Derivatives of inverse trigonometric functions

Like many functions, the trigonometric functions also have inverse.  Just like how we can find the derivative of trigonometric functions, we can also find the derivatives of inverse trigonometric functions. In this article we shall take a quick look at inverse trig functions derivatives. To be able to find the standard formulae for derivatives of inverse trig functions, we would need the formula for derivative of any general inverse function.

That would be like this:
f’(x) = 1/g’(f(x)), where f and g are the inverse functions of each other.
Let us first try finding the inverse trig function derivatives of the sine function. The sine function inverse is written as arc sin (x). Therefore the function would look like this: y = arcsin(x),
Therefore, x= sin (y) for –pi/2 ≤ y ≤ pi/2. We fix this domain for the sine function to ensure that our inverse exists. If we don’t restrict the domain, then y could have multiple values for same value of x. For example, sin (pi/4) = 1/sqrt(2) and sin (3pi/4) is also 1/sqrt(2). With that in mind, we can write the relationship between sin and arc sin as follows:
Sin(arc sin (x)) = x and arc sin(sin(x))  = x
Thus here our f(x) = arc sin(x) and g(x) = sin (x), then using the derivative of inverse formula that we stated above we have:
f’(x) = 1/f’(g(x)) = 1/cos (arc sin (x))
This formula may to be applicable in practice. So let us make a few changes in there. We had earlier x = sin y so y = arc sin (x). Using that, the denominator of our derivative would become
Cos(arc sin (x)) = cos y
Next we use our primary trigonometric identity which was:
Cos^2 (x) + sin^2 (x) = 1
Thus, Cos^2 (y) + sin^2 (y) = 1 from this we have
Cos (y) =√( 1- sin^2 (y)) = √(1 – (sin y)^2). But we have already established that sin y = x. So replacing that we get,
Cos (y) = √(1 – x^2). So now plugging all that back to our equation of derivative of inverse trig functions for sine f’(x), we have:
f’(x) = 1/cos(arc sin(x)) = 1/cos y = 1/√(1-x^2)
Therefore we see that derivative of arc sin (x) = 1/√(1-x^2).
The derivatives of other trigonometric inverse functions of arc cos, arc tan etc can be derived in a similar way. They are:
d/dx arc cos (x) = -1/√(1-x^2) and d/dx arc tan (x) = 1/(1+x^2)

All about the hypotenuse of a right triangle

Like every triangle, a right angled triangle would also have three sides. However, in a right triangle one of the angles is a right angle. That means one of the angle measures 90 degrees (or 𝛑/2 radians). Since the sum of angles in any triangle has to be 180 degrees, in a right triangle as one angle is already 90 degrees, the sum of the other two angles have to be 90 degrees. That means that the other two angles are compliments of each other. It also means that the other two angles have to be acute angles. A typical right triangle would look as follows:

The longest side is called the hypotenuse. The side that is adjacent to the know angle is called the adjacent side and the side opposite to the known angle is called the opposite side. By default the hypotenuse will always be the side opposite the right angle.

The adjacent and opposite sides together are also called the legs of the right triangle. The length of the hypotenuse of a right angled triangle can be found using different methods, depending on what part of the triangle is given to us.

To find the hypotenuse of a right triangle given the length of the legs:
If the hypotenuse = c and the legs are ‘a’ and b. If a’ and b’ are known, then we can calculate the length of the hypotenuse using the Pythagorean rule as follows:
C^2 = a^2 + b^2
Finding hypotenuse of a right triangle given one of the angles and the adjacent side:
In the picture below,

Suppose the angle marked in red is x and the adjacent side = a, then the length of the hypotenuse H can be given by the formula:
H = a/Cos (x)
Formula for the hypotenuse of a right triangle given one of the angles and its opposite side:
Again from the picture above, if we are given the opposite = b instead of the adjacent side. Then the formula for the hypotenuse can be written as follows:
H = b/sin (x), where x is again the angle marked in red.
Thus as we saw above there are more than one ways to find the length of the hypotenuse of a right triangle.

Trapezoidal Numerical Integration

Introduction to numerical integration:  As we have seen, the ideal way to evaluate a definite integral a to b f(x) dx is to find a formula F(x) for one of the antiderivatives of f(x) and calculate the number F(b) – F(a). But some anti derivatives are hard to find, and still others, like the antiderivatives of (sin x)/x and sqrt(1 + x^4), have no elementary formulas. We do not mean merely that no one has yet succeeded in finding elementary formulas for the anti-derivatives of (sin x)/x and sqrt(1 + x^4). We mean it has been proved that no such formulas exist. Whatever, the reason, when we cannot evaluate a definite integral with an anti-derivative, we turn to numerical methods such as the trapezoidal rule and Simpson’s rule.
Numerical Integration Methods are Trapezoidal rule and Simpson’s rule. Let us now describe Trapezoidal numerical integration.

Trapezoidal Numerical Integration: When we cannot find a workable anti-derivative for a function f that we have to integrate, we partition the interval of integration, replace f by a closely fitting polynomial on each sub interval  integrate the polynomials and add the result to approximate the integral of f. The higher the degrees of the polynomials for a given partition, the better the results are.  For a given degree, the finer the partition, the better the results, until we reach the limits imposed by round-off and truncation errors.

The polynomials do not need to be of high degree to be effective. Even line segments (graphs of polynomials of degree 1) give good approximations if we use enough of them  .To see why, suppose we partition the domain [a, b] of f into n subintervals of length delta x = h = (b – a)/n and join the corresponding points on the curve with line segments.

The vertical lines from the ends of the segments to the partition points create a collection of trapezoids that approximate the region between the curve and the x-axis. We add the areas of the trapezoids, counting area above the x-axis as positive and area below the axis as negative.
T = ½ (y0 + y1) h + ½ (y1 + y2)h + ….. + ½ (yn-2 + yn-1) h + ½ (yn-1 + yn)h= h (1/2y0 + y1 + y2 + ….. + yn-1 + ½ yn) = h/2 (y0 + 2y1 + 2y2 + …. + 2yn-1 + yn).Where, Y0 = f (a), y1 = f(x1), Yn-1 = f(x n-1), yn = f(b).

The trapezoidal rule says: use T to estimate the integral of f from a to b. Numerical Integration in R As we have seen , the definition of the Riemann integral is not very efficient way to prove that a function is Riemann integral . However once it is known that a function f is Riemann integral on some interval [a, b] a modification of the definition makes it possible to evaluate the integral of simple limit.

Volume of sphere

Introduction to Sphere : A tennis ball and a fully blown football  are some familiar objects which bring to our mind the concept of a sphere .A sphere is a three dimensional geometrical object  which can be defined as follow The set of all points in space which are equidistant from a fixed point , is called a sphere.

The fixed point is called the center of the sphere and the constant distance is called its radius. A line segment through the center of a sphere, and with the end points on the sphere . All diameters of a sphere  are of constant length , being equal to twice the radius of the sphere .Thus , if d is the length of a diameter of a sphere   of radius r  then d= 2r . The length of diameter is also called the diameter of sphere .The solid sphere is the region in sphere, bounded by sphere .

Also every point whose distance from the center is less than or equal to the radius is a point of the solid sphere .A sphere can also be considered as a solid obtained on rotating a circle about its diameter.

Volume of a sphere: The volume of a sphere (v) of radius r is given by v = 4/3 ∏r3 cubic units .Let us take an example how to volume of sphere.

Find the sphere volume  of radius 7 cm We know that formula for volume of a sphere of radius r is given by v = 4/3 ∏r3  cubic units here r = 7 cm  therefore v = 4/3 x 22/7 x 7 x 7 x 7 cm3 another example of volume of a sphere.

Calculate volume of a sphere whose surface area is 154 square cm. Let the radius of the sphere be r cm .then ,
Surface area = 154 cm2 => 4∏r2  = 154 => 4 x 22/7 x r2 = 154 => r2 = 154 x 7 / 4 x 22 = 49/4 => r = 7/2 cm , so let v be the volume of sphere .
We will use volume of a sphere equation

That is , v= 4/3∏r3   = 4/3 x 22/7 x 7/2 x 7/2 x 7/2 cm3 = 179.66cm3.

Let us take more example volume of a sphere. A sphere of diameter 6 cm is draped in a right circular cylinder vessel partly filled with water .The diameter of the cylindrical vessel is 12 cm. If the sphere is completely submerged in water, by how much will the level of water rise in the cylindrical vessel?

We have radius of sphere = 3cm, volume of a sphere formula = 4/3 ∏ r3  cm3 = 4/3 ∏ (3)³ cm³ = 36cm³, radius of cylindrical vessel = 6cm.
Suppose of water level rises by h cm and radius 6 cm = (∏x 62 x h) cm 3 = 36∏h cm3 , clearly volume of the water displaced by the sphere is equal to the volume of  the  sphere =36∏h = > h=1 cm , hence water level rises by 1 cm

Difference (Newton) Quotient Made Simple

To set up a difference quotients for a given function requires an understanding of a function notation. Given the function f(x)=4x^2-3x-7. This notation is read as “f of x equals..” This implies that the value of the function, that is the y-value depends upon the replacement for x. We get the numerical value for the function by substituting a number for ‘x’. If a non-numerical quantity is substituted for ‘x’, we get an expression rather than a numerical value. One important point to be remembered is careful use of parenthesis which is essential. For instance, f(x)=4x^2-3x-7; f(3)=4(3)^2 – 3(3)-7=36-9-7=20

Derivative Quotients  at x for a function f is given by, [f(x+h)-f(x)]/h. Sometimes it is written using delta(x) for the change in x and delta(y) for change in y; delta(x)=h and delta(y)=f(x+delta(x))- f(x). The Difference Quotient is so called as it involves the operations subtraction and division. The common forms of Difference Quotient are as follows:
1. [f(x+h) – f(x)]/h
2. [f(a+h) – f(a)]/h
3. [f(x+delta(x)) – f(x)]/delta(x)

Simplifying Difference Quotient
The difference quotient is simplified to get a h or delta(x) in the denominator which can be canceled to get the final value. Let us consider a difference quotient example to understand the step involved in simplifying difference quotient ; f(x) = 4x^2-3x-7. First we need to find the function f(x+h),which we can get by substituting (x+h) in all x in the given function.  f(x+h)= 4(x+h)^2 – 3(x+h) – 7 = 4(x^2+2xh+h^2) – 3x – 3h – 7= 4x^2+8xh + 4h^2-3x – 3h – 7. Subsituting f(x+h) in the difference quotient, we get,
Difference Quotient = [f(x+h) – f(x)]/h
       = {[4x^2+8xh+4h^2-3x-3h-7] – [4x^2-3x-7]}/h
        = 4x^2+8xh+4h^2-3x-3h-7 -4x^2+3x+7]/h  (opening the parenthesis)
On simplification, we get
        = [8xh+4h^2-3h]/h  
        = h[8x +4h-3]/h            (taking h common)
        = (8x +4h-3)     (canceling h)

Difference Quotient Example
Given function, f(x) = 2x^2-1
First we need to calculate f(x+h) which is got by substituting (x+h) in all x of the function
f(x+h) = 2(x+h)^2 -1 = 2(x^2+2xh+h^2) -1 = 2x^2+ 4xh+2h^2 -1
Next we substitute f(x+h) and f(x) in difference quotient
 [f(x+h) – f(x)]/h ={[2x^2 +4xh +2h^2-1]- [2x^2 -1]}/h
= [2x^2+4xh +2h^2-2x^2+1]/h
= [4xh+2h^2]/h  (combining like terms)
= h[4x +2h]/h      (taking h common)
= [4x +2h] (canceling h)

Newton Quotient
The difference quotient is attributed to Sir Isaac Newton and hence given the name Newton Quotient. The slope of a line through the points [(x+h),f(x+h)] and [x, f(x)] is given by [f(x+h) – f(x)]/h. This expression is the Newton Quotient or Newton’s difference quotient.
Newton Quotient  =[f(x+h) – f(x)]/h

Anti Derivatives and their Rules

What are Anti derivatives?
Anti derivative is nothing but indefinite integral or primitive integral in calculus. If there is a function h, then the anti-derivative of this function will be a differential function, say H. The derivative of H will be equal to h.
H’ = h

The anti derivates are solved by a process called indefinite integration or anti differentiation, which is the opposite process of differentiation that finds the derivative.

Rules of Anti derivatives
The rules of anti derivatives are generally the reverse of the rules of the derivatives. We can say that the anti derivative of a function is equal to the sum of the derivative of the function and a constant, so it is just one step and simple.

Constant Rule
Consider a constant “a”, which has to be multiplied with the function g(y). The value obtained by multiplying the constant with the anti derivative of the function g(y) for all values of y will be equal to the anti derivative of the function g(y), which is calculated after multiplying the constant with the function g(y) for all values of y.

Sum Rule
If there are two functions f(y) and g(y), the anti derivative of the sum of the two functions will be equal to the sum of the anti derivative of the function f(y) and the anti derivative of the function g(y).

Difference Rule
If there are two functions f(y) and g(y), the anti derivative of the difference of the two functions will be equal to the difference of the anti derivative of the function g(y) from the anti derivative of the function f(y).

Reverse Rules
Listed below are the anti derivative rules that form the reverse of derivative rules:
• The anti derivative of cos y is given by sin y + a, where “a” is a constant. Thus we can say, the anti derivative of the derivative of y will result in y + a.
• The anti derivative of sin y is the negative of the sum of cos y and a.
• The anti derivative of the square of sec y is the sum of tan y and a.
• The anti derivative of the square of cosec y is the negative of the sum of cot y and a.
• The anti derivative of the product of sec y and tan y is given by the sum of sec y and a.
• The anti derivative of the product of cosec y and cot y is given by the negative of the sum of cosec y and a.

Power Rule
The power rule of the anti derivative is the reverse of the power rule of the derivative. The power rule of the derivative usually comprises of two steps as the power is brought in the front to be multiplied with the derivative and then the power is reduced by 1 and then it is simplified.

In case of power rule for anti derivative, the power rule comprises of the following two steps.
Step 1: The value of the power in the function is increased by 1. Say for example, g(y) = 6y^2. Then by increasing the power value by 1, the function will become g(y) = 6y^3.

Step 2: Divide the function g(y) obtained from step 1 by the new power value. In this example, it will be: (6 divided by 3) (y^3), which will result in g(y) = 2y^3.

Thus anti derivative of this example is given by 2y^3 + a, where “a” is a constant, as any anti derivative is the sum of the derivative and a constant “a”.

Calculus: Rules of Integration

Calculus Integration Rules
Following are the Rules of Integration where a, b, c, n are some constants and u=f(x);v=g(x) and w=h(x)
General Integration Rules
1.Integral [a dx]= ax +c
2. Integral [a f(x) dx]= a Integral [f(x) dx]
3.Integral [x^n dx]= x^(n+1)/(n+1) +c
4.Integral [f(x)+g(x)+h(x)]dx = Integral[f(x)dx]+ Integral[g(x)dx]+ Integral[h(x)dx]
5. Integral[f(x)-g(x)-h(x)]dx = Integral[f(x)dx] – Integral[g(x)dx] – Integral[h(x)dx]
6. Integration by parts: Integral [u dv] = uv – Integral [v du]
7.Integral[F(u)dx] = Integral [F(u)/u’] du
8. Integral[1/x dx] =ln|x| +c
9. Integral [1/(x^2+a^2)]dx =1/a tan^-1[x/a] +c
10.Integral [1/(x^2-a^2)]dx = [1/2a ]ln |x-a/x+a| +c

Integral Rules of Exponential Functions
1.Integral [e^x dx]= e^x +c
2. Integral [a^x dx]= a^x/ln a +c
3.Integral [ln x dx]= x(ln x -1) +c
4.Integral[log base a of x]dx= (x/ln a)(ln x -1) +c
5. Integral[x e^(ax)]dx= [e^(ax)/a^2](ax-1) +c
6. Integral[e^(ax)/x] dx = ln|x| + summation(i=1 to infinity) [(ax)^i/i.i!] +c
7.Integral[x^2 e^(ax)]dx= e^(ax)[(x^2/a – 2x/a^2 + 2/a^3)] +c
8. Integral[x^n e^(ax)]dx = (1/a)x^n e^(ax) – (n/a)Integral [x^(n-1) e^(ax)]dx
9.Integral[e^(ax)/x^n] dx = [1/(n-1)][- e^(ax)/x^(n-1) + a Integral e^(ax)/x^(n-1)]dx
10.Integral[x^n ln x] dx= [x^(n+1)]/(n+1)^2 [(n+1)lnx – 1] + c

Integration Rules of Trigonometric Functions
1.Integral [sin x dx]=  - cos x +c
2. Integral [cos x dx]= sin x +c
3. Integral [tan x dx]= ln |sec x|+ c
4. Integral [cot x dx]= ln|sin x| + c
5. Integral [sec^2(x) dx]= tan x +c
6. Integral [csc^2(x) dx]= - cot x +c
7. Integral [tan^2(x) dx]= tan x – x +c
8. Integral [cot^2(x) dx]= cot x – x +c
9.Integral [sec x tan x dx]= sec x +c
10. Integral [csc x cot x dx] = - csc x +c
11. Integral [sec x dx]= ln |sec x +tan x| +c
12.Integral [cos^2(x) dx] = x/2 + ¼(sin 2x) +c
13. Integral[sin ^n(x) dx]= (-1/n)sin^(n-1) x cos x + (n-1)/n .Integral [sin^(n-2) x dx]
14. Integral [cos^n(x)dx]= (1/n)cos^(n-1) x sin x + (n-1)/n. Integral[cos^(n-2) dx]

Integration Rules of Hyperbolic Functions
1.Integral[sinh x dx] = cosh x +c
2.Integral[cosh x dx]= sinh x +c
3. Integral[tanh x dx]= ln cosh x +c
4. Integral[coth x dx]= ln |sinh x| + c
5. Integral[sech x dx] = sin^-1[tanh x] +c
6. Integral[csch x dx] = ln tanh (x/2) +c
7. Integral[sinh^2(x)dx] = (sinh 2x)/4 – (x/2) + c
8. Integral[cosh^2(x)dx]= (sinh 2x)/4 + (x/2) +c
9. Integral[sech^2 (x) dx]= tanh x +c
10. Integral[csh^2(x)dx]= -coth x +c
11. Integral[tanh^2(x) dx]= x – tanh x +c
12. Integral[coth^2(x)dx]= x – coth x +c
13.Integral[sechx tanh x] dx =  - sech x +c
14. Integral[csch x coth x]dx = -csch x +c