The Chain Rule. Mathematics 11: Lecture 18. Dan Sloughter. Furman University. October 10, 2007

Transcription

1 The Chain Rule Mathematics 11: Lecture 18 Dan Sloughter Furman University October 10, 2007 Dan Sloughter (Furman University) The Chain Rule October 10, / 15

2 Example Suppose that a pebble is dropped in a calm pond, creating a circular wave which expands so that after t seconds the radius is r = 4 t centimeters. Dan Sloughter (Furman University) The Chain Rule October 10, / 15

3 Example Suppose that a pebble is dropped in a calm pond, creating a circular wave which expands so that after t seconds the radius is r = 4 t centimeters. If A is the area of the circle, then A = πr 2, and da dr = 2πr cm2 /cm. Dan Sloughter (Furman University) The Chain Rule October 10, / 15

4 Example Suppose that a pebble is dropped in a calm pond, creating a circular wave which expands so that after t seconds the radius is r = 4 t centimeters. If A is the area of the circle, then A = πr 2, and da dr = 2πr cm2 /cm. Also, dr dt = 4 2 t = 2 cm/sec. t Dan Sloughter (Furman University) The Chain Rule October 10, / 15

5 Example (cont d) For example, when t = 9, r = 12 cm, da dr = 24π cm 2 /cm and dr r=12 dt = 2 t=9 3 cm/sec. Dan Sloughter (Furman University) The Chain Rule October 10, / 15

6 Example (cont d) For example, when t = 9, r = 12 cm, da dr = 24π cm 2 /cm and dr r=12 dt = 2 t=9 3 cm/sec. We should expect that da dt = da ( ) dr 2 t=9 dr r=12 dt = (24π) = 16π cm 2 /sec. t=9 3 Dan Sloughter (Furman University) The Chain Rule October 10, / 15

7 Example (cont d) For example, when t = 9, r = 12 cm, da dr = 24π cm 2 /cm and dr r=12 dt = 2 t=9 3 cm/sec. We should expect that da dt = da ( ) dr 2 t=9 dr r=12 dt = (24π) = 16π cm 2 /sec. t=9 3 That is: the rate of change of the area of the circle with respect to time should be the product of the rate of change of the area with respect to the radius and the rate of change of the radius with respect to time. Dan Sloughter (Furman University) The Chain Rule October 10, / 15

8 Example (cont d) We can check this directly by noting that A = πr 2 = π(4 t) 2 = 16πt, from which it follows that da dt = 16π cm 2 /sec. t=9 Dan Sloughter (Furman University) The Chain Rule October 10, / 15

9 Idea of the chain rule If y is a function of u and u is a function of x, then we should expect that dy dx = dy du du dx. Dan Sloughter (Furman University) The Chain Rule October 10, / 15

10 Idea of the chain rule If y is a function of u and u is a function of x, then we should expect that dy dx = dy du du dx. Note: this would be obvious if these were really ratios (which they are not). Dan Sloughter (Furman University) The Chain Rule October 10, / 15

11 Idea of the chain rule If y is a function of u and u is a function of x, then we should expect that dy dx = dy du du dx. Note: this would be obvious if these were really ratios (which they are not). Note: if y = f (u) and u = g(x), then this says that (f g) (x) = dy dx = dy du du dx = f (g(x))g (x). Dan Sloughter (Furman University) The Chain Rule October 10, / 15

12 The chain rule If f and g are both differentiable, then (f g) (x) = f (g(x))g (x). Dan Sloughter (Furman University) The Chain Rule October 10, / 15

13 Proof A full proof of this statement involves a number of technicalities, but the general idea is as follows. Dan Sloughter (Furman University) The Chain Rule October 10, / 15

14 Proof A full proof of this statement involves a number of technicalities, but the general idea is as follows. We first note that (f g) f g(x + h) f g(x) (x) = lim h 0 h f (g(x + h)) f (g(x)) = lim h 0 h f (g(x + h)) f (g(x)) = lim h 0 g(x + h) g(x) g(x + h) g(x), h provided g(x + h) g(x) 0 for all h 0, which we will assume to be the case. Dan Sloughter (Furman University) The Chain Rule October 10, / 15

15 Proof (cont d) Now let u = g(x + h) g(x), and notice that g(x + h) = g(x) + u and u goes to 0 as h goes to 0 (since g is assumed differentiable, and hence continuous). Dan Sloughter (Furman University) The Chain Rule October 10, / 15

16 Proof (cont d) Now let u = g(x + h) g(x), and notice that g(x + h) = g(x) + u and u goes to 0 as h goes to 0 (since g is assumed differentiable, and hence continuous). Then (f g) f (g(x + h)) f (g(x)) g(x + h) g(x) (x) = lim h 0 g(x + h) g(x) h f (g(x) + u) f (g(x)) g(x + h) g(x) = lim lim u 0 u h 0 h = f (g(x))g (x). Dan Sloughter (Furman University) The Chain Rule October 10, / 15

17 Example Consider h(x) = (x 2 + 1) 10. Dan Sloughter (Furman University) The Chain Rule October 10, / 15

18 Example Consider h(x) = (x 2 + 1) 10. We may write h(x) = f (g(x)) where f (x) = x 10 and g(x) = x Dan Sloughter (Furman University) The Chain Rule October 10, / 15

19 Example Consider h(x) = (x 2 + 1) 10. We may write h(x) = f (g(x)) where f (x) = x 10 and g(x) = x Now f (x) = 10x 9 and g (x) = 2x, so h (x) = f (g(x)g (x) = 10(x 2 + 1) 9 (2x) = 20x(x 2 + 1) 9. Dan Sloughter (Furman University) The Chain Rule October 10, / 15

20 General example The previous example is a special case of the following: if n is a rational number and h(x) = (g(x)) n, then or, using Leibniz notation, h (x) = n(g(x)) n 1 g (x), d dx (g(x))n = n(g(x)) n 1 g (x). Dan Sloughter (Furman University) The Chain Rule October 10, / 15

21 General example The previous example is a special case of the following: if n is a rational number and h(x) = (g(x)) n, then or, using Leibniz notation, Example: if f (x) = h (x) = n(g(x)) n 1 g (x), d dx (g(x))n = n(g(x)) n 1 g (x). 4 (x 3 + x) 8, then f (x) = 32(x 3 + x) 9 (3x 2 + 1) = 32(3x 2 + 1) (x 3 + x) 9. Dan Sloughter (Furman University) The Chain Rule October 10, / 15

22 Example If g(x) = x 5 4x 2 + 1, then g (x) = x 5 ( 1 2 (4x 2 + 1) 1 2 (8x) ) + 5x 4 4x = 4x 6 4x x 4 4x Dan Sloughter (Furman University) The Chain Rule October 10, / 15

23 Examples If f (x) = sin(5x), then f (x) = cos(5x)(5) = 5 cos(5x). Dan Sloughter (Furman University) The Chain Rule October 10, / 15

24 Examples If f (x) = sin(5x), then f (x) = cos(5x)(5) = 5 cos(5x). If f (x) = sin 2 (x), then f (x) = 2 sin(x) cos(x). Dan Sloughter (Furman University) The Chain Rule October 10, / 15

25 Examples If f (x) = sin(5x), then f (x) = cos(5x)(5) = 5 cos(5x). If f (x) = sin 2 (x), then f (x) = 2 sin(x) cos(x). If g(t) = 5 cos 4 (6t), then g (t) = 20 cos 3 (6t)( sin(6t))(6) = 120 cos 3 (6t) sin(6t). Dan Sloughter (Furman University) The Chain Rule October 10, / 15

26 Examples If f (x) = sin(5x), then f (x) = cos(5x)(5) = 5 cos(5x). If f (x) = sin 2 (x), then f (x) = 2 sin(x) cos(x). If g(t) = 5 cos 4 (6t), then g (t) = 20 cos 3 (6t)( sin(6t))(6) = 120 cos 3 (6t) sin(6t). If h(u) = 6 sec 4 (3u), then h (u) = 24 sec 3 (3u)(sec(3u) tan(3u))(3) = 72 sec 4 (3u) tan(3u). Dan Sloughter (Furman University) The Chain Rule October 10, / 15

27 Leibniz notation If y = f (u) and u = g(x), then, in Leibniz notation, dy dx = (f g) (x) = f (g(x))g (x) = dy du du dx. Dan Sloughter (Furman University) The Chain Rule October 10, / 15

28 Leibniz notation If y = f (u) and u = g(x), then, in Leibniz notation, dy dx = (f g) (x) = f (g(x))g (x) = dy du du dx. Suppose y = u and u = 4 x. Then dy du = 2u and du dx = 4 x 2. Dan Sloughter (Furman University) The Chain Rule October 10, / 15

29 Leibniz notation If y = f (u) and u = g(x), then, in Leibniz notation, dy dx = (f g) (x) = f (g(x))g (x) = dy du du dx. Suppose y = u and u = 4 x. Then dy du = 2u and du dx = 4 x 2. To find dy dx, we have, noting that u = 1 when x = 4, x=4 dy dx = dy ( du x=4 du u=1 dx = (2) 4 ) = 1 x= Dan Sloughter (Furman University) The Chain Rule October 10, / 15

30 Example Suppose the area A of a circle of radius r is increasing at a rate of 10 square centimeters per second and we want to find the rate of growth of the radius of the circle when the radius is 4 centimeters. Dan Sloughter (Furman University) The Chain Rule October 10, / 15

31 Example Suppose the area A of a circle of radius r is increasing at a rate of 10 square centimeters per second and we want to find the rate of growth of the radius of the circle when the radius is 4 centimeters. We know that A = πr 2, so, using the chain rule, da dt = 2πr dr dt. Dan Sloughter (Furman University) The Chain Rule October 10, / 15

32 Example Suppose the area A of a circle of radius r is increasing at a rate of 10 square centimeters per second and we want to find the rate of growth of the radius of the circle when the radius is 4 centimeters. We know that A = πr 2, so, using the chain rule, da dt = 2πr dr dt. We are given that at any time t. da dt = 10 cm2 /sec Dan Sloughter (Furman University) The Chain Rule October 10, / 15

33 Example (cont d) Hence, when r = 4, we have 10 = 2π(4) dr dt, r=4 so dr dt = 5 r=4 4π cm/sec. Dan Sloughter (Furman University) The Chain Rule October 10, / 15