CARDIFFBUSINESSSCHOOL
MACROECONOMICS (BS1652)
Spring Semester 2005-06
Maths Macro Lecture 5
Differentiating with respect to time
In many respects, differentiating a variable with respect to time is no different from differentiating it with respect to any other variable. For example, if
y = 6.t + t2where t is time
thendy/dt = 6 + 2.t
But what ify = e10t
In this situation we clearly need to know how to differentiate e10t with respect to t .
One of the ‘nice’ properties of Euler’s constant, e , referred to in Maths Macro Lecture 4, is that the derivative with respect to time of et is et :
d (et) / dt = et
This property can be used to determine derivatives of more complex functions of e :
For example, ify = eatwhere a is a constant
what is the derivative of y with respect to t ?
The product rule of differentiation says that if y = f(x) and x = g (t)
then dy/dt = (dy/dx) (dx/dt)
Clearly y = eat can be thought of as y = ex where x = at
Hence, using the product rule of differentiation:
dy/dt = (dy/dx) (dx/dt)
d (eat) /dt = [ d (ex ) / dx ] [ d (at) / dt ]
dy/dt = ex . a
= a.eat(since x = at )
Therefore, ify = e10t
dy/dt = 10. e10t
What happens if the original function is in log form?
Considery = ln v
thendy/dt = (dy/dv).(dv/dt)(product rule of differentiation)
Sinced (ln v) / dv = 1/v(from the properties of logs)
thendy/dt = ( 1/v ) (dv/dt)
or, to put it in words, the derivative of y with respect to time is equal to the rate of growth of the variable v.
Thus, returning to the example used at the end of Maths Macro Lecture 4:
V = A.ert
then this expression can be re-written as:
ln V = ln A + rt
Differentiating with respect to t gives us:
(1/V) (dV/dt) = r (note ln A disappears since dA/dt = 0)
i.e. the rate of growth of V is given by r .
Another way of deriving this result would have been to differentiate the original function with respect to time, and then divide through by V as follows:
V = A.ert
Differentiating with respect to time gives us:
dV/dt = A. r. ert
Dividing through by V gives us:
(1/V).(dV/dt) = A. r. ert / A.ert = r
Hence, we get the same answer as before, namely that r is the instantaneous rate of growth of V.
Clearly it does not matter which method is chosen, since they both give the same result. Depending upon the situation faced, however, one method may prove more straightforward than another. If the original function is more complicated (e.g. comprises several variables multiplied together), and especially where the concern is with the growth of variables over time, then conversion to natural logarithms may prove to make life easier. For example consider equilibrium in the money market. This can be represented through the quantity theory of money equation, which states that:
M.V = P.Q
i.e. the level of aggregate demand (P.Q) is consistent with the stock of money M multiplied by the velocity of its circulation V .
It is usual to assume that the velocity of circulation of money, V , is constant, so the equation can be written in the form:
M = k. (P.Q)where k is a constant
If we differentiate this function with respect to time then we get:
dM/dt = k. [ (dP/dt).Q + P.(dQ/dt) ]
[note that since P and Q are both functions of time, to obtain the expression d(PQ)/dt we need to differentiate the first variable and multiply it by the second variable, and then add this to the first variable multiplied by the derivative of the second variable.]
If we now divide through by M we get:
(1/M) (dM/dt) = k. [ (dP/dt).Q + P.(dQ/dt) ] / k.(P.Q)
(1/M) (dM/dt) = [(1/P) (dP/dt) + (1/Q) (dQ/dt) ]
Rearranging:
(1/P) (dP/dt) = (1/M) (dM/dt) - (1/Q) (dQ/dt)
That is, the rate of growth of prices (i.e. the rate of inflation) is equal to the rate of growth of the money supply minus the rate of growth of output. Alternatively, if the rate of growth of the money supply exceeds the rate of growth of output, then the ‘excess’ feeds through into inflation. However, if the rate of growth of money supply is exactly equal to the rate of growth of output, then inflation is zero and the price level does not change.
The same result could have been derived using natural logarithms:
M.V = P.Q
Taking logs:ln (M.V) = ln (P.Q)
ln M + ln V = ln P + ln Q
Assuming V is constant (i.e. dV/dt = 0 ), then differentiating through by t we get:
(1/M) (dM/dt) = (1/P) (dP/dt) + (1/Q) (dQ/dt)
and, as before:
(1/P) (dP/dt) = (1/M) (dM/dt) - (1/Q) (dQ/dt)
If we denote the growth of M by t , that of P by t and that of Q by gt , then we can re-write this expression as:
t = t - gt
Further material on rates of growth
Throughout both macroeconomics and microeconomics, much use is often made of the Cobb-Douglas production function, not least for certain nice mathematical properties which the function possesses. In general form the function is given as:
Y = A.K.L
where Y represents output, K the capital stock and L the labour force employed.
In neoclassical growth theory, this functional form proves extremely useful. If we express the function in natural logarithmic form, we get:
ln Y = ln A + .ln K + .ln L
If we now differentiate with respect to time, we get:
(1/Y) (dY/dt) = .(1/K) (dK/dt) + .(1/L) (dL/dt)
(note that ln A disappears since A is a constant and hence dA/dt = 0 ).
That is, the rate of growth of output is a weighted average of the rates of growth of capital and labour, i.e. it is equal to times the rate of growth of capital and times the rate of growth of labour.
Elasticity
You should remember from Microeconomics that the price elasticity of demand is defined as (dQ/Q) / (dP/P). Assume that Q = a. Pb , then it follows that
log Q = log a + b. log P.
Remembering thatd log Q / dQ = 1 / Qand d log P / dP = 1 / P
Thend log Q = dQ / Qand d log P = dP / P
That is:elasticity of demand = (dQ/Q) / (dP/P) = d log Q / d log P
The price elasticity of demand is therefore the derivative of log Q with respect to log P . Hence, given our demand curve Q = a. Pb , it follows that the elasticity is:
(dQ/Q) / (dP/P) = d log Q / d log P = b
This implies that, if a function can be written in logarithmic form, then the elasticity of the dependent variable with respect to an explanatory variable can be found by differentiating its log with respect to the log of the explanatory variable. For example,
ifY = A.K
Then, sincelog Y = log A + .log K
The elasticity of Y with respect to K is given by:
(dY/Y) / (dK/K) = d log Y / d log K =
that is, the coefficient of K in the original equation, , represents the elasticity of Y with respect to K .
Let us examine a further example, this time based on the demand for money.
Let the demand for money function be represented by the equation:
Md = P. i .y
where Md represents the demand for money, P is the price level, i is the nominal rate of interest and y is income.
The demand for money function can be re-written as
log Md = log P + .log i + .log y
If we wish to know the elasticity of demand for money with respect to income, then we simply need to differentiate log Md with respect to log y:
( dMd /Md )/ (dy / y ) = d log Md / d log y =
that is, the elasticity of demand for money with respect to income is simply the coefficient of the income variable, i.e. , in the original demand for money function.
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