# 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

=          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

=          100 – 8X - Y
dX

=          -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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