Lecture 19: Differentiation Rules, Rolle’s Theorem, and the Mean Value Theorem

18.100A: Complete Lecture Notes Lecture 19: Differentiation Rules, Rolle’s Theorem, and the Mean Value Theorem Theorem 1 Let f : I → R, g : I → R be differentiable at c ∈ I. Then, 1. (Linearity) ∀α ∈ R, (αf + g)0 (c) = αf 0 (c) + g 0 (c) . 2. (Product rule) (f g)0 (c) = f 0 (c)g(c) + f (c)g 0 (c). 3. (Quotient rule) If g(x) 6= 0 for all x ∈ I, then  0 f 0 (c)g(c) − f (c)g 0 (c) f (c) = . g (g(c))2 Proof : 1. We can compute this directly: lim x→c (αf + g)(x) − (αf + g)(c) f (x) − f (c) g(x) − g(c) = lim α + = αf 0 (x) + g 0 (x). x→c x−c x−c x−c 2. We first write f (x)g(x) − f (c)g(c) f (x) − f (c) g(x) − g(c) = · g(x) + f (c) · x−c x−c x−c and use the fact that limx→c g(x) = f (c). 3. The quotient rule is left as an exercise to the reader. Theorem 2 (Chain Rule) Let I1 , I2 be two intervals, g : I1 → I2 be differentiable at c ∈ I1 , and f : I2 → R differentiable at g(c). Then, f ◦ g : I1 → R is differentiable at c and (f ◦ g)0 (c) = f 0 (g(c))g 0 (c). Proof: Let h(x) = f (g(x)) and d = g(c). We want to prove that h0 (c) = f 0 (d)g 0 (c). Define the following    f (y)−f (d) y 6= d  g(x)−g(c) x 6= c y−x x−c u(y) = and v(y) = . f 0 (d) g 0 (d) y=d x=c Then, lim u(y) = lim y→d y→d f (y) − f (d) = f 0 (d) = u(d). y−d 1 Similarly, g(x) − g(c) = g 0 (c) = v(c). x−c In other words, u is continuous at d and v is continuous at c. Now, lim v(x) = lim x→c x→c f (y) − f (d) = u(y)(y − d) g(x) − g(c) = v(x)(x − c) =⇒ h(x) − h(c) = f (g(x)) − f (d) = u(g(x))(g(x) − g(c)) = u(g(x))v(x)(x − c). Therefore, h(x) − h(c) = lim u(g(x))v(x) x→c x→c x−c lim = u(g(c))v(c) = f 0 (g(c))g 0 (c). Mean Value Theorem Definition 3 (Relative Maximum/Minimum) Let S ⊂ R and f : S → R. Then, f has a relative maximum at c ∈ S if ∃δ > 0 such that for all x ∈ S, |x − c| < δ =⇒ f (x) ≤ f (c). The definition for relative maximum is analogous. Theorem 4 If f : [a, b] → R has a relative max or min at c ∈ (a, b) and f is differentiable at c, then f 0 (c) = 0. Proof : If f has a relative maximum at c ∈ (a, b) then ∃δ > 0 such that (c − δ, c + δ) ⊂ (a, b) and ∀x ∈ (c − δ, c + δ), f (x) ≤ f (c). Let δ ∈ (c − δ, c). xn = c − 2n Then, xn → c so f (xn ) − f (c) f 0 (c) = lim ≥ 0. n→∞ xn − c 2 Now let yn = c + δ ∈ (c, c + δ). 2n Then, yn → c so f 0 (c) = lim n→∞ f (yn ) − f (c) ≤ 0. yn − c Therefore, f 0 (c) = 0. The proof for relative minimum is similar and thus left to the reader. Theorem 5 (Rolle) Let f : [a, b] → R be continuous and differentiable on (a, b). If f (a) = f (b), then ∃c ∈ (a, b) such that f 0 (c) = 0. Remark 6. Are the hypotheses all necessary? This is left to the reader to figure out. Proof : Let K = f (a) = f (b). Since f is continuous on [a, b], ∃c1 , c2 ∈ [a, b] such that f achieves an absolute maximum at c1 and absolute minimum at c2 . If f (c1 ) > K =⇒ c1 ∈ (a, b). Therefore, f 0 (c1 ) = 0 by the previous theorem. Similarly, if f (c2 ) < K, then c2 ∈ (a, b) =⇒ f 0 (c2 ) = 0. If f (c1 ) ≤ K ≤ f (c2 ) =⇒ f (x) = K ∀x ∈ [a, b] =⇒ f 0 (c) − 0 for any c ∈ (a, b). Theorem 7 (Mean Value Theorem) Let f : [a, b] → R be continuous, and let f be differentiable on (a, b). Then, ∃c ∈ (a, b) such that f (b) − f (a) = f 0 (c)(b − a). Remark 8. The Mean Value Theorem is sometimes denoted MVT. Proof : Define g : [a, b] → R with g(x) = f (x) − f (b) + 3 f (b) − f (a) (b − x). b−a Then, g(a) = g(b) = 0. Thus, by Rolle’s theorem, ∃c ∈ (a, b) with g 0 (c) = 0, and hence 0 = f 0 (c) − f (b) − f (a) . b−a We now look at some useful applications of the MVT. Theorem 9 If f : I → R is differentiable and f 0 (x) = 0 for all x ∈ I, then f is constant. Proof : Let a, b ∈ I with a < b. Then, f is continuous on [a, b] and differentiable on (a, b). Therefore, ∃c ∈ (a, b) such that f (b) − f (a) = (b − a)f 0 (c) = 0. Hence, f (b) = f (a) for all a, b ∈ I such that a < b. Theorem 10 Let f : I → R be differentiable. Then, 1. f is increasing if and only if f 0 (x) ≥ 0 for all x ∈ I and 2. f is decreasing if and only if f 0 (x) ≤ 0 for all x ∈ I. Proof : 1. ( ⇐= ) Suppose f 0 (x) ≥ 0 for all x ∈ I. Let a, b ∈ I with a < b. Then, by MVT, ∃c ∈ (a, b) such that f (b) − f (a) = (b − a)f 0 (c) ≥ 0 =⇒ f (a) ≤ f (b). ( =⇒ ) Suppose f is increasing. Let c ∈ I and let {xn } be a sequence in I such that xn → c such that ∀n, (c) xn < c. Then, for all n, f (xn ) − f (c) ≤ 0 =⇒ f (xxnn)−f ≥ 0. Therefore, −c f 0 (c) = lim n→∞ f (xn ) − f (c) ≥ 0. xn − c Now let {xn } be a sequence in I such that xn → c such that ∀n, xn > c. Then, for all n, f (xn ) − f (c) ≥ (c) ≥ 0. Therefore, 0 =⇒ f (xxnn)−f −c f (xn ) − f (c) f 0 (c) = lim ≥ 0. n→∞ xn − c Hence, in either case, f 0 (c) ≥ 0. 2. Notice that f is decreasing if and only if −f is increasing, and apply 1. to −f . 4