Quantitative Techniques (Finance) 203 Derivatives for Functions with Multiple Variables Felix Chan October 2006 1 Introduction In the previous lecture, we discussed the concept of derivative as approximation to the slope of a function at a given point. We have also introduced various ways to derive the derivatives for different functions as well as method to derive the tangent line for a function at a given point. This topic will extend the results and methodology developed thus far to functions that involve more than one independent variable. We will discuss the concept of partial derivatives and total derivatives as well as the concept of implicit differentiation. 2 Functions with Multiple variables Definition 1. A function F : R n R with y = F(x 1,..., x n ) defines a function with n- variables. Essentially, the definition above defines F as a function that takes n inputs and return a scalar y as an output. An example of function with many variables would be a production function π = F(K, L), 1
where productivity π is a function of both labour input L and capital input K. An explicit example for the production function is the famous Cobbs-Douglas production function which is defined as follows: π = AK α L β. In the case such as the one above, we would often want to know the marginal impact on production if we change our inputs by a small amount. However, unlike the univariate case where there is only one input variable, we have two in this case. So we could potentially derive three sets of marginal effects, namely, 1. Marginal impact on π with respect to labour input L. 2. Marginal impact on π with respect to capital input k. 3. Marginal impact on π with respect to both labour and capital inputs. The first two marginal impacts can be derived using partial derivatives whereas the last marginal impact requires total derivatives. 3 Partial Derivatives Let y = F(x 1,..., x n ) be a function with n variables. The partial derivative of F with respect to x i is F(x 1,..., x i + h i,..., x n ) F(x 1,..., x i,..., x n ) = lim, i = 1,..., n. (1) hi 0 h i In practice, the partial derivative of F with respect to x i is derived by differentiating the function F with respect to x i holding all the other input variables constant. Since there are n variables in the function, we will end up with a set of n partial derivatives. Example 1. Find all the partial derivatives for the Cobbs-Douglas production function: π = 4L 0.5 K 0.5. 2
Since there are two input variables for the Cobbs-Douglas function, so there will be two partial derivatives. The partial derivative of π with respect to K is π = 4L 0.5K 0.5 π K = (0.5)(4)L0.5 K 0.5 1 = 2L 0.5 K 0.5. Notice in the derivation above, L is treated as a constant. Similarly, the partial derivative of π with respect to L is π = 4L 0.5 K 0.5 π L = (0.5)(4)L0.5 1 K 0.5 = 2L 0.5 K 0.5. where K is treated as a constant this time. Example 2. Find all the partial derivatives for the following function: y = ax 1 x 2 + bx 1 + cx 2. Again, there are two input variables here, so we will end up with 2 partial derivatives, namely x 1 and x 2. For the derivative of y with respect to (w.r.t.) x 1 : 3
y = ax 1 x 2 + bx 1 + cx 2 x 1 = ax 2 + b The last term vanished in the derivative as we are treating x 2 as constant in this case and the derivative of a constant is 0. Similarly, y = ax 1 x 2 + bx 1 + cx 2 x 1 = ax 1 + c. Similar to the standard derivative, the partial derivative is itself a function with multiple variables. This implies there will be n second order partial derivatives from each first order partial derivatives, which makes a total of n 2 second order partial derivatives. To illustrate this point, let s consider the second order derivatives for the Cobbs-Douglas production function. Example 3. From Example 1, we derived the two first order derivatives as follows: π K = 2K 0.5 L 0.5 π L = 2K0.5 L 0.5. Notice that each partial derivative is a function of both K and L and therefore there will be two second order derivatives by differentiating each first order derivative with respect to K and L, namely: 4
L L L K K K K L L 2 L K K 2 K L Now, deriving each of these derivatives we get L L L 2 = ( 0.5)(2)L 0.5 1 K 0.5 = L 1.5 K 0.5 L K L K = (0.5)(2)L 0.5 1 K 0.5 = L 0.5 K 0.5 K K K 2 = ( 0.5)(2)L 0.5 K 0.5 1 = L 0.5 K 1.5 5
K π L K L = (0.5)(2)L 0.5 K 0.5 1 = L 0.5 K 0.5 An interesting observation from the example above is that the cross derivatives actually equal to each other. In fact, if a function is well behaved then its second order cross derivatives will always equal to each other. Formally, Theorem 1. (Clairaut s theorem /Schwarz s theorem/young s Theorem) Let F : R n R with y = F(x 1,..., x n ), if F is a twice differentiable function with continuous second partial derivatives at every point then 2 F = 2 F i, j = 1,..., n. (2) x j x j 4 Total Derivative Let F : R n R with y = F(x 1,..., x n ), then the partial derivative with respect to x i measures the marginal impacts on y with respect to a small change in x i, i = 1,.., n holding x k constant k i. A basic assumption on the partial derivative is that we are ignoring the impact on other variables as we change x i. The following example would demonstrate this point more clearly: Example 4. Consider the following Cobbs-Douglas type production function where Q(K(t), L(t), t) = t 1 K(t) 0.5 L(t) 0.5. Notice this production function is no longer a function of just Capital and Labour input but also a function of time. Moreover, the Capital and Labour inputs are themselves changing over time. Therefore changing in time will have two impacts on the production function - (i) 6
a direct impact on the productivity Q as time changes and (ii) indirect impact on productivity, Q through the changes in capital and labour inputs as a result of time changes. So the total derivative of Q with respect to time, t should therefore be dq dt = Q + Q dk }{{} t K dt + Q dl }{{ L dt} Direct Effect Indirect Effect Notice we use dy/dx to denote the total derivative of y with respect to x and we use / x to denote the partial derivative of y with respect to x. If dk/dt = 0 and dl/dt = 0, that is, if neither capital input or labour input are functions of time, then the partial derivative equals to the total derivative as there is no indirect effect. In general, let F : R n R with y = F(x 1,..., x n ) then the total derivative of y with respect to x i is dy dx i = + j=1 i j x j dx i. (3) Example 5. Find the total derivative of Q with respect to t for the following Cobbs-Douglas production function: Q(K(t), L(t), t) = t 1 K(t) 0.5 L(t) 0.5 L(t) = t 2 + 1 K(t) = t + 1 In this case, the total derivative of Q with respect to t is Now, dq dt = Q t + Q dk K dt + Q L dl dt 7
Q t = t 2 K(t) 0.5 L(t) 0.5 Q K = 0.5t 1 K(t) 0.5 L(t) 0.5 dk dt = 0.5t 0.5 Q L = 0.5t 1 K(t) 0.5 L(t) 0.5 dl dt = 2t So the total derivative of Q with respect to t is dq dt = t 2 K(t) 0.5 L(t) 0.5 + 0.5t 0.5 (0.5t 1 K(t) 0.5 L(t) 0.5 ) + 2t(0.5t 1 K(t) 0.5 L(t) 0.5 ) = t 2 K(t) 0.5 L(t) 0.5 + 0.25t 1.5 K(t) 0.5 L(t) 0.5 + K(t) 0.5 L(t) 0.5 The concept of total derivative can be redefined in terms of differentials: dy = dx i (4) The differential dy can be interpreted as (though not very rigorously) the changes in y, and / dx i can be interpreted as (again not very rigorously) the marginal impact on y per small amount changes in x i multiply by the total changes in x i - that is, the total changes in y due to the changes in x i. Note that under certain conditions, the ratio between the differentials dy and is equivalent to the total derivative dy/. Thus, the total derivative of y with respect to x j is dy = dx i. Recall that / = 1 and hence the expression above can be rewritten as 8
dy = + x j i j dx i, which is equivalent to the definition of total derivative we had earlier. 5 Derivatives for Implicit Functions The dependent variable in all the functions we have encountered so far can be expressed in terms of the other variables using simple algebraic manipulation. However, there exists functions which are algebraically impossible to express the dependent variables in terms of the other variables. In such cases, it is still possible to compute the derivatives by using equation (4). Consider F(y, x 1,..., x n ) = c where c is a constant, then df = dy + dx i. However, since F = c which is a constant, this implies df = 0 and therefore dy + dx i = 0. (5) Now if we are interested in the derivatives of y with respect to x j then equation (5) will become x j + dx i = 0 + = 0 x j x j = / x j x j / dx i = 0 i j 1 i = j 0. The following example will provide an application for the result above: 9
Example 6. Find the derivative of y with respect to x for the following function: x + x 2 y + xy 2 = 0. F(x, y) = x + x 2 y + xy 2 = x2 + 2xy = 1 + 2xy + y2 x x = / x / 1 + 2xy + y2 = x 2 + 2xy 10