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M104 Notes: Lecture 1
Aaron McMillan
August 6, 2008
1 Differentiation
This section will focus on functions f : R → R. We’ll frequently use the following notation: limx→a f (x) = b if for all sequences (xn) ⊆ (R \ {a}) such that (xn) → a, the sequence (f (xn)) converges to b. Alternatively (and equivalently), limx→a f (x) = b if and only if every B(b) contains the image of some Bδ(a) neighborhood.
Definition 1.1. A function f : R → R is differentiable at a ∈ R if
lim x→a
f (x) − f (a) x − a
exists.
We denote the value of this limit by f ′(a) ∈ R and call it the derivative of f at a.
Thus we see, by our second characterization of limits above, f is differ- entiable at a ∈ R with derivative L, if and only if ∀ > 0, there exists δ > 0 such that
if |x − a| < δ then |
f (x) − f (a) x − a
− L| <
Proposition 1.2 (Differentiation Rules). Suppose f, g : R → R.
(i) If f is differentiable, then f is continuous.
(ii) If f, g are differentiable at x ∈ R, then so is f + g. Furthermore, (f + g)′(x) = f ′(x) + g′(x).
(iii) If f, g are differentiable at x ∈ R, then so is f · g. Furthermore, (f g)′(x) = f ′(x)g(x) + f (x)g′(x).
(iv) If f (x) = c is constant, then f ′(x) = 0 for all x ∈ R.
(v) If f, g are differentiable at x ∈ R and g(x) 6 = 0, then so is (f /g). Furthermore, (f /g)′(x) =
f ′(x)g(x) − f (x)g′(x) g(x)^2
Proof: I’ll include proofs of (i) and (v) – otherwise, see Ross. (i) A function is continuous at x ∈ R, if limt→x f (t) = f (x). For t 6 = x we have
f (t) =
(f (t) − f (x) t − x
(t − x) + f (x)
So we see that limt→x f (t) = f ′(x) · 0 + f (x) = f (x). (v) Note that
(f /g)(t) − (f /g)(x) =
f (t) g(t)
f (x) g(x)
f (t)g(x) − f (x)g(t) g(x)g(t)
Since g is continuous at x (by (i)), and g(x) 6 = 0, there is a small interval around x where g(t) 6 = 0, and we can therefore take the limit. Thus we see that (f /g)(t)−(f /g)(x) x−t =^
f (t)g(x)−f (x)g(t) g(x)g(t)(x−t)
= (^) g(x^1 )g(t)^ f^ (t)g(x)−f^ (x)g(x(x)+−ft^ )(x)g(x)−f^ (x)g(t)
= (^) g(x^1 )g(t) (g(x)f^ (t) t−−fx^ ( x)− f (x)g(t()x−−gt()x ))
Taking the limit as t → x, and noting that limt→xg(t) = g(x) because g is continuous, the result follows.
Corollary 1.3. If p(x) = anxn^ +... + a 1 x + a 0 is a polynomial, then p′(x) = nanxn−^1 +... + 2a 2 x + a 1.
Theorem 1.4 (Chain Rule). If f is differentiable at x and g is differentiable at f (x), then (g ◦ f ) is differentiable at x and the derivative is (g ◦ f )′(x) = g′(f (x))f ′(x).
Proof: To prove the chain rule, we note that for some small interval around x, we can write f (t) − f (x) = (t − x)(f ′(x) + u(t))
Theorem 1.10 (Mean Value Theorem). Suppose f : [a, b] → R is continuous on [a, b] and differentiable on (a, b). Then ∃θ ∈ (a, b) such that f (b) − f (a) = f ′(θ)(b − a).
Proof: Define the function φ(x) = f (x) − kx, where k = f^ (b b)−−fa^ ( a). Thus we see that φ(a) = φ(b). By Rolle’s Theorem, ∃θ ∈ (a, b) such that φ′(θ) = 0. However, φ′(x) = f ′(x) − k giving us the desired result.
Corollary 1.11. Suppose f : [a, b] → R is continuous. Then if f ′(x) = 0 for all x ∈ (a, b), then f is constant.
Corollary 1.12. If f : [a, b] → R is continuous, and f ′(x) > 0 for all x ∈ (a, b), then f is strictly increasing on [a, b].
Theorem 1.13 (Intermediate Value Property of Derivatives). If f is dif- ferentiable on (a − , b + ), then f ′(x) takes every value between f ′(a) and f ′(b).
Proof: Suppose f ′(a) < f ′(b), and pick f ′(a) < η < f ′(b). Define φ(x) = f (x) − ηx. We want to find a c such that φ′(c) = 0. (Note: φ′(a) < 0 and φ′(b) > 0) Since φ is continuous on [a, b], there exists c ∈ [a, b] such that φ(c) is a minimum. If φ′(c) > 0, then c 6 = a and ∃x 1 ∈ (a, b) such that φ(x 1 ) < φ(c), which contradicts that φ(c) is a minimum. Similarly, if φ′(c) < 0, then c 6 = b and ∃x 2 ∈ (a, b) such that φ(x 2 ) < φ(c), which again is a contradiction. Thus φ′(c) = 0.
Corollary 1.14. Suppose f : (a, b) → R is differentiable. If f ′^ is discontin- uous at c, then either limx→c− f ′(x) or limx→c+ f ′(x) cannot exist. In other words, f ′^ cannot have a jump discontinuity at c.
1.2 Inverse Function Theorem
Theorem 1.15. Suppose f : [a, b] → R is continuous, and differentiable on (a, b), with f ′(x) 6 = 0 for all x ∈ (a, b). Then the inverse function f −^1 exists and is differentiable on its domain, with derivative (f −^1 )′(x) = (^) f ′(f −^11 (x)).
Proof Sketch: First one shows that f −^1 exists. Since f ′(x) 6 = 0 and f ′^ has the intermediate value property, it follows that either (1) f ′(x) > 0 for all x, in which case f is strictly increasing, or (2) f ′(x) < 0 for all x, in which case f is strictly decreasing. In either case, this proves f is injective (1 to 1),
and so f −^1 exists, with domain [f (a), f (b)] (if f is increasing) or [f (b), f (a)] (if f is decreasing) – we are implicitly using intermediate value theorem here (how?). Second, we need to show that f −^1 is continuous. To do this, fix a point x 0 ∈ [f (a), f (b)] (we assume f is increasing), and an > 0. Choose δ < min{x 0 − f (f −^1 (x 0 ) − ), f (f −^1 (x 0 ) + ) − x 0 } and use the fact that f −^1 is also strictly increasing to show this choice of δ works (draw a picture). Third, to show it’s differentiable, just calculate the derivative of f −^1 using the definition, and use the fact that f −^1 is continuous to justify that if x → x 0 , then f −^1 (x) → f −^1 (x 0 ).
Example 1.16. We can use the inverse function theorem to define log(x) from ex. If f (x) = ex, we know that f ′(x) > 0 for all x ∈ R. So if y = ex, then
(f −^1 )′(y) =
f ′(f −^1 (y)
f ′(x)
ex^
y
Thus we see that the derivative of log(x) is (^1) x.
Example 1.17. If f (x) = x^3 , then f is invertible, with continuous inverse given by f −^1 (x) = x (^13)
. However, f −^1 is not differentiable at x = 0, and this is because f ′(0) = 0.
1.3 Taylor’s Theorem
Fix an f : I → R that’s at least (n + 1)-times differentiable on some open interval I, and two points x, a ∈ I. We will define
Fn(t) = f (t) + f ′(t)(x − t) +
f ′′(t) 2!
(x − t)^2 +... +
f (n)(t) n!
(x − t)n
We can apply the Mean Value theorem to Fn (check that it satisfies the hypotheses!) on the interval [a, x], and obtain the existence of some θ ∈ (a, x) such that Fn(x)−Fn(a) = F ′(θ)(x−a). Now, we want to decipher both sides of this equality. First, using the product rule, we have
F (^) n′(t) = f ′(t)+f ′(t)(−1)+f ′′(t)(x−t)+
f ′′(t) 2
(x−t)(−2)+.. .+
f (n+1)(t) n!
(x−t)n
Note that all the terms cancel except for the last term, giving that
F (^) n′(t) =
f (n+1)(t) n!
(x − t)n
