Sunday, July 28, 2019

Differential Calculus: Derivative Formulas (Part 1)


* Differential Calculus

- Branch of mathematics that deals with the rate at which a variable or function is changing.

Formula:
 
1.  d(C)/dx


- The derivative of any constant (C) is always equal to zero.

Examples:

d(3) / dx = 0

- read as "derivative of 3 with respect to x is equal to 0"


* d(e) / dx = 0


*d(𝝿) / dx = 0


* d(1,000,000) / dx = 0




2. d(u + v)/dx
  = du/dx + dv/dx

- Derivative can be distributed among the separated functions.

Example:

* d(3x + 4) / dx
 = d(3x)/dx + d(4)/dx


= 3 + 0


= 3




3. d(x
n) / dx = nxn-1

- The derivative of a variable (x) raised to exponent (n) is equal to the product of the exponent (n) and the variable (x) raised to the value of exponent (n) minus 1.

Examples:

* d(3x
2) / dx


= (2)(3)(x)2-1

              
= 6x 


* d(4x
- 2x+ 5x - 10)/dx 


d(4x3)/dx - d(2x)/dx + d(5x)/dx d(10)/dx


= (3)(4)(x)3-1 - (2)(2)(x)2-1 + 5 - 0


= 12x2 - 4x + 5


* d(5x3) / dx 


= (3)(5)(x)3-1


= 15x2


* d[(x+ 2x + 1)/(x + 1)]/dx 

d[(1)/(x + 1)]/dx

= d(x + 1)/dx

= d(x)/dx + d(1)/dx

= 1




4. d(Cu)/dx = C

- The derivative of the product of a constant (C) and a variable (u) raised to 1 is equal to the value of the constant.

- Derived from formula no. 3

* d(Cx)/dx


= (1)(C)x1-1


= (1)(C)(x0)


= (1)(C)(1)


= C


Examples:

* d(3x)/dx
 = 3


* d(5y)/dy = 5


* d(x)/dx = 1




5. d(uv)dx = ud(v)/dx + vd(u)/dx


Example:

* d[(x + 2)(3x2)]


= (x + 2) * d(3
x2)/dx + (3x2) * d(x+2)/dx


= (x + 2)(6x) + (3x2)(1)


= 6x2 + 12x + 3x2


= 9x2 + 12x


* d[(5
x- 4)(2x + 3)] / dx


= (5x2 - 4) * d(2x + 3)/dx
 + (2x + 3) * d(5x2 - 4)/dx


= (5x2 - 4)(2) + (2x + 3)(10x)


= 10x2 - 8 + 20x2 + 30x


= 30x2 + 30x - 8




6. d(un) / dx = nun-1 . d(u)/dx 

- If a non-monomial quantity (u) is raised to an exponent (n) not equal to zero but greater than or less than 1, the exponent will be multiplied to the derivative of the quantity (u) and the quantity raised to its exponent (n) minus by 1.

Example:

* d
(7x - 4)2 / dx = (2)(7x - 4)2-1 . d(7x - 4)/dx


= (2)(7x - 4) . 7


= 14(7x - 4)


= 98x - 56

Thursday, July 11, 2019

Methods of Integration: Powers of Tangent and Secant - Indefinite Integral


  When the integrands are Tangent and Secant together that are raised with a particular power it is better to know the techniques on simplifying it.

Here’s how the given problem should look like,

General form:

    ∫ tanmu secnudu

Where m and n are exponents (constant)

Here are the rules that will help us:

Rule I. If m is any number and n is even.

    Use:
      ∫ tanmu secn-2u sec2udu

Rule II. If m is a positive odd integer and n is any number.

    Use:
        ∫ tanm-1u secn-1u secu tanudu

Rule III. If m is a positive integer and n is zero.
 
    Use:
        ∫ tanm-2u tan2udu



Examples:

Rule I

1. ∫ tan3x sec4xdx = ∫ tan3x sec4-2x sec2xdx

= ∫ tan3x sec2x sec2xdx

= ∫ tan3x (tan2x + 1) sec2xdx

= ∫ [tan5xsec2x + tan3xsec2xdx]dx

= ∫ tan5xsec2xdx + ∫ tan3xsec2xdx

Let:

u = tanx
du = sec2xdx

= tan6x/6 + tan4x/4 + C


Rule II

2. ∫ tan3x sec5xdx

= ∫ tan3-1x sec5-1x secxtanx dx

= ∫ tan2x sec4x secxtanx dx

= ∫ (sec2x - 1) sec4x secxtanx dx

= ∫ [sec6x secxtanx - sec4x secxtanx]dx

= ∫ sec6x secxtanxdx - ∫ sec4x secxtanxdx

Let:

u = secx
du =secxtanxdx

= sec7x/7 - sec5x/5 + C


Rule III

3. ∫ tan3xdx

= ∫ tan3-2x tan2xdx

= ∫ tanx tan2xdx

= ∫ tanx (sec2x - 1)dx

= ∫ [tanxsec2x - tanx]dx

= ∫ tanxsec2xdx - ∫ tanxdx

= tan2x/2 - ln(secx) + C

Note:
    - These rules can also be used in a problem having cosecant and cotangent together as an integrand.

Slope, Angle of Inclination and Angle Between Two Lines - Analytic Geometry


* Slope of a Line

Formula:
 
(Slope Intercept Form)

y = mx + b

m = rise/run
 = (y2 - y1) / (x2 - x1)


Where:

x1 and y1 = coordinates of first point
x2 and y2 = coordinates of second point

Note:
- In solving for the slope of a line, there must be first and second (points needed) since there are uncountable points that forms a line. Terms such as "first" and "second" points are just used to avoid confusion because we have to be consistent on what values to be substituted.

Example:

Given:

Two points = (2, 4) and (5, 6)

Required: m

Solution:

First let what to be the first and second point. Whether we let (2, 4) or (5, 6) as our first point, it doesn't matter because it will not affect the answer.

Let:

P1(2,4)
P2(5,6)

m = 
(y2 - y1) / (x2 - x1)

m = 
(6 - 4) / (5 - 2)

m = 2/3



(Solution 2)

Let:

P1(5,6)
P2(2,4)

m = 
m = (y2 - y1) / (x2 - x1)

m = 
(4 - 6) / (2 - 5)

m = -2 / -3

m = 2/3

- We just got the same answer!








* Angle of Inclination

- It is the angle formed by a line with respect to the positive x-axis.

Formula:

tanϴ = m

Where:

ϴ = angle of inclination
m = slope of the line

Examples:

* Determine the angle of inclination of a line passing through (-3, -2) and (5, 1).

Given: 

P1(-3, -2)
P2(5, 1)

Required: ϴ

Solution:

tanϴ = m

ϴ = tan-1 [(y2 - y1) / (x2 - x1)]

ϴ = tan-1 [(1 - (-2)) / (5 - (-3))] 

ϴ = tan-1 [(1 + 2) / (5 + 3)] 

ϴ = tan-1(3/8)

ϴ = 20.36 degrees



* Find the slope of the line having an angle of inclination equal to 45 degrees.

Given:

ϴ = 45 

Required: m

Solution:

tanϴ = m

m = tan(45 degrees)

m = 1


* The equation of the line is 3x - y + 4 = 0. Find its angle of inclination.

Given:

3x - y + 4 = 0

Required: ϴ

Solution:

Make in the form:

y = mx + b

3x - y + 4 = 0

y = 3x + 4

tanϴ = m

ϴ = tan-1(3)

ϴ = 71.57 degrees








* Angle between Two Lines

    - angle formed by two intersecting lines.

Formula:

ø = ϴ2 - ϴ1

tanø = (
tanϴ2 - tanϴ1) / (1 + tanϴ1tanϴ2)

tanø = 
(m2 - m1) / (1 + m1m2)


Where:

ϴ1 = angle of the first line
ϴ 2 = angle of the second line
m1 = slope of the first line
m2 = slope of the second line
ø = angle between the two lines

Note:
    - If you get a negative angle just get its absolute. Perhaps, m2 and m1 have just been interchanged.

Example:

* Two intersecting lines have slopes of m = 2 and m = 3. Find the angle between them.

Given:

Let:

m1 = 2
m2 = 3

Required: ø

Solution:

tanø = 
(m2 - m1) / (1 + m1m2)

tanø = (3 - 2) / (1 + (2)(3)) 

ø = tan-1(1/7)

ø = 8.13 degrees