Partial Differentiation AGEC 317 In economic analysis, it is common to deal with models in which several variables are present. Partial derivatives are used in this case, which is an extension of the singlevariable calculus. 1. Consider the function y = f(x1, x2, . . . , xn) where the variables xi (i= 1, 2, . . . , n) are independent of each other. Independence allows variable xi to change without affecting the remaining n - 1 variables. For example, this allows x1 to undergo a change, ∆x1, while x2, x3, . . ., xn remain fixed. We are interested in the relative change in y, ∆y for a change in x1, ∆x1: y f (x 1 x 1 , x 2 ,, x n ) f (x 1 , x 2 ,, x n ) . x 1 x 1 If we take the limit of ∆y/∆x as ∆x1→0, this is a derivative by definition. It is called a partial derivative because only one of the variables is changing, x1. All other variables are held fixed or constant. It is a partial change in y. A similar process could be used to find partial with respect to the other variables in the function, y. Partial derivatives are denoted by the symbol, ∂, which is a variant of the Greek letter lower case delta. The partial derivative is written as ∂y/∂xi, which is read the partial derivative of y with respect to xi. Partial derivatives are also denoted as y x1 , y x 2 , or f x1 , f x 2 ,. 2. Partial differentiation differs from the single-variable differentiation in that additional variables are present, but these n - 1 variables are held constant and only one variable is allow to change. You have learned how to handle constants in differentiation in MATH 142 and the brief single-variable differentiation review. Therefore, the rules of differentiation you learned for single-variable functions hold for functions of multiple variables. The only difference is n - 1 variables are treated as constants. Rules of differentiation that must be memorized for AGEC 317. Constant function rule if g(x) = c, dg/dx = 0. Derivative of a constant is zero. Power function rule if g(x) = cxn, dg/dx = cnxn-1. Sum rule if y = f(x) + g(x), dy dx f ' (x) g' (x) . Difference rule if y = f(x) - g(x), dy dx f ' (x) g' (x) . Product rule if y = f(x) g(x), dy dx f ' (x)g(x) f (x)g' (x) . Quotient rule if y f (x) f ' ( x )g ( x ) f ( x )g ' ( x ) , dy / dx . g( x ) (g( x )) 2 Chain rule if z f ( y) where y g( x ) then dz / dx dz dy . dy dx Log rule (natural or base 10) if y = ln f(x), dy dx f ' (x) f (x) Exponential rule if y e f ( x ) , dy / dx e f ( x ) f ' ( x). To find the first order partial derivative of the following cubic function, y 3x 12 x 1 x 2 4x 22 x 13 x 32 , both the power-rule and sum rule must be applied. The sum rules states we can take the derivative of each component and sum them. The power rule indicates the exponent is placed as a constant and the exponent because the original exponent minus 1. Two partials can be taken, with respect to x1 and x2. y 2 3x 121 x 2 0 3x 131 x 32 6 x 1 x 2 3x 12 x 32 x 1 y 0 x 1 2 4 x 221 3x 13 x 321 x 1 8x 2 3x 13 x 22 x 2 3. Because the first order partial derivative is, in general, a function of the variable the variable xi, the first order partial derivative is itself differentiable with respect to xi, if it is continuous and smooth. Higher order partial derivatives are the partial derivative of the partial derivatives. For example, the second order partial derivative is the partial derivative of the first order partial derivative with y x 2y i 2 f x i x i ( x ). Note, higher order partial derivatives are respect to xi: x i x i denoted by a superscript number, which is the same as the order of the partial derivative. The same rules of differentiation used to find the first order partial apply to higher order partial derivatives. Continuing the above example, the second order partial derivatives are 2y 6 0 2 3x 121 x 32 6 6x 1 x 32 2 x 1 4. 2y 0 8 2 3x 13 x 221 8 6x 13 x 2 2 x 2 In general, the first order partial derivatives are functions of all variables in the original function, xi (i= 1, 2, . . ., n). Because ∂y/∂xi is a function of all n variables, the partial of each first derivative can be taken with respect to the other variables. Continuing the example from 2, the following higher order partials can be taken: y 2 x 1 y y 2 31 1 6x 12 x 22 x1x 2 f x1x 2 0 1 2 3x 1 x 2 x 2 x 1x 2 . y 2 x 2 y y 31 2 2 2 x 2 x1 f x 2 x1 1 0 2 3x 1 x 2 1 6 x 1 x 2 x 1 x 2 x 1 These are called the cross partial derivatives. Cross partial derivatives measure the rate of change of one first-order partial derivative with respect to the other variable. As long as the two cross partial derivatives are continuous, the order of differentiation does not matter. That is, the two cross partial derivatives will be equal. 5. Analogous to single variable functions, relative extreme points can be found for functions of more than one variable. The first order conditions, FOC, state the first order partial derivatives are set equal and the resulting system of equations is solved. The FOC are a necessary, but not sufficient condition. By setting the partials equal to zero and solving the system of equations, we are finding the point were the slopes are simultaneously equal to zero. The second order conditions, SOC, examine the second order and cross partials to determine the shape of the curve at the extreme point. The SOC indicate whether the point is a maximum or minimum. An incomplete test for an extreme point is: Conditions for extreme point: y= f(x, z) Condition Maximum FOC fx = fz = 0 SOC fxx < 0, fzz < 0, and fxx fzz > (fzx)2 Minimum fx = fz = 0 fxx > 0, fzz > 0, and fxx fzz > (fzx)2 All second order and cross partial derivatives are to be evaluated at the stationary point (satisfies FOC) given by the FOC. These are necessary, but not sufficient conditions. A function may violate the cross partial condition, but still be an extreme point. For example, a function may be characterized by fxx fzz = (fzx)2 and still be an extreme point. More on this point latter in class, if time allows. A saddle point is a stationary point, which is characterized by fxx fzz < (fzx)2. A saddle point is analogous to an inflection point on a single-variable function. Find the extreme point(s) of the following function: y = f(x, z) = xz + x2 + z2 + 3z Relevant partials are: fx = z + 2x fz = x + 2z + 3 fxx = 2 fzz = 2 fxz = fzx = 1 FOC - set first order partials equal to zero and solve fx = z + 2x = 0 fz = x + 2z + 3 = 0 Solving one obtains z = -2 and x = 1 SOC - evaluated at z = -2 and x = 1 fxx = 2 > 0 fzz = 2 > 0 fxxfzz > (fzx)2 => 2 * 2 > 12 Find y y = f(1, 2) = 1(-2) + (1)2 +(-2)2 + 5(-2) = -7 Have a minimum at y = -7, x = 1, and z = -2. 6. To find a change in the function y = f(x), we found the derivative dy/dx and regarded this ratio as a single entity. Now, consider finding the change in y, ∆y, y x . This states the change in y can be found for a change in x, ∆x, as y x once the rate of change in y, ∆y/∆x, and the variation in x, ∆x, are known. Letting y x . By definition, the the change in x approach zero, we obtain lim y lim x 0 x 0 x first term on the right side of the equivalent sign is the first order derivative. Therefore, we have dy dy dx f ' ( x )dx . dx The symbols dy and dx are called differentials of y and x. Another way of driving the differentials is to treat the derivative ratio as two separate components, that is, treat dy/dx as two separate entities. Taking the identity, dy/dx = dy/dx = dy dx f ' ( x )dx . f’(x) and dividing each side by dx, we obtain dy dx Find the differential of the function dy = 3x2 + 7x -5 dy =f’(x)dx = (2(3x2-1) + 7) dx = (6x + 7) dx 7. Similar to a single-variable function, we can find the total differential of a multivariable function through the process of differentiation. Consider the following function, y = f(x, z). From number 6, we can infer a small change in y caused by y dx . For the same a small change in x can be represented by the expression x y dz . The reason, a small change in y for a small change in z is represented by z total change in y for small changes in x and z is equal to dy y y dx dz . x z The expression dy, being the sum of change from both sources is called the total differential of the function y. The process of finding the total differential is called total differentiation. Find the total differential of the function y = f(x, z) = xz + x2 + z2 + 5z dy y y dx dz (z 2x )dx ( x 2z 5)dz x z