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Math 63: Winter 2021 Lecture 17 Math 63: Winter 2021 Lecture 17 Dana P. Williams Dartmouth College Monday, February 15, 2021 Dana P. Williams Math 63: Winter 2021 Lecture 17 Getting Started 1 We should be recording. 2 Remember it is better for me if you have your video on so that I don't feel I'm just talking to myself. 3 Our midterm will be available Friday after class and due Sunday by 10pm. It will cover through Today's lecture. More details soon. 4 Time for some questions! Dana P. Williams Math 63: Winter 2021 Lecture 17 Rules To Differentiate With The following are routine consequences of the definition. Lemma If f (x) = c for all x 2 R, then f 0(x) = 0 for all x 2 R. If f (x) = x for all x 2 R, then f 0(x) = 1 for all x 2 R. Theorem Suppose that f and g are differentiable at x0. Then so are f ± g, f fg, and if g(x0) 6= 0, g . Futhermore, 0 0 0 1 (f ± g) (x0) = f (x0) ± g (x0), 0 0 0 2 (fg) (x0) = f (x0)g(x0) + f (x0)g (x0), and 0 f 0(x )g(x )−f (x )g 0(x ) 3 f 0 0 0 0 (x0) = 2 . g g(x0) Dana P. Williams Math 63: Winter 2021 Lecture 17 More Formulas Corollary Suppose that f is differentiable at x0 and c 2 R. Then cf is 0 0 differentiable at x0 and (cf ) (x0) = cf (x0). Corollary Suppose that n 2 Z and f (x) = xn. Then f 0(x) = nxn−1. Remark We are being a bit sloppy here. It is implict that if n < 0 then f is not defined at 0, and if n = 0, we are just claiming that f 0 = 0. Proof. We have already dealt with the cases that n = 0 and n = 1. If d n n−1 n ≥ 1, and we assume dx (x ) = nx , then d n+1 d n n n−1 n dx (x ) = dx (x · x ) = x + x(nx ) = (n + 1)x so the result holds for all n ≥ 0 by induction. Dana P. Williams Math 63: Winter 2021 Lecture 17 Proof Proof Continued. d n d 1 −nx−n−1 n−1 If n < 0, then dx (x ) = dx x−n = − x−2n = nx . Proposition (The Chain Rule) Suppose that U and V are open subsets of R. Let f : U ! V and g : V ! R be functions such that f is differentiable at x0 2 U and g is differentiable at f (x0) 2 V . Then g ◦ f is differentiable at x0 and 0 0 0 (g ◦ f ) (x0) = g (f (x0))f (x0): Dana P. Williams Math 63: Winter 2021 Lecture 17 Proof Proof. ( g(y)−g(f (x0)) if y 2 V and y 6= f (x0), and Let A(y) = y−f (x0) . Then 0 g (f (x0)) if y = f (x0). 0 limy!f (x0) A(y) = g (f (x0)). It follows that A is continuous at f (x0). Therefore limx!x0 A(f (x)) = A(f (x0)) since the composition of continuous functions is continuous. Therefore 0 g(f (x)) − g(f (x0)) (g ◦ f ) (x0) = lim x!x0 x − x0 f (x) − f (x ) = lim A(f (x)) 0 x!x0 x − x0 0 0 = g (f (x0))f (x0): Dana P. Williams Math 63: Winter 2021 Lecture 17 Extreme Values Proposition Suppose that U is open in R and that f : U ! R has either a maximum or minimum value at x0 2 U. Then either f is not 0 differentiable at x0 or f (x0) = 0. Proof. 0 Assume that f (x0) exists. Then ( f (x)−f (x0) if x 6= x0, and g(x) = x−x0 0 f (x0) if x = x0 0 is continuous at x0. Hence if f (x0) > 0, then is a δ > 0 such that 0 f (x0) 0 < jx − x0j < δ implies g(x) > 2 > 0. But if f (x0) is either a maximum or a minimum, then f (x) − f (x0) is either always non-positive or always non-negative. Since g(x) 6= 0 if x 6= x0, it follows that f (x) − f (x0) is always either always positive or always negative. But x − x0 can be both negative or positive. Hence g(x) 0 can't always be positive. Thus f (x0) ≤ 0. But a similar argument 0 implies f (x0) ≥ 0. Dana P. Williams Math 63: Winter 2021 Lecture 17 Rolle's Theorem Lemma (Rolle's Theorem) Suppose a < b in R, that f :[a; b] ! R is continuous and differentiable on (a; b). If f (a) = f (b), then there is a c 2 (a; b) such that f 0(c) = 0. Proof. By the Extreme-Value Theorem, f attains its maximum and minimum on [a; b]. If both these values occur at the endpoints a and b, then f is constant and f 0(c) = 0 for all c 2 (a; b). Otherwise, either the maximum or minimum must occur at some c 2 (a; b). By the previous proposition, f 0(c) = 0. Dana P. Williams Math 63: Winter 2021 Lecture 17 The Mean Value Theorem Theorem (The Mean Value Theorem) Suppose that a < b in R and that f :[a; b] ! R is continuous and differentiable on (a; b). Then there is a c 2 (a; b) such that f (b) − f (a) f 0(c) = : b − a Remark In my calculus courses, I emphasize that the MVT says that there is always a point where the instantaneous rate of change equals the average rate of change. (Picture) Dana P. Williams Math 63: Winter 2021 Lecture 17 Proof Proof. f (b)−f (a) Let F (x) = f (x) − f (a) − (b−a) (x − a). Then F is continuous on [a; b] and differentiable on (a; b). Since F (b) = F (a) = 0, there is a c 2 (a; b) such that F 0(c) = 0. The result follows from this. Corollary Suppose that f :(c; d) ! R is differentiable and f 0(x) = 0 for all x 2 (c; d). Then f is constant. Proof. It suffices to see that if a; b 2 (c; d), then f (a) = f (b). But we may assume a < b, then the MVT implies f (b)−f (a) 0 f (b) − f (a) = b−a (b − a) = f (c)(b − a) = 0. Corollary If f and g are differentiable real-valued functions on an open interval such that f 0 = g 0, then f and g differ by a constant. Dana P. Williams Math 63: Winter 2021 Lecture 17 Increasing and Decreasing Functions Definition A real-valued function on a set U ⊂ R is called increasing(strictly increasing) if a < b in U implies f (a) ≤ f (b)(f (a) < f (b)). On the other hand, f is called decreasing(strictly decreasing) if a < b in U implies f (a) ≥ f (b)(f (a) > f (b)). Proposition Suppose that a real-valued function on an open interval is such that f 0(x) > 0 for all x in the interval. Then f is strictly increasing on that interval. Proof. Suppose that a < b in the interval. Then f (b)−f (a) 0 f (b) − f (a) = b−a (b − a) = f (c)(b − a) > 0. Dana P. Williams Math 63: Winter 2021 Lecture 17 Break Time Time for a break and some questions. Dana P. Williams Math 63: Winter 2021 Lecture 17 Higher Order Derivatives Remark If f is a real-valued function which is differentiable on an open set U ⊂ R, then we can consider the function f 0 : U ! R. If f 0 is differentiable, then we say that f is twice differentiable and variously write f 00 or f (2) for the second derivative (f 0)0. Of course, we can continue the madness and consider higher order derivatives f 000 = f (3), and so on|if the derivatives exist. If f is n-times dnf (n) differentiable, we also write dxn for f . To make some formulas work, we agree that f (0) = f (the 0th-derivative). Notation We will also employ the factorial notation: 0! := 1 and (n + 1)! = (n + 1)n! for n ≥ 1. Or less pedantically, n! = 1 · 2 ··· n. Dana P. Williams Math 63: Winter 2021 Lecture 17 A Lemma It will be convenient to establish the following technical result. Lemma Suppose that f : U ! R is (n + 1)-times differentiable. If b 2 U, define a function on Rn(b; ·): U ! R by f 00(x) f (b) = f (x) + f 0(x)(b − x) + (b − x)2+ 2! f (n)(x) ··· + (b − x)n + R (b; x): n! n Then d f (n+1)(x) R (b; x) = − (b − x)n: dx n n! Dana P. Williams Math 63: Winter 2021 Lecture 17 Proof Proof Continued. Without the unusual structure, we have n X (b − x)j R (b; x) = f (b) − f (x) − f (j)(x) : n j! j=1 Therefore Rn(b; ·) is differentiable and using the product and chain rules, we have n d Xh (b − x)j−1 (b − x)j i R (b; x) = −f 0(x) + f j (x) − f (j+1)(x) dx n (j − 1)! j! j=1 which telescopes to (b − x)n = −f (n+1)(x) : n! Dana P. Williams Math 63: Winter 2021 Lecture 17 Taylor's Theorem Theorem (Taylor's Theorem) Suppose that U ⊂ R is open and that f : U ! R is (n + 1)-times differentiable on U.
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