# 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.

__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
__

__dπ__= -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.**

__Derivative of a Product of the two function__

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)**

__Derivative of the Quotient of the Two Functions__

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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**Other topics under Managerial Economics: Nature, Scope and Optimization Techniques:**

- Basic Model of the Firm and Role of Profits
- Demand and Demand Function
- Derivative of Function of a Function (Chain Rule)
- Macro Economics Policy
- Managerial Assessment Making Procedure
- Marginal and Incremental Analysis
- Maximisation by Marginal Examination
- Restrained Maximisation
- Restrained Maximisation: Substitution Method