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## Differentiation in Economics

Most undergrad level core micro and macro involves fairly simple differentiation, you will do a lot of optimisation and use the chain rule and product rules a lot. One thing you will have to get used to in economics is seeing things written as functions and differentiating them.

You are always differentiating to find ‘marginals‘. The concept of ‘marginals’ (marginal revenue, marginal product, marginal cost) etc is about the most important concept in microeconomics, because all decisions are taken ‘at the margin’. Do you increase production by another unit or just produce at the level you are doing? Well if your marginal revenue (the amount of revenue you will earn by producing another unit of output) is higher than your marginal cost (the amount it will cost you to produce another unit) then go for it. If your marginal cost is higher then you don’t. As you produce more your MR will fall and your MC will rise so you will maximise profits by producing where MR = MC. Basic golden rule of micro!

Because MR is basically the ‘change in revenue over the change in output’ you find it by differentiating total revenue with respect to output. Total revenue is price x quantity.

So you have $TR = PQ$, $MR = \frac{d(TR)}{dQ}$ so $MR = \frac{d(PQ)}{dQ}$

Just a note here, don’t get confused by the fact that PQ is P times Q, and TR and MR are ‘total revenue’ and ‘marginal revenue’ you aren’t doing T times R or M times R, its just the abbreviation.

So if $MR = \frac{d(PQ)}{dQ}$ you use the product rule to say $MR = P\frac{dQ}{dQ}+Q\frac{dP}{dQ}$ so $MR = P+Q\frac{dP}{dQ}$

You also have to use the chain rule, and recognise how to do this in terms of the notation. Lets consider the question ‘what wage do you pay if you want to maximise profit?’

To show this example you first have to understand the concept of the production function. This will express Q as a function of K and L.

So we know that TR is a function of P and Q, and Q is a function of K and L. We have a ‘function of a function’ so the chain rule is coming up when we have to differentiate.

Now in terms of costs, we usually use w to refer to the cost (per unit) of labour (ie the wage) and r to refer to the cost of capital. If the capital was say machinery, then you aren’t having to pay the machinery a wage, but it is costing you, either you are renting it for a fee, or if you bought the machines outright, there is an ‘opportunity cost’ (ie you could have been investing the money you spent on the machines at the market interest rate, r, so if your machines are earning you less than r, you would have been better not buying them and investing your money instead)

So your total costs are going to be the wage times the number of units of labour you hire, plus the cost of capital times the number of units of capital you are using, ie $TC = wL + rK$

Your profits (usually $\pi$ is used to symbolise profit in economics, are $\pi = TR - TC$ so $\pi = TR(Q(K,L)) - wL - rK$. What this notation is saying is that total revenue is a function of Q (its also a function of P, but we don’t need that here as we are looking at labour which comes directly into the function for Q not P), and Q is a function of K and L.

To maximise profits with respect to the cost of our inputs, K and L, its an optimisation problem, we set the two partial derivatives of profit by K and L to zero to find the optimising points, ie $\displaystyle \frac{\partial \pi}{\partial K} = 0$ and $\displaystyle \frac{\partial \pi}{\partial L} = 0$.

We use the chain rule to do these partials. $\displaystyle \frac{\partial \pi}{\partial K} = \frac{d(TR)}{dQ}\frac{dQ}{dK}-r$ and $\displaystyle \frac{\partial \pi}{\partial L} = \frac{d(TR)}{dQ}\frac{dQ}{dL}-w$. When you use the chain rule you do the derivative of the outside function times the derivative of the inside function.

Remember that $\frac{dQ}{dK} = MR$ so $\displaystyle \frac{\partial \pi}{\partial K} = MR\frac{dQ}{dK}-r$ and $\displaystyle \frac{\partial \pi}{\partial L} = MR\frac{dQ}{dL}-w$.

We can also introduce a couple of concepts here called ‘marginal productivity’. The marginal productivity of capital is the amount of extra output you get by adding one more unit of capital, ie $\frac{dQ}{dK}$. The marginal productivity of labour is the amount of extra output you get by adding one more unit of labour, ie $\frac{dQ}{dL}$.

So we can now say $\displaystyle \frac{\partial \pi}{\partial K} = MR(MP_K)-r$ and $\displaystyle \frac{\partial \pi}{\partial L} = MR(MP_L)-w$.

Remember that to find the profit maximising point we had to set both partials to zero, ie $\displaystyle \frac{\partial \pi}{\partial K} = MR(MP_K)-r=0$ and $\displaystyle \frac{\partial \pi}{\partial L} = MR(MP_L)-w=0$ so $MR(MP_K)=r$ and $MR(MP_L)=w$

So the profit maximising wage is to pay each worker the multiple of his marginal productivity multiplied by the marginal revenue. In an environment of perfect competition, $P=MR$ so if we had perfect competition, $P(MP_L)=w$. This is quite an important result in micro which you use in questions to do with factor markets, because it shows that two things will drive wages up – either a rise in price of the good being produced, or an increase in marginal productivity of labour (eg workers becoming more skilled, technological advances making them more productive, or simply adding more capital so each worker has more capital to work with). Fairly logical results but that is the economics behind them.

For another example of using differentiation and algebraic manipulation, check out this little exercise in deriving the elasticity of demand for a firm in a competitive market.