# Derivative of Function Of A Function (Chain Rule)

According to chain rule if Y = F(U) and U = g(X), the derivative of Y with respect to X can be procured by multiplying mutually the derivative of Y with respect to U and the derivative of U with respect to X.

Let us see an illustration of this type.

__Illustration 5__

Presume Y = U^3 + 30 and U = 6X^2. Compute using the Chain rule.

__Solution:__

__dY__ = __dY__ * __dU__

dX dU dX

= __d(U^3
+ 30)__ * __d(6X^2)__

dU dX

Derivative of U^3 + 30 = 3*U^3-2 = 3U^2

Derivative of 6X^2 = 12X^2-1 = 12X^1 = 12X

Derivative of dU and dX = 1

= 3U^2 * 12X

Substituting U = 6X^2 in the above, we get the following.

__dY__ = 3(6X^2)^2
* 12X

dX

= 18X^4 * 12X

** = 216X^5**

__Application of Derivatives to Optimisation Problems__

__Use in Profit Maximisation__

__Illustration 6__

The profit function is as follows: π = -50 + 80Q – 20Q^2. π is the profit and Q is the units of productivity. Compute the derivative function setting it to zero.

__Solution__

__dπ__ = 80 – 40Q

dQ

80 – 40Q = 0

40Q = 80

Q = 80 / 40

** Q = 2.**

Hence, at 8 units of productivity profits will be optimum.

Substituting the value of Q = 2 in the primary function gives the following.

π = -50 + 80(2) – 20(2)^2

= -50
+ 160 – 20*4

= 110 – 80

** = 30**

**Therefore, the productivity level of 2 units, profits are equal to 30.**

__Minimisation Problem__

__Illustration 7__

Let us assume the average function of a firm is given as below:

AC = 10,000 – 90Q + 0.5Q^2

Ascertain the level of productivity to minimise the average cost.

__Solution__

To derivate the Average Cost Function with respect to productivity Q is attained by parity to zero. Therefore,

__d(AC)__ = -90
+ 1Q

d(Q)

Defining the equation by setting to zero procures,

-90 + Q = 0

**Hence, Q = 90**

**By applying the second condition to make certain whether it is really minimum,
we consider the second derivation of AC function**

** d^2AC = 1**

**dQ^2**

**As the second condition derivative AC function is positive and d^2AC > 0, **

**dQ^2**

**productivity of 90 units of output is the one that minimises the average cost of manufacturing.**

__Multivariate Maximisation__

__Illustration 8__

Let us assume the following two product function of a firm. π = 100X – 4X^2 – XY – 8Y^2 + 150Y, where X and Y are two separate variables denoting the levels of productivity of two products.

Ascertain the levels of productivity of the two products that optimises profits.

__Solution__

__dπ__ = 100 – 8X
- Y

dX

__dπ__ = -X – 16Y
+ 150

dY

For optimising profits we should the derivatives to zero.

100 – 8X – Y = 0 …..Equation (1)

150 – X – 16Y = 0 …..Equation (2)

To solve the equation, we can multiply the Equation (1) by -16, resulting the following

-1600 + 128X + 16Y = 0 …..Equation (3)

__+
150 - X - 16Y = 0__ …..Equation
(2)

-1450 + 127X = 0

127X = 1450

X = 1450 / 127

** X = 11.42**

Substituting the value of X = 11.42 in the Equation (1) we get the following.

100 – 8*11.42 – Y = 0

** Y = 8.64**

**Therefore, the firm will maximise profits of its productivity and sells 11.42
units of products X and 8.64 units of product Y.**

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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
- Macro Economics Policy
- Managerial Assessment Making Procedure
- Marginal and Incremental Analysis
- Maximisation by Marginal Examination
- Optimisation Concepts and Techniques
- Restrained Maximisation
- Restrained Maximisation: Substitution Method