# Optimisation Concepts And Techniques

Rules of Differentiation

The procedure of ascertaining the derivative of a function is termed as differentiation. The derivative of a function denotes the variation in the dependant variable due to a microscopically small variation in the independent in the independent variable and is written as dY / dX for a function Y = f(X). A sequence of decrees has been derived for differentiating several types of functions. Let us discuss each one of them.

1. Derivative of a Sum or Difference of Two Functions

The derivative of a sum the two functions parties the sum of the derivatives procured independently of the two functions.

Let us consider an illustration of this model.

Illustration 1

Presume, Y = 3X^2 + 4X. Calculate the derivative of a sum two functions.

Solution

dY       =          [2*3(X)^2-1]   +          [1*4(X)^1-1]
dX

=          2*3X   +          1*4X^0

=          6X       +          4

Therefore, Y  =          6X + 4

Foot note: Anything to the power zero is equal to 1.

Illustration 2

Now let us see an illustration containing three terms representing a function.

Presume π = -80 + 280Q – 20Q^2, where π stands for profit and Q for level of productivity. Compute the derivative of π with respect to Q.

Solution

=          -80*0 + 280*(1*Q^1-1) – 20*2*Q^2-1
dQ

=          0 + 280*1 – 40*Q^1

=          280 – 40Q

Foot note: The derivative of a constant i.e. (-80) is zero.

1. Derivative of a Product of the two function
2. If suppose Y is the product of the two individual functions f(X) and g(X), then Y = f(X)*g(X). The derivative of the second function in addition to the second function multiplied by the derivative of the first function, therefore, dY / dX = (f(X)*dg(X) / dX) + (g(X)*df(X) / dX).

Let us see an illustration in this model.

Illustration 3

Assume Y = 20X^2 (4X + 6). Compute the derivative of the product of the two functions.

Solution

dY       =          20X^2 * d(4X+6)       +          (4x+6) * d(20X^2)
dX                                      dX                                                          dX

derivative of 4X+6 = 1*4X^1-1 = 1*4X^0 = 4 and

derivative of constant 6 = 0

derivative of X = 1*X^1-1 = 1*X^0 = 1

derivative of 20X^2 = 2*20X^2-1 = 40X^1 = 40X

Thus,                                       20X^2 * 4 + (4X+6) * 40X

=          80X^2 + 160X^2 + 240X

=          240X^2 + 240X

=          240(X^2 + X)

1. Derivative of the Quotient of the Two Functions
2. If suppose the variable Y parities to the quotient of the two functions f(X) and g(X). That is Y = f(X) / g(X), the derivative of the quotient of the two functions parities to the denominator times the derivative of the nominator minus the nominator times the derivative of the denominator and the whole divided by the denominator whole squared. Thus the following function is obtained.

dY       =          g(X) * df(X) – f(X) * dg(X)
dX                               d(X)                d(X)
[g(X)]^2

Let us see an illustration of this model.

Illustration 4

Assume the following function of Y = 10X + 4 / X-2. Compute the derivative quotient of the two functions.

Solution

f(X) = 10X+4 and g(X) = X-2, therefore,

dY       =          (X-2) * d(10x+4) – (10X+4) * d(X-2)
dX                                   d(X)                            d(X)
(X-2)^2

Derivative of 10x+4 = 10

Derivative of X-2 = 1

Derivative of X = 1

Therefore,

=          (X-2) * 10       -           (10X+4) * 1
(X-2)^2

=          (10X – 20)       -           (10X + 4)
(X-2)^2

=          10X – 10X      +          (4-20)
(X-2)^2

=             - 16
(X-2)^2

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