MATH 19520/51 Class 5

MATH 19520/51 Class 5 Minh-Tam Trinh University of Chicago 2017-10-04

1 Definition of partial derivatives. 2 Geometry of partial derivatives. 3 Higher derivatives. 4 Definition of a partial differential equation (PDE). 5 Cobb Douglas production function revisited.

Partial Derivatives What is... The partial derivative of x y with respect to x? The partial derivative of log(ab) with respect to a? The partial derivative of 1 xy+xz+yz with respect to x?

Partial Derivatives What is... The partial derivative of x y with respect to x? yx y 1 The partial derivative of log(ab) with respect to a? The partial derivative of 1 xy+xz+yz with respect to x?

Partial Derivatives What is... The partial derivative of x y with respect to x? yx y 1 The partial derivative of log(ab) with respect to a? 1 a The partial derivative of 1 xy+xz+yz with respect to x?

Partial Derivatives What is... The partial derivative of x y with respect to x? yx y 1 The partial derivative of log(ab) with respect to a? 1 a The partial derivative of 1 xy+xz+yz with respect to x? y+z (xy+xz+yz) 2

In general, let f be a function of n variables x 1,..., x n.

In general, let f be a function of n variables x 1,..., x n. Formally, the partial derivative of f with respect to x i is (1) lim ɛ 0 f (x 1,..., x i + ɛ,..., x n ) f (x 1,..., x i,..., x n ). ɛ Informally, it s the derivative we get by treating the other variables x 1,..., x i 1, x i+1,..., x n as constants.

There are several different notations for partial derivatives: (2) f x, xf, D x f, f x all mean the partial derivative of f with respect to x. Notice that we use instead of d.

Example If f (x, y, z) = x yz, then what is f z (e, e, e)?

Example If f (x, y, z) = x yz, then what is f z (e, e, e)? Use the chain rule: (3) f z = (xyz log x) z (yz ) = (x yz log x)(y z log y). Then f z (e, e, e) = (e ee log e)(e e log e) = e ee e e = e ee +e.

Example ( Let f (x, y) = sin x 1 + y ). Evaluate x f and y f at (x, y) = (π, 2).

Example ( Let f (x, y) = sin x 1 + y We compute ( ) f x = cos x (4) 1 + y ). Evaluate x f and y f at (x, y) = (π, 2). ( x ) x 1 + y = ( ) 1 1 + y cos x 1 + y. (5) ( ) f y = cos x 1 + y ( x ) y 1 + y ( x = (1 + y) 2 cos x 1 + y ).

Example ( Let f (x, y) = sin x 1 + y We compute ( ) f x = cos x (4) 1 + y ). Evaluate x f and y f at (x, y) = (π, 2). ( x ) x 1 + y = ( ) 1 1 + y cos x 1 + y. (5) ( ) f y = cos x 1 + y ( x ) y 1 + y ( x = (1 + y) 2 cos x 1 + y ). Thus, f x (π, 2) = 1 3 cos π 3 = 1 6 and f y(π, 2) = π 9 cos π 3 = π 18.

Geometry of Partial Derivatives In 1-variable calculus, df dx (a) measures the rate of change of f in the positive x-direction at a. This is the slope of the tangent line to the graph of f at a.

Geometry of Partial Derivatives In 1-variable calculus, df dx (a) measures the rate of change of f in the positive x-direction at a. This is the slope of the tangent line to the graph of f at a. In 2-variable calculus, f f x (a, b) and y (a, b) respectively measure the rate of change of f in the positive x- and y-directions at (a, b). These are the slopes of tangent lines to the graph of f at (a, b). Which tangent lines?

http://calcnet.cst.cmich.edu/faculty/angelos/m533/lectures/pderv.htm

Higher Derivatives A higher derivative is when we apply several partial derivatives in succession. Stewart s notation: 2 f x 2 = ( ) f 2 f x x y x = ( ) f y x (6) and 2 f x y = ( ) f x y 2 f y 2 = ( ) f y y (7) f xx = (f x ) x f yx = (f y ) x f xy = (f x ) y f yy = (f y ) y

f xx, f xy, etc. are called 2 nd -order partial derivatives. In 2 variables, there are 4 kinds of 2 nd -order partial derivatives we can take: xx, xy, yx, yy.

f xx, f xy, etc. are called 2 nd -order partial derivatives. In 2 variables, there are 4 kinds of 2 nd -order partial derivatives we can take: xx, xy, yx, yy. In 3 variables, how many kinds of 2 nd -order partial derivatives can we take?

f xx, f xy, etc. are called 2 nd -order partial derivatives. In 2 variables, there are 4 kinds of 2 nd -order partial derivatives we can take: xx, xy, yx, yy. In 3 variables, how many kinds of 2 nd -order partial derivatives can we take? xx, xy, xz, yx, yy, yz, zx, zy, zz.

Example Find all 2 nd -order partial derivatives of f (x, y) = x 2 + 2y 2 + 3xy.

Example Find all 2 nd -order partial derivatives of f (x, y) = x 2 + 2y 2 + 3xy. Compute: (8) f xx = 2 f xy = 3 f yx = 3 f yy = 4 Notice that f xy = f yx.

Theorem (Clairaut) Let f be a differentiable function of x and y. If f x and f y exist and are continuous on a disk around (a, b), then (9) f xy (a, b) = f yx (a, b).

Theorem (Clairaut) Let f be a differentiable function of x and y. If f x and f y exist and are continuous on a disk around (a, b), then (9) f xy (a, b) = f yx (a, b). Warning! If f x and f y are not continuous at (0, 0), then Clairaut doesn t apply. If f (x, y) = xy(x2 y 2 ), then we can check that f x 2 +y 2 x and f y are discontinuous at (0, 0). We compute f xy (0, 0) = 1 1 = f yx (0, 0). https://calculus.subwiki.org/wiki/failure_of_clairaut s_theorem_where_ both_mixed_partials_are_defined_but_not_equal

Similarly, we can talk about 3 rd -order partial derivatives.

Similarly, we can talk about 3 rd -order partial derivatives. Example Find g xyz where g(x, y, z) = e xyz.

Similarly, we can talk about 3 rd -order partial derivatives. Example Find g xyz where g(x, y, z) = e xyz. Compute z ( y ( x (e xyz ))) = z ( y (e xyz yz)) (10)

Similarly, we can talk about 3 rd -order partial derivatives. Example Find g xyz where g(x, y, z) = e xyz. Compute (10) z ( y ( x (e xyz ))) = z ( y (e xyz yz)) = z (e xyz xz yz + e xyz z) = z (e xyz (xyz 2 + z))

Similarly, we can talk about 3 rd -order partial derivatives. Example Find g xyz where g(x, y, z) = e xyz. Compute (10) z ( y ( x (e xyz ))) = z ( y (e xyz yz)) = z (e xyz xz yz + e xyz z) = z (e xyz (xyz 2 + z)) = e xyz xy (xyz 2 + z) + e xyz (2xyz + 1) = e xyz (x 2 y 2 z 2 + 3xyz + 1).

Partial Differential Equations A partial differential equation (PDE) is any equation involving the partial derivatives of a function. The order of the equation is the highest order among the partial derivatives involved.

Partial Differential Equations A partial differential equation (PDE) is any equation involving the partial derivatives of a function. The order of the equation is the highest order among the partial derivatives involved. Example Laplace s equation for a function u(x, y) is a 2 nd -order PDE: (11) u xx + u yy = 0. Think of the function u as a variable in its own right.

Solution to PDEs are functions, not numbers. Just as an ordinary equation can have many numbers as solutions, a PDE can have many functions as solutions.

Solution to PDEs are functions, not numbers. Just as an ordinary equation can have many numbers as solutions, a PDE can have many functions as solutions. Example The function u(x, y) = x solves Laplace s equation: (12) (x) xx + (x) yy = 0 + 0 = 0. But so does u(x, y) = x 2 y 2 : (13) (x 2 y 2 ) xx + (x 2 y 2 ) yy = 2 2 = 0.

Cobb Douglas Revisited Recall the Cobb Douglas production function: (14) P(L, K) = bl α K 1 α.

Cobb Douglas Revisited Recall the Cobb Douglas production function: (14) P(L, K) = bl α K 1 α. It is a solution to the system of equations P(L, 0) = P(0, K) = 0 (15) P L = α P L P K = β P K α + β = 1 P P L is marginal productivity of labor and K capital. is marginal productivity of

What do these equations mean? P(L, 0) = P(0, K) = 0 P L = α P L P K = β P K α + β = 1

What do these equations mean? P(L, 0) = P(0, K) = 0 To be productive, you need both labor and capital. P L = α P L P K = β P K α + β = 1

What do these equations mean? P(L, 0) = P(0, K) = 0 P L = α P L To be productive, you need both labor and capital. The rate at which labor increases productivity is proportional to the ratio of productivity to labor. P K = β P K α + β = 1

What do these equations mean? P(L, 0) = P(0, K) = 0 P L = α P L P K = β P K To be productive, you need both labor and capital. The rate at which labor increases productivity is proportional to the ratio of productivity to labor. The rate at which capital increases productivity is proportional to the ratio of productivity to capital. α + β = 1

What do these equations mean? P(L, 0) = P(0, K) = 0 P L = α P L P K = β P K α + β = 1 To be productive, you need both labor and capital. The rate at which labor increases productivity is proportional to the ratio of productivity to labor. The rate at which capital increases productivity is proportional to the ratio of productivity to capital. We want scaling factors to work: P(aL, ak) = ap(l, K).