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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How to Learn Critical Point Problems

In the previous post we have discussed about  How to Define Inflection Point and In today's session we are going to discuss about Critical Point Problems, A critical point is a most important concept of calculus in mathematics. According to mathematical definition, a critical point can be define as a variable for any function with real variable where function has their derivative equal to zero or  function is non-differentiable. If any real value of a function exists in a critical point situation then that value consider as a critical value of function. The concept of critical point can be represented by using the graph of function f(). Here the graph should not be admit as tangent or graph’s tangent should be a line in horizontal or vertical form.

The concept of critical point can be defined as an expression if any function f (s) is in the critical point f' (s) = 0 and the value of the co domain of ‘f’ for which the derivative of the function is zero are known as the critical value that means value of ‘x’ on which differentiation of the function is not possible is known as the critical value.
We can define it by a simple example as if there is a function f (s) = s² then function gives the zero when s = 0 that means when s = 0 then differentiation of the function is not possible, so s = 0 define as the critical point of the function f (s).
It is also define as the points on the graph that is defined on the basis of the given function for which the value of the derivative is equal to zero or the derivation of the function is not possible. It is define by a simple graph plotted on the basis of defined function in which critical point is defined as when tangent is drawn on the critical point either it does not exist on the curve or the tangent line does not cross it.
Differentiation Strategy can be define as set of rules and concept to perform the differentiation on the function. indian science engineering eligibility test books is a syllabus that contains the detail description with topics of general ability subject which helps in the student to make their preparation in a better way.

How to Define Inflection Point

In the previous post we have discussed about Implicit Differentiation and In today's session we are going to discuss about Inflection point, It is a term which is generally used under the theory of the differential calculus in the subject of the math. We all must have heard about the point of the inflection but many of us do not know its exact meaning. So in this article we will discuss some of the facts about the inflection point.
We will now begin with the definition of the inflection point. The point of the inflection is generally defined as a point existing on the curve where the sign of the concavity or the sign of the curvature changes from the positive to the negative or from the negative to the positive. By the above definition of the point of the inflection we mean to say that at the point of the inflection the curve either changes from the concave upwards which is known as the positive curvature to the concave downwards which is known as the negative curvature or can change from the concave downwards that is the negative curvature to the concave upwards that is the positive curvature.
We have now discussed about the definition of the point of the inflection in the last paragraph so let us now take an easy, simple and a practical example in this paragraph to understand about the point of the inflection more precisely. Suppose a person is driving a car along a road which is of winding type then the point of the inflection there is that point where the wheel of the steering straight for a moment when the car is just turned from the right to the left or from the left to the right.
In order to get more help on the topics: Inflection Point, Formula for Area of a Triangle and cbse previous year question papers class 12, you can use various Online Portals.

Implicit Differentiation

In the previous post we have discussed about inverse trig functions and In today's session we are going to discuss about Implicit Differentiation. Hello friends today we will be discussing a new topic Implicit Differentiation Calculator, firstly we will talk about the normal differentiation then we will move to  implicit differentiation, if we talk about normal differentiation then in normal differentiation we differentiate with respect to one variable  as if have to differentiate  y = 7x+y  with respect to x then  the differentiation will be dy /dx = 7 ,  as we are differentiating with respect to x so all the variable which are not having x with them the differentiation  of them will be 0.  But in implicit differentiation we will differentiate with respect to both the function x and y.  As in the above case the differentiation of y was zero but if we do implicit differentiation then the differentiation of y will not be zero it will be dy/dx.
Step 1: in this step we have to separate the terms, means the term containing x one side and term containing y one side.  (know more about Implicit and explicit functions , here)
Step 2: the term which don’t have x and y will be treated as constant and we have to differentiate the other terms..
Step 3: now find the value of dy/dx and that will be the solution for the given problem.
Now we will see one example in which we will do the implicit differentiation of a function.
Example: find the implicit differentiation of
Y = 3x + 4y2
Solution
Step 1: we have to separate the terms as,
Y – 4y2 =3x
Step 2 :  dy /dx – 8y* dy/dx = 3
Take dy/dx as common
Dy/dx (1 – 8y) =3
Step 3:
Dy/dx = 3/(1 -8y)
This is the required solution for the given problem.
In this way we can find the implicit differentiation of any function.
If you are appearing in Karnataka board then go through the topic What is Median , it is very important from exam point of view.