Derivatives of Power Functions of e Explained

What is the Derivative of Ex

Ex here is the natural exponential function which is written as e^x, where x is any real number. To find the derivative of e^x first let us understand what a derivative is. Derivative can be defined as the instantaneous rate of change of a function with respect to one of its variables which in simple words is the slope of the tangent line to the function at a point. The derivative of a function f(x) with respect  to ‘x’ is the function f’(x) given by , f’(x) = lim(h?0) [f(x+h) – f(x)]/h where f’(x) is read as ‘f prime of x’. Let us now find what is the derivative of Ex. We need to  find the derivative of e^x to get the derivative of Ex. Let us take y= e^x, which is the inverse of y= ln x, for which we can obtain the derivative.  y= e^x implies ln y = ln e^x = x. Taking derivatives on both sides of lny = x, we get d[ln y]/dx = d[x]/dx  and applying the chain rule to ln y, we get, 1/y. y’ = 1 which gives, y’ = y . So, in this case we can see that the derivative of y is y which means derivative of e^x is itself. We can conclude that the Derivative of e^x is e^x

Let us now find the derivative of e with a functional exponent. Let y = e^u(x), applying the chain rule, we get, d[e^u(x)]/dx=  d[e^u(x)]/dx. du(x)/dx which gives us e^u(x). du(x)/dx. Using the derivative of e with a functional exponent, let us find the derivative of E 4x. Here u=4x, d[e^4x]/dx would be, e^u. d[u]/dx. du/dx = d(4x)/dx = 4. So, substituting u=4x and du/dx = 4, we get, derivative of E 4x as 4e^4x.

Derivative of E Ax would be the derivative of e^Ax. Taking  g(x)=Ax where A is some constant, let us apply the chain rule to find the derivative of  E Ax. As per the chain rule, f(g(x)’ = f’(g(x)).g’(x). Here f(x) = e^Ax and g(x)=Ax. Let us find the derivatives of each functions, f’[g(x)] and g’(x); f’g(x) = e^Ax and g’(x)=A. Substituting these values we get Derivative of E Ax as A.e^Ax

Derivative of E -1 is written as derivative of e^-1. e^-1 is something like e^x where x = -1. We know that the derivative of e^x is e^x and hence the derivative of e^-1 is e^-1. So, Derivative of E -1 is e^-1

Standard deviation of the mean

Standard deviation of the mean: If the values of x or f are large, the calculation of AM by the direct method is quite tedious and time consuming, because calculations involved are lengthy. In such a case to minimize the time involved in calculation, we take deviations from an arbitrary point as discussed below. Let x1, x2, x3, …., xn be values of a variable x with corresponding frequencies f1, f2, f3,…, fn respectively. Taking deviations about an arbitrary point ‘A’, we have

Taking the deviation d of the mid-point of class interval from the mean, squaring it to get d2, multiplying this by the frequency of the class, i.e., fd2, adding all the items, i.e., Sfd2, taking the average, Sfd2/n, and then taking the square root, we get, Standard deviations from the mean is given by S.D. = s = sqrt(Sfd2/n)

Standard deviation from the mean: To understand the concept of standard deviation from mean, we should know The population standard deviation, or s, is simply the square root of the population variance. Because the variance is the average of the squared distances of the observations from the mean, the standard deviation is the square root of the average of the squared distances of the observations from the mean. While the variance is expressed in the square of the units used in the data, the standard deviation is in the same units as those used in the data.
The standard deviation of the mean formula is
s = sqrt(s2) = sqrt(S(x – µ)2/N) = sqrt((Sx2/n) – µ2)
where x = observation, µ = population mean, N = total number of elements in the population, S = sum of all values (x – µ)2, or all the values x2, s = population standard deviation,s2 = population variance.The square root of a positive number may be either positive or negative because a2 = (-a)2. When taking the square root of the variance to calculate the standard deviation, however, statisticians consider only the positive square root.

Standard deviation of the mean example
Example: Find out the standard deviation of the following items: 8, 10, 12, 14, 16, 18, 20, 22, 24, 26.
Solution: Calculation of standard deviation.
Size of items (x) Deviation from mean = 17(d) (d2)
8 -9 81
10 -7 49
12 -5 25
14 -3 9
16 -1 1
18 +1 1
20 +3 9
22 +5 25
24 +7 49
26 +9 81
Sx = 170 Sd2 = 330

Arithmetic average or  = Sx/n = 170/10 = 17
Standard deviation or s = sqrt(Sd2/n) = sqrt(330/10) = sqrt(33) = 5.74

Geometry of circle

A circle is a conic section. When a cone is cut by a plane that is exactly perpendicular to the axis of the cone, the cross section we get is a circle. All points on the circle are equidistant from a fixed point in the circle. This fixed point is called the centre.

Radius of a circle:

The distance between the centre of a circle and any point on the circle is called the radius of that circle. The radius is half the diameter of the circle. So if we denote the radius by r and the diameter by d then, r = d/2

Circle formula:

We can measure the circumference of a circular object by winding a piece of fine string around the curved surface of the object exactly once and then measuring the length of the string with a meter scale. On measuring the circumference of a number of objects we find that the value of the ratio Circumference/diameter in each case is almost the same. This would always be some number between 3.1 and 3.2. This constant ratio is named by the Greek letter pi (pronounced as pi). Therefore we can write the formula for circumference of a circle as : C = pid, where d = diameter of the circle.

Area of a circle is given by the formula: A = pi*r^2 = pi * (d/2)^2 = (pi/4)d^2

Area of a semicircle:

We know that a semicircle is formed when a diameter divides a circle into two equal halves. So obviously the area of a semicircle is exactly half the area of the circle of the same diameter. Mathematically it is written like this,
A.S. = (pi/2)*r^2, where A.S. = area of semi circle, r = radius of the semi circle.

Area of a quadrant of a circle = (1/4)* pi*r^2

Circle Geometry:

Consider a unit circle centered at the origin(O) of a co-ordinate axis. A point Q on the circle is such that the line segment OQ makes an angle of h with the positive x axis. Then the co-ordinates of the point Q would be (cos(h), sin(h)). If the radius of the circle is r, then the co-ordinates of the point Q would be (r*cos(h), r*sin(h)).

From the above figure we see that triangle OQS is a right triangle. The radius of the circle is 1, so OQ = 1. Therefore adjacent side to angle h = cos(h) and opposite side to angle h = sin(h).

When and How to Use Law of Sines and Cosines

What type of triangles use Law of Cosines and Sines 
In a right triangle, we can find the unknown sides or angles using the Pythagorean Theorem. But if the given triangle is not right triangle and is an oblique triangle then how do we go about? In such triangles we use the Law of Cosines and Sines to solve triangles. Sine and Cosine Laws are given as follows: Law of Sines: a/Sin(A)=b/Sin(B)=c/Sin(C)
Law of Cosines: a^2=b^2+c^2-2bc Cos A; b^2=c^2+a^2-2ca Cos B ; c^2=a^2+b^2-2ab Cos C

In a given oblique triangle, when do we use  Law of Sines and Law of Cosines
The Law of Sines are used when we know two sides and one opposite side or when we know two angles and one opposite side of an oblique triangle.  The Law of Cosines are used when we know two sides and the included angle or given the three sides of an oblique triangle.

 Law of Sines and Cosines
The Law of Sines help to establish a relationship between the side lengths and angles of a triangle ABC. There are three Sines and hence the relationship explains the plural ‘s’ of Law of Sines.
 The Law of Sines are, a/Sin[A] = b/Sin[B] =c/Sin[C] or we can even write them as Sin[A]/a=Sin[B]/b=Sin[C]/c ;  where a,b,c are the side lengths and A, B and C are the opposite angles of the respective sides a,b,c in the  oblique triangle ABC
The Law of Cosines is used most widely than the Law of Sines. When we know two sides of a triangle and their included angle, then Law of Cosines enables us to find the third side. The plural‘s’ in the law of Cosines is used as there are three cosines and hence by rotation similar formulas are valid for other angles. Law of Cosines are, a^2=b^2+c^2-2bc Cos A; b^2=c^2+a^2-2ca Cos B ; c^2=a^2+b^2-2ab Cos C.  The Law of Sines and Cosines, are also known as the Sine Rule and the Cosine Rule.

Solving  Law of Sines and Cosines Word Problems
Let us solve some Law of Sines and Cosines Problems with the given side lengths and angles
Example: Peter wants to measure the height of a tree. He walks 100ft from the base of the tree and looks up. The angle of elevation found is 33 degrees. This particular tree grows at an angle of 83 degrees with respect to the ground rather than vertically. Calculate the height of the tree.
Solution:
Angle B=83 degrees , Angle A= 33 degrees, c=100ft [two angles and included side]
Angle C= 180-[angle A+angleB] = 180-116= 64 degrees
Using Sine Rule,  a/SinA = c/SinC
              a = c . SinA/SinC = 100. Sin(33)/Sin(64) = 100. (0.606) = 60.6
So, the height of the tree calculated by Peter is 60.6 ft

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