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(xⁿ) / dx = nx^n-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(3x²) / dx

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

              

= 6x 


* d(4x³ - 2x² + 5x - 10)/dx 

= d(4x³)/dx - d(2x² )/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[(x² + 2x + 1)/(x + 1)]/dx 

= d[(x + 1)² /(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)x^1-1

= (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(3x²)/dx + (3x²) * d(x+2)/dx

= (x + 2)(6x) + (3x²)(1)

= 6x² + 12x + 3x²

= 9x² + 12x

* d[(5x² - 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 = nu^n-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

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:

    ∫ tan^mu 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:
      ∫ tan^mu sec^n-2u sec²udu

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

    Use:
        ∫ tan^m-1u sec^n-1u secu tanudu

Rule III. If m is a positive integer and n is zero.
 

    Use:
        ∫ tan^m-2u tan²udu

Examples:

Rule I

1. ∫ tan³x sec⁴xdx = ∫ tan³x sec^4-2x 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

= ∫ tan^3-1x sec^5-1x 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

= ∫ tan^3-2x 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.