6 Multivariate calculus
6.5 Total differentiation and implicit differentiation
In the previous section, we derived the relationship between the marginal rate of sub-stitution and the marginal utility, as in Equation (6.10). Here we will derive the same relationship by using a different mathematical technique. This section also deals with some more new mathematical notions that we will use later in solving the consumer’s utility maximisation problem.
6.5.1 Total differential and total differentiation
Recall that, in Chapter 4, we defined the differential of a single-variate function C(q) as the following:
dC= C(q)dq.
It means the effect of a tiny change in q upon C is given by the product of the derivative of the function C(q) and the differential dq.
A similar idea for a multivariate function u= U(x1, x2) can be expressed: if we have a very tiny change in x1, which we denote by dx1, the consequent change in the value of u is given by: if it held, would be written as:
du
dx1 = U1, (6.12)
which relates to our statement at the end of Section 5.2: the derivative (in this case, the partial derivative on the RHS) of a function can be expressed as the ratio of the differentials.
In Equation (6.12), the partial derivative U1 is expressed as the ratio of the differentials duand dx1. However, Equation (6.11) (or (6.12)) will not generally hold. That is, the ratio of the differentials, du and dx1, and the partial derivative U1are different, with some exceptions. I show why it is the case in the following.
Since there are two arguments to the function u= U(x1, x2), in addition to a change in x1, we need to consider a small change in x2. Let us denote this change in x2 as dx2 (the differential of x2). Multiplying this differential by ∂U(x1, x2)
∂x2
gives us the effect of the change in x2on u, and together with the effect of the change in x1on u, we obtain the total effect on u:
du= ∂U(x1, x2)
∂x1 dx1+∂U(x1, x2)
∂x2 dx2. (6.13)
We call du in Equation (6.13) the total differential of the function u= U(x1, x2). It is the sum of the marginal changes in all the arguments of the function, multiplied by the corresponding partial derivatives. The technique of taking the total differential of a function is called total differentiation.
Now let us divide both sides of Equation (6.13) by dx1: du
The LHS of Equation (6.14) is called the total derivative of the function u= U(x1, x2) with respect to x1. It is the ratio of the two differentials du and dx1. As we can see, it equals the RHS of Equation (6.14), which consists of two terms. The first term is the
161 6.5 Total differentiation and implicit differentiation
0 u
A A′ u u
x1 dx1
dx2
x1 x1
x2
x2
1 2
( , ) u=U x x
Figure 6.8 The total derivative.
partial derivative of u with respect to x1, ∂U(x1, x2)
∂x1 , and the second is ∂U(x1, x2)
∂x2 dx2 dx1. What this equation says is important. It says that the effect of a tiny change in x1 upon ucan be decomposed into two things. The first term shows the ‘direct effect’ of a tiny change in x1 upon u (because when we take the partial derivative, ∂U(x1, x2)
∂x1
, we are holding x2constant), whereas the second term shows an “indirect effect” of a tiny change in x1 upon u through a change in x2. Therefore, so long as the indirect effect is not zero, the total derivative, i.e. the ratio of the differentials, and the partial derivative are different.
Let us see intuitively what we have done. For this purpose I will use Equation (6.13) and Figure 6.8 (which is essentially the same as Figure 6.6).
The (total) differential du measures the change in the level of utility (or the change in the altitude of where you are standing on the 3D object), as a consequence of changing consumption by an infinitesimally small amount in any direction. That is, here, we are no longer constrained to move only in the direction of x1 or in the direction of x2. For example, we can move somewhere in between the directions of x1and x2.
6.5.2 Implicit differentiation
Total differentiation is very useful: it can be used to differentiate single-variate functions, which would be rather cumbersome to handle in the usual way. For example, to calculate
dy
dx for an implicit function:
y2+ x = 0, (6.15)
we can write this as:
x= −y2. (6.16)
Hence:
An alternative approach is called implicit differentiation. It involves applying the total differentiation to the implicit function and goes as follows. First, we regard y2+ x as a function of x and y, which we express by using the following notation:
f = F (x, y) = y2+ x = 0. (6.19)
Taking the total differentials (recall Equation (6.13)) gives the following:
df = ∂F(x, y)
∂x dx+ ∂F(x, y)
∂y dy. (6.20)
In this equation, df has to be zero because f always equals zero (remember, df is a tiny movement in f . If f = 0 always hold, then df has to be zero). Therefore:
0= ∂F(x, y)
∂x dx+∂F(x, y)
∂y dy. (6.21)
This expression boils down to:
dy
which is the same as what we have got in Equation (6.18).
Now, think about the indifference curve corresponding to a certain level of utility, ¯u, that is:
u= U(x1, x2)= ¯u. (6.23)
Suppose we have tiny changes in x1and x2along this indifference curve (meaning that we are not moving off the curve). Since we consider changes along the same indifference curve, u= ¯u, and so du = 0:
This expression boils down to:
− dx2
dx1 = ∂U /∂x1
∂U /∂x2. (6.25)
163 6.5 Total differentiation and implicit differentiation
0
A
I dx2
−dx1
B x2
x2
x1 x1
Figure 6.9 Obtaining MRS by using implicit differentiation.
Using the fact that the marginal rate of substitution (MRS12) is the absolute value of the slope of the indifference curve at the point of consumption (see Figure 6.9), we can conclude that:
MRS12= ∂U /∂x1
∂U /∂x2. (6.26)
Now let us go through a question on the implicit differentiation to consolidate the understanding.
Question Consider x2− y2 = 6. Obtain dy dx. Solution Let f be the following:
f = F (x, y) = x2− y2= 6.
Taking the total differentials (recall Equation (6.13)) gives the following:
df = ∂F(x, y)
∂x dx+ ∂F(x, y)
∂y dy.
In this equation, df has to be zero because f always equals 6. Therefore:
0= ∂F(x, y)
∂x dx+∂F(x, y)
∂y dy.
This expression boils down to:
dy dx = −
∂F(x,y)
∂x
∂F(x,y)
∂y
= − 2x (−2y) = x
y. Exercise 6.5 Implicit differentiation.
0 A
x1 x2
x1* x2*
f (x1*, x2*)
f (x1, x2)
Figure 6.10 Local maximum.