan^-1 p = 1 / 1 + p²;

Now, assume that the function for finding derivative tan inverse ‘p’.

Let function f (p) = tan^-1 p,

If we put the value of p = tan ⊖;

On putting the value we get:

= f (tan ⊖) = ⊖;

If we differentiate it then, we get:

= f’ (tan ⊖) sec² ⊖= 1;

We can rewrite it as:

= f’ (tan ⊖) = 1 / sec² ⊖…… (1);

We can write the sec² ⊖ as:

Sec² ⊖= tan² ⊖ + 1;

Now assume q = tan ⊖ then we can write it as:

→ sec² ⊖= 1 + q²……… (2);

After this, put the value of equation (2) in the equation (1);

On putting the value in equation 1 we get:

= f’ (tan ⊖) = 1 / sec² ⊖…… (1);

= f’ (tan ⊖) = 1 / 1 + p²;

Now, we can write the derivative of tan inverse p as:

= d / dp tan^-1 p = 1 / 1 + p²;

One condition is given for this inverse derivative function as:

When we put the limit of p is +∞ then we get the value of derivative of tan inverse p is 0. (know more about arctan, here)   

= d / dp tan^-1 p = 0;  

This is how we proved derivative of tan inverse 'p'. Tangential Acceleration is a rate in which the velocity of a body change with time. The cbse syllabus for class 9th 2013 is necessary for 9^th class student.

arcsin

In mathematics, trigonometric functions deals with the angles and length of a triangles and also we will study different types of trigonometric functions. Let’s talk about the derivative of a function f (p) which is given by:

d / dp (f (p) ) = lim _{h →0} f (p + h) – f(p) / h.

This function is also said to be derivative or we can say differentiation with respect to ‘p’. In some cases, the differentiation of a function f (p) is known as differentiation coefficient of f (p). Now we will discuss how to find the inverse sine function or arcsin? To find this first first we will see the range and domain of arcsin function. (know more about arcsin, here)

Here the Sin^-1 p is a function and its domain value lies in between [-1, 1] and with range as [-⊼/2, ⊼/2].

Now differentiate sin^-1 p;

So we can write it as in differentiate form:

= d / dp sin^-1 p,

Let f (y) = sin^-1 y then,

F (sin⊖) = ⊖;

If we differentiate we get:

F’ (sin⊖) cos ⊖ = 1;

We can also write it as:

F’ (sin⊖) = 1 / cos⊖

Let y = sin⊖;

As we know, cos ⊖ = √ (1 – sin²âŠ–);

We know that, value of sin ⊖ is ‘y’;

So plug in value ‘y’ in place sin ⊖;

=> Cos ⊖ = √ (1 – y²);

And we know that,

= f’(sin⊖) = 1 / cos⊖

Sin ⊖ = y;

And value of cos ⊖ is √ (1 – y²);

So put the value of sin ⊖ and cos ⊖ in the given formula:

F’ (sin⊖) = 1/ cos⊖

On putting value we get:

F’ (y) = 1/√ (1 – y²),

So the differentiation of sin^-1 p is:

d / dp sin^-1 = 1 / √ (1 – p²).

This is how we can sole the inverse of arc sin function. The Constant Acceleration Equations is given by: v = v₀ + at. To get more information about acceleration then free download cbse books. It is given more systematic in cbse books and In the next session we will discuss about arctan. 

Derivative of Tanx

In the previous post we have discussed about Derivative of Cos and In today's session we are going to discuss about Derivative of Tanx, The equation forms derivatives that are in relation with ordinary and partial derivatives are said to be differential equation. Let’s talk about the derivative of a function f (a) which is given by: d/da (f(a)) = limh →0 f (a + h) – f(a) / h; This given function is also known as derivative or we can say differentiation with respect to‘a’. In some cases the differentiation of a function f (a) is known as differentiation coefficient of f (a). Now we will talk about the Derivative of Tanx. The derivative of tan x is given by:

d / dx = tan x = sec2 x.

Let’s discuss the proof of derivative of tanx. If we want to find the derivative of tan x than it is necessary to find the derivative of sin x and cos x because we know that tan x is written in the form of sin x / cos x; so by using quotient rule: So we can write it as: Tan x = sin x / cos x; In the derivative form we can write it as:

d/dx Tan x = d/dx (sin x / cos x);

On further solving we get:

(cos (x) d/dx sin (x) – sin (x) d/dx  cos (x)) / cos2 (x);

As we know that differentiation of sin (x) is cos (x) and differentiation of cos (x) is – sin (x). So put the value of sin (x) and cos (x) in the above expression we get.

(cos (x) cos (x) + sin (x) sin (x)) / cos2 (x); We can also write it as:

= 1 + tan2 (x);

We know that 1 + tan2 (x) = sec2 x;

Now we will discuss What is a Function in Algebra.  We know that function is used to show the relationship between the given set of elements that is known as domain of a function and the given set of elements is said to be co-domain of a function. To get more information please go through the cbse syllabus for class 9th.

Derivative of Cos

Trigonometry is a branch of mathematics that is used to show the relation between the angles of triangle to the lengths of sides of a triangle. In trigonometry we will study the different types of derivatives but here we will see the Derivative of Cos. Now we will see the how to find the derivative of cos (s). The derivative of cos (s) is mention below:
Cos (s) = - sin (s); Now the prove of derivative of cos (s);
First we write the cos (s) in the derivative form:
d / ds cos (s) = - sin (s);
Here we have to apply the chain rule to find the derivative of cos (s).
In mathematics, Chain rule is used to differentiating compositions of functions. The chain rule is given by: D f (g(s)) = f’ (g (s)) g’(s); on applying we can write it as:
→ cos (s) = sin (s + π / 2); in the derivative form we can write it as:
→ d / ds cos (s) = d / ds sin (s + π / 2); On further solving the derivative we get:
→ d / ds sin (u) * d / ds ( s + π / 2) (put u = s + π / 2);
We know that differentiation of sin s = cos s, so put in the above expression.
= cos (u) * 1 = cos (s + π / 2);
On further solving we get – sin (s). This is how to solve the derivative of cos (s). And also there are different ways to Solving Equations with Fractions. We know that any number that is written in the form of i/o is known as fraction. Before entering in the examination hall please focus on icse sample papers 2013 so that we can easily face the problem which is occur in the exam and In the next session we will discuss about Derivative of Tanx.
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