From Basic to Involved Mathematics
Rules for Differentiation
Hopefully, you now have an understanding of what is happening when you are asked to differentiate something. There are a number of rules to remember that will help you in this process. If you are really keen consult a pure maths textbook to get a proof of all these formulas but if not just try to remember them!
The Constant Function Rule:
Given a function f(x) = k, where k is a constant:
f1
(x) = 0
The Linear Function Rule:
Given a function y = ax + b - a linear function:
dy/dx = a
The Power Function Rule:
Given a function y = axn
dy/dx = naxn-1,
The Sums and Differences Rule:
Given a function y = a(x) +/- b(x):
dy/dx = da(x)/dx +/- db(x)/dx - the sum of the differentials of each function
The Product Rule:
Given a function y = a(x)b(x):
dy/dx = a(x)db(x)/dx + b(x)da(x)/dx
The Quotient Rule:
Given a function y = a(x) / b(x):
| dy/dx = | b(x)da(x)/dx - a(x)db(x)/dx |
| [b(x)]2 |
The Generalised Power Function Rule:
Given a function y = [a(x)]n:
dy/dx = n[a(x)]n-1 da(x)/dx
The Chain Rule:
Given a function y = f(a) and a = f(x):
dy/dx = dy/da . da/dx
Note, this last rule is also referred to as the 'function of a function' rule. In the example, the value of y is dependent on the value of a and in turn the value of a is dependent on the value of x.
Using the rules of differentiation
The following examples relate to particular cases in economics where you may be expected to use the different rules of differentiation.
Marginal Cost
Take the total cost function TC = 5q2 + 8q + 12. Calculate the marginal cost for the values q = 5 and q = 8.
We know that marginal cost is the change in total cost as a result of a small change in output. When differentiating the Total Cost curve therefore we are attempting to find the change in total cost as a result of an infinitessimally small change in one or more of the factors affecting total cost. Marginal Cost (MC) is the derivative of total cost with respect to output (dTC/dq). Given the function, the rule we must use is the Power Function rule (dy/dx = anxn-1,). In this example MC = dTC/dq = 10q + 8.
If q = 5, therefore, MC = 10(5) + 8 = 58
If q = 8, MC = 10(8) + 8 = 88
Question:
Calculate marginal revenue (MR) from the following total revenue (TR) function: TR = 70q - 4q3
Profit Maximisation
Profit maximisation occurs where the difference between Total Cost (TC) and Total Revenue (TR) is at its greatest. It can also be found at the output level where MC = MR. Let's put together some of the work covered so far to see how we might arrive at a solution to a problem.
Question:
A firm faces a demand schedule given by p = 320 - 3q and a cost schedule TC = 15 + 0.80q2 where output is in 000s and price in £s. Calculate the output required to maximise profit (π) and state what the profit level will be at this output.
Solution:
Calculate MR:
- MR = dTR/dq
- TR = p x q
- TR = (320 - 3q) x q
- TR = 320q - 3q2
- MR = 320 - 6q
Calculate MC:
- MC = dTC/dq
- MC = 1.6q
Profit maximisation occurs where MR = MC, therefore = 320 - 6q = 1.6q
- 320 = 7.6q
- πmax output = 42.1
The profit level at this output will be:
- π = TR - TC
- π = (320q - 3q2) - (15 + 0.8q2)
- π = 320q - 3q2 - 15 - 0.8q2
- π = 320q - 3.8q2 - 15
- π = 320(42.1) - 3.8(42.12) - 15
- π = 13,472 - 3.8(1772.41) - 15
- π = 13,472 - 6735.16 - 15
- π = £6,721.84
Calculating Maxima and Minima
There are a number of instances where you will have to find out where the maximum or minimum value occurs. The most obvious example is in calculating the point where total revenue is at a maximum and where average cost is at a minimum.
Look at the diagrams below which show a total revenue curve and an average cost curve.
You will notice that the TR curve reaches a maximum when the gradient of the curve becomes horizontal and in the same way, AC reaches a minimum when the gradient of the AC curve is horizontal. What does this mean in 'calculus speak'?
A horizontal gradient has a value of 0. The gradient of a function at a particular point is found through differentiation, therefore if we set the derivative of a function equal to zero then we will find the point at which a function is at its maxima or minima. Note, we are dealing here with relatively simple quadratic functions - more complex functions with different shaped curves present slightly different problems that we will not deal with at this stage.
Question:
Take the total revenue function TR = 264q - 0.6q2
To calculate the gradient we differentiate the function thus:
dTR/dq = 264 - 1.2q
We can set this result to zero to find the output level at which TR will be maximised.
264 - 1.2q = 0
264 = 1.2q
q = 220
The same principle would be used for calculating the point where the AC curve reached a minimum.