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 ∈ R, then f 0(x) = 0 for all x ∈ R. If f (x) = x for all x ∈ R, then f 0(x) = 1 for all x ∈ 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 ∈ R. Then cf is 0 0 differentiable at x0 and (cf ) (x0) = cf (x0).

Corollary Suppose that n ∈ 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 ∈ U and g is differentiable at f (x0) ∈ 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 ∈ 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 ∈ 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 < |x − x0| < 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 ∈ (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 ∈ (a, b). Otherwise, either the maximum or minimum must occur at some c ∈ (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 ∈ (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 ∈ (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 ∈ (c, d). Then f is constant.

Proof. It suffices to see that if a, b ∈ (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 ∈ 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. Then for any a, b ∈ U, we have

f 00(a) f (b) = f (a) + f 0(a)(b − a) + (b − a)2 + ··· 2! f (n)(a) f (n+1)(c) ··· + (b − a)n + (b − a)n+1 n! (n + 1)!

where c lies between a and b. (In the degenerate case a = b, we can let c = a.)

Remark In the case n = 1, this reduces to a milder form of the Mean Value Theorem.

Dana P. Williams Math 63: Winter 2021 Lecture 17 Proof

Proof. If a = b, the result is trivial. The awkward statement of the last lemma is so that we can see that we just need to show that, in the notation of that lemma,

f (n+1)(c) R (a, b) = (b − a)n+1 n (n + 1)! for some c between a and b. Since a 6= b, there is a real number K such that (b − a)n+1 R (a, b) = K . n (n + 1)!

Dana P. Williams Math 63: Winter 2021 Lecture 17 Proof

Proof Continued. Define ϕ : U → R by

(b − x)n+1 ϕ(x) = R (b, x) − K . n (n + 1)! Clearly ϕ is differentiable on U and ϕ(a) = ϕ(b) = 0. Therefore ϕ restricted to the interval [a, b] if a < b or [b, a] if b < a satisfies the hypotheses of Rolle’s Theorem. Thus there is a c between a and b such that ϕ0(c) = 0. But by our lemma,

f (n+1)(x) (b − x)n ϕ0(x) = − (b − a)n + K . n! n! (n+1) f (n+1)(c) n+1 Thus K = f (c) and Rn(a, b) = (n+1)! (b − a) as claimed.

Dana P. Williams Math 63: Winter 2021 Lecture 17 An Interpretation

Remark Suppose that f : U → R is (n + 1)-times differentiable on an open set U. If a ∈ U, let

f (n)(a) P (a, x) = f (a) + f 0(a)(x − a) + ··· + (x − a)n. n n! th We call Pn(a, x) the n -degree Taylor polynomial for f about x = a. Note that P1(a, x) is just the linear approximation to f (x) near a given by the derivative. That is P1(a, x) is the unique 0 0 polynomial such that P1(a, a) = f (a) and P1(a, a) = f (a). It is not so hard to check that Pn(a, x) is the unique polynomial of (k) (k) degree at most n such that Pn (a, a) = f (a) for all 0 ≤ k ≤ n. Then f (x) = Pn(a, x) + Rn(a, x) and Rn(a, x) is just the difference f (x) − Pn(a, x). Thus |Rn(a, x)| is just the error in approximating f (x) with Pn(a, x).

Dana P. Williams Math 63: Winter 2021 Lecture 17 Example

Example 1 (n) n n! Let f (x) = x and let a = 1. Then f (x) = (−1) xn+1 . Thus f (j)(1) j j! = (−1) and the Taylor polynomial of degree n about x = 1 is

2 n n Pn(1, x) = 1 − (x − 1) + (x − 1) − · · · + (−1) (x − 1) .

It is fun to use Mathematica to draw some pictures. Note that if x ∈ [1, b], then

(x − 1)n+1 |R (1, x)| = ≤ |b − 1|n+1 n cn+2 since c ∈ (1, b).

Dana P. Williams Math 63: Winter 2021 Lecture 17 Enough

1 That is enough for today.

Dana P. Williams Math 63: Winter 2021 Lecture 17