Partial Derivatives October 2013
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1 Partial Derivatives October 2013
2 Derivative in one variable. Recall for a function of one variable, f (a) = lim h 0 f (a + h) f (a) h slope f (a + h) f (a) h a a + h
3 Partial derivatives. For a function of two variables z = f (x, y): Partial derivatives f x (a, b) = f f (a + h, b) f (a, b) (a, b) = lim x h 0 h f y (a, b) = f f (a, b + h) f (a, b) (a, b) = lim y h 0 h
4 Example: Limit definition. f (x, y) = x 2 y 3 : f x = lim f (x + h, y) f (x, y) h 0 h (x + h) 2 y 3 x 2 y 3 = lim h 0 h x 2 y 3 + 2xhy 3 + h 2 y 3 x 2 y 3 = lim h 0 h 2xhy 3 + h 2 y 3 = lim h 0 h = lim 2xy 3 + hy 3 h 0 = 2xy 3. So f x = 2xy 3. Exercise: f y = 3x 2 y 2.
5 Geometric meaning. Say C is the vertical trace of the graph of f on the plane y = b. C is given by z = f (x, b), y = b Tangent line to C at point (a, b) has slope given by the ordinary derivative of the single-variable function z = f (x, b), at x = a That is the same as the partial derivative f x (a, b)
6 Geometric meaning, y. Similarly let C be the vertical trace of the graph of f on the plane x = a. C is given by z = f (a, y), x = a Tangent line to C at point (a, b) has slope given by the ordinary derivative of the single-variable function z = f (a, y), at y = b That is the same as the partial derivative f y (a, b)
7 Summary of geometric meaning. f (a, b) is the slope of the tangent line to the vertical x trace of the graph of f on y = b Treating y = b as a constant, leaving x as a variable f (a, b) is the slope of the tangent line to the vertical y trace of the graph of f on x = a Treating x = a as a constant, leaving y as a variable
8 Computing partial derivatives. To compute partial derivatives: To compute f : treat y as a constant, differentiate with x respect to x To compute f : treat x as a constant, differentiate with y respect to y Example: For f (x, y) = x 3 + xy + y 2 (x + 1), f x = f y = =
9 Computing partial derivatives. To compute partial derivatives: To compute f : treat y as a constant, differentiate with x respect to x To compute f : treat x as a constant, differentiate with y respect to y Example: For f (x, y) = x 3 + xy + y 2 (x + 1), f x = 3x 2 + y + y 2 f y = =
10 Computing partial derivatives. To compute partial derivatives: To compute f : treat y as a constant, differentiate with x respect to x To compute f : treat x as a constant, differentiate with y respect to y Example: For f (x, y) = x 3 + xy + y 2 (x + 1), f x = 3x 2 + y + y 2 f y = 0 + x + 2y(x + 1) =
11 Computing partial derivatives. To compute partial derivatives: To compute f : treat y as a constant, differentiate with x respect to x To compute f : treat x as a constant, differentiate with y respect to y Example: For f (x, y) = x 3 + xy + y 2 (x + 1), f x = 3x 2 + y + y 2 f y = 0 + x + 2y(x + 1) = x + 2y(x + 1)
12 Clicker Question: Suppose that the price P (in dollars), to purchase a used car is a function of C, its original cost (also in dollars), and its age A (in years). So P = f (C, A). The sign of P is C A. Positive, and I am very confident B. Positive, but I am not very confident C. Negative, but I am not very confident D. Negative, and I am very confident receiver channel: 41 session ID: bsumath275
13 Clicker Question: If f x = f y everywhere, then f (x, y) is constant: A. True, and I am very confident B. True, but I am not very confident C. False, but I am not very confident D. False, and I am very confident receiver channel: 41 session ID: bsumath275
14 Clicker Question: There exists a function f (x, y) with f x = 2y and f y = 2x: A. True, and I am very confident B. True, but I am not very confident C. False, but I am not very confident D. False, and I am very confident receiver channel: 41 session ID: bsumath275
15 Second order partial derivatives. f xx = (f x ) x aka ( ) f, or x x 2 f x 2
16 Second order partial derivatives. f xx = (f x ) x aka ( ) f, or x x 2 f x 2 f xy = (f x ) y aka ( ) f, or y x 2 f y x
17 Second order partial derivatives. f xx = (f x ) x aka ( ) f, or x x 2 f x 2 f xy = (f x ) y aka ( ) f, or y x 2 f y x
18 Second order partial derivatives. f xx = (f x ) x aka ( ) f, or x x 2 f x 2 f xy = (f x ) y aka ( ) f, or y x 2 f y x f yx = (f y ) x aka ( ) f, or x y 2 f x y
19 Second order partial derivatives. f xx = (f x ) x aka ( ) f, or x x 2 f x 2 f xy = (f x ) y aka ( ) f, or y x 2 f y x f yx = (f y ) x aka ( ) f, or x y 2 f x y f yy = (f y ) y aka ( ) f, or y y 2 f y 2
20 Example: Second order partial derivatives. Example: f (x, y) = f x = = = x x y = x(x y) 1 and f y =
21 Example: Second order partial derivatives. Example: f (x, y) = x x y = x(x y) 1 f x = (x y) 1 x(x y) 2 = = and f y =
22 Example: Second order partial derivatives. Example: f (x, y) = x x y = x(x y) 1 f x = (x y) 1 x(x y) 2 = x y x (x y) 2 = and f y =
23 Example: Second order partial derivatives. Example: f (x, y) = x x y = x(x y) 1 f x = (x y) 1 x(x y) 2 = x y x (x y) 2 = y = y(x y) 2 (x y) 2 and f y =
24 Example: Second order partial derivatives. Example: f (x, y) = x x y = x(x y) 1 f x = (x y) 1 x(x y) 2 = x y x (x y) 2 = y = y(x y) 2 (x y) 2 and f y = x(x y) 2 = x (x y) 2
25 Example, continued. f x = y (x y) 2 = y(x y) 2, f y = x(x y) 2 = x (x y) 2 f xx = f xy = f yx = f yy = Observation: f xy = f yx. Is this a coincidence?
26 Example, continued. f x = y (x y) 2 = y(x y) 2, f y = x(x y) 2 = f xx = 2y(x y) 3 = f xy = 2y (x y) 3 x (x y) 2 f yx = f yy = Observation: f xy = f yx. Is this a coincidence?
27 Example, continued. f x = y (x y) 2 = y(x y) 2, f y = x(x y) 2 = f xx = 2y(x y) 3 = 2y (x y) 3 f xy = 1(x y) 2 2y(x y) 3 = x y (x y) 3 f yx = x (x y) 2 f yy = Observation: f xy = f yx. Is this a coincidence?
28 Example, continued. f x = y (x y) 2 = y(x y) 2, f y = x(x y) 2 = f xx = 2y(x y) 3 = 2y (x y) 3 f xy = 1(x y) 2 2y(x y) 3 = x y (x y) 3 f yx = (x y)2 2x(x y) (x y) 4 = x y (x y) 3 f yy = x (x y) 2 Observation: f xy = f yx. Is this a coincidence?
29 Example, continued. f x = y (x y) 2 = y(x y) 2, f y = x(x y) 2 = f xx = 2y(x y) 3 = 2y (x y) 3 f xy = 1(x y) 2 2y(x y) 3 = x y (x y) 3 f yx = (x y)2 2x(x y) (x y) 4 = x y (x y) 3 f yy = 2x (x y) 3 Observation: f xy = f yx. Is this a coincidence? x (x y) 2
30 Another example. g(x, y) = sin(x + 3y): g x = g y = g xx = g xy = g yx = g yy =
31 Another example. g(x, y) = sin(x + 3y): g x = cos(x + 3y) g y = g xx = g xy = g yx = g yy =
32 Another example. g(x, y) = sin(x + 3y): g x = cos(x + 3y) g y = 3 cos(x + 3y) g xx = g xy = g yx = g yy =
33 Another example. g(x, y) = sin(x + 3y): g x = cos(x + 3y) g y = 3 cos(x + 3y) g xx = sin(x + 3y) g xy = g yx = g yy =
34 Another example. g(x, y) = sin(x + 3y): g x = cos(x + 3y) g y = 3 cos(x + 3y) g xx = sin(x + 3y) g xy = 3 sin(x + 3y) g yx = g yy =
35 Another example. g(x, y) = sin(x + 3y): g x = cos(x + 3y) g y = 3 cos(x + 3y) g xx = sin(x + 3y) g xy = 3 sin(x + 3y) g yx = 3 sin(x + 3y) g yy =
36 Another example. g(x, y) = sin(x + 3y): g x = cos(x + 3y) g y = 3 cos(x + 3y) g xx = sin(x + 3y) g xy = 3 sin(x + 3y) g yx = 3 sin(x + 3y) g yy = 9 sin(x + 3y)
37 Clairaut s Theorem. Theorem (Clairaut s Theorem) If f xy and f yx exist and are continuous in a neighborhood of (a, b), then f xy (a, b) = f yx (a, b). Examples: If f (x, y) is a polynomial or rational function then the partial derivatives are still polynomial/rational, hence continuous, so Clairaut s Theorem applies automatically (as long as denominator 0). f (x, y) = x + y : partial derivatives don t exist Example where f xy f yx : see textbook, pg. 805
38 Worksheet. Worksheet #1 2
39 Clicker Question: The figure below shows the surface z = f (x, y). What are the signs of f xx (P) 133. andthe f yy (P)? figure below shows the surface z = f(x, y). W f yy (P )? A. f (a) f xx (P ) > 0, f yy (P ) 0 xx (P) > 0, f yy (P) 0 C. f xx (P) 0, f yy (P) 0 (b) f xx (P ) > 0, f yy (P ) < 0 B. f xx (P) > 0, f yy (c) (P) f xx < (P 0 ) 0, D. f yy f(p xx (P) ) < 0 0, f yy (P) > 0 receiver channel: 41 (d) f xx (P ) < 0, f yy (P session ) > 0 ID: bsumath275
Example: Limit definition. Geometric meaning. Geometric meaning, y. Notes. Notes. Notes. f (x, y) = x 2 y 3 :
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