Differentiation Mean Value Theorem and Taylor's Theorem 1 Differentiation Differentiability of functions Rules of differentiation Second and higher derivatives 2 Mean Value Theorem and Taylor's Theorem Mean value theorem L'Hopital's rule Taylor's theorem MATH1010 University Mathematics Differentiability of functions Differentiation Rules of differentiation Mean Value Theorem and Taylor's Theorem Second and higher derivatives Definition (Differentiability) Let f (x) be a function. Denote f (a + h) − f (a) f 0(a) = lim h!0 h and we say that f (x) is differentiable at x = a if the above limit exists. We say that f (x) is differentiable on (a; b) if f (x) is differentiable at every point in (a; b). The above limit can also be written as f (x) − f (a) f 0(a) = lim : x!a x − a MATH1010 University Mathematics Differentiability of functions Differentiation Rules of differentiation Mean Value Theorem and Taylor's Theorem Second and higher derivatives Theorem If f (x) differentiable at x = a, then f (x) is continuous at x = a. Differentiable at x = a ) Continuous at x = a Proof. Suppose f (x) is differentiable at x = a. Then f (x) − f (a) lim(f (x) − f (a)) = lim (x − a) x!a x!a x − a f (x) − f (a) = lim lim(x − a) x!a x − a x!a = f 0(a) · 0 = 0 Therefore f (x) is continuous at x = a. Note that the converse of the above theorem does not hold. The function f (x) = jxj is continuous but not differentiable at 0. MATH1010 University Mathematics Differentiability of functions Differentiation Rules of differentiation Mean Value Theorem and Taylor's Theorem Second and higher derivatives Example eh − e0 eh − 1 1 f (x) = ex : f 0(0) = lim = lim = 1. h!0 h h!0 h ln(1 + h) − ln 1 ln(1 + h) 2 f (x) = ln x: f 0(1) = lim = lim = 1. h!0 h h!0 h sin h − sin 0 sin h 3 f (x) = sin x: f 0(0) = lim = lim = 1. h!0 h h!0 h MATH1010 University Mathematics Differentiability of functions Differentiation Rules of differentiation Mean Value Theorem and Taylor's Theorem Second and higher derivatives Example ( 4x − 1; if x ≤ 1 Find the values of a; b if f (x) = is differentiable at ax 2 + bx; if x > 1 x = 1. Solution Since f (x) is differentiable at x = 1, f (x) is continuous at x = 1 and we have lim f (x) = f (1) ) lim (ax 2 + bx) = a + b = 3: x!1+ x!1+ Moreover, f (x) is differentiable at x = 1 and we have f (1 + h) − f (1) (4(1 + h) − 1) − 3 lim = lim = 4 h!0− h h!0− h f (1 + h) − f (1) a(1 + h)2 − b(1 + h) − 3 lim = lim = 2a + b h!0+ h h!0+ h ( ( a + b = 3 a = 1 Therefore ) : 2a + b = 4 b = 2 MATH1010 University Mathematics Differentiability of functions Differentiation Rules of differentiation Mean Value Theorem and Taylor's Theorem Second and higher derivatives Definition (First derivative) Let y = f (x) be a differentiable function on (a; b). The first derivative of f (x) is the function on (a; b) defined by dy f (x + h) − f (x) = f 0(x) = lim : dx h!0 h Theorem Let f (x) and g(x) be differentiable functions and c be a real number. Then 0 0 0 1 (f + g) (x) = f (x) + g (x) 0 0 2 (cf ) (x) = cf (x) MATH1010 University Mathematics Differentiability of functions Differentiation Rules of differentiation Mean Value Theorem and Taylor's Theorem Second and higher derivatives Theorem d n n−1 + 1 x = nx , n 2 , for x 2 dx Z R d x x 2 e = e for x 2 dx R d 1 3 ln x = for x > 0 dx x d 4 cos x = − sin x for x 2 dx R d 5 sin x = cos x for x 2 dx R MATH1010 University Mathematics Differentiability of functions Differentiation Rules of differentiation Mean Value Theorem and Taylor's Theorem Second and higher derivatives d Proof ( xn = nxn−1) dx n Let y = x . For any x 2 R, we have dy (x + h)n − xn = lim dx h!0 h (x + h − x)((x + h)n−1 + (x + h)n−2x + ··· + xn−1) = lim h!0 h = lim ((x + h)n−1 + (x + h)n−2x + ··· + xn−1) h!0 = nxn−1 + Note that the above proof is valid only when n 2 Z is a positive integer. MATH1010 University Mathematics Differentiability of functions Differentiation Rules of differentiation Mean Value Theorem and Taylor's Theorem Second and higher derivatives d Proof ( ex = ex ) dx x Let y = e . For any x 2 R, we have dy ex+h − ex ex (eh − 1) = lim = lim = ex : dx h!0 h h!0 h (Alternative proof) dy d x 2 x 3 x 4 = 1 + x + + + + ··· dx dx 2! 3! 4! 2x 3x 2 4x 3 = 0 + 1 + + + + ··· 2! 3! 4! x 2 x 3 = 1 + x + + + ··· 2! 3! = ex In general, differentiation cannot be applied term by term to infinite series. The second proof is valid only after we prove that this can be done to power series. MATH1010 University Mathematics Differentiability of functions Differentiation Rules of differentiation Mean Value Theorem and Taylor's Theorem Second and higher derivatives Proof d 1 ln x = Let f (x) = ln x. For any x > 0, we have dx x h dy ln(x + h) − ln x ln 1 + x 1 = lim = lim = : dx h!0 h h!0 h x d cos x = − sin x Let f (x) = cos x. For any x 2 R, we have dx h h dy cos(x + h) − cos x −2 sin x + 2 sin 2 = lim = lim = − sin x: dx h!0 h h!0 h d sin x = cos x Let f (x) = sin x. For any x 2 R, we have dx h h dy sin(x + h) − sin x 2 cos x + 2 sin 2 = lim = lim = cos x: dx h!0 h h!0 h MATH1010 University Mathematics Differentiability of functions Differentiation Rules of differentiation Mean Value Theorem and Taylor's Theorem Second and higher derivatives Definition Let a > 0 be a positive real number. For x 2 R, we define ax = ex ln a: Theorem Let a > 0 be a positive real number. We have x+y x y 1 a = a a for any x; y 2 R d 2 ax = ax ln a. dx Proof. 1 ax+y = e(x+y) ln a = ex ln aey ln a = ax ay d d 2 ax = ex ln a = ex ln a ln a = ax ln a dx dx MATH1010 University Mathematics Differentiability of functions Differentiation Rules of differentiation Mean Value Theorem and Taylor's Theorem Second and higher derivatives Example Let f (x) = jxj for x 2 R. Show that f (x) is not differentiable at x = 0. Proof. Observe that f (h) − f (0) −h lim = lim = −1 h!0− h h!0− h f (h) − f (0) h lim = lim = 1 h!0+ h h!0+ h Thus the limit f (h) − f (0) lim h!0 h does not exist. Therefore f (x) is not differentiable at x = 0. MATH1010 University Mathematics Differentiability of functions Differentiation Rules of differentiation Mean Value Theorem and Taylor's Theorem Second and higher derivatives Figure: f (x) = jxj MATH1010 University Mathematics Differentiability of functions Differentiation Rules of differentiation Mean Value Theorem and Taylor's Theorem Second and higher derivatives Example 0 Let f (x) = jxjx for x 2 R. Find f (x). Solution When x < 0, f (x) = −x 2 and f 0(x) = −2x. When x > 0, f (x) = x 2 and f 0(x) = 2x. When x = 0, we have f (h) − f (0) −h2 − 0 lim = lim = 0 h!0− h h!0− h f (h) − f (0) h2 − 0 lim = lim = 0 h!0+ h h!0+ h Thus f 0(0) = 0. Therefore 8 −2x; if x < 0 <> f 0(x) = 0; if x = 0 :>2x; if x > 0 = 2jxj: Note that f 0(x) = 2jxj is continuous at x = 0. MATH1010 University Mathematics Differentiability of functions Differentiation Rules of differentiation Mean Value Theorem and Taylor's Theorem Second and higher derivatives Figure: f (x) = jxjx MATH1010 University Mathematics Differentiability of functions Differentiation Rules of differentiation Mean Value Theorem and Taylor's Theorem Second and higher derivatives Example Let 8 1 <x sin ; if x 6= 0 f (x) = x : :0; if x = 0 1 Find f 0(x) for x 6= 0. 2 Determine whether f (x) is differentiable at x = 0. Solution 1. When x 6= 0, 1 1 1 f 0(x) = sin − cos : x x x 2. We have f (h) − f (0) h sin 1 1 lim = lim h = lim sin h!0 h h!0 h h!0 h does not exist. Therefore f (x) is not differentiable at x = 0. MATH1010 University Mathematics Differentiability of functions Differentiation Rules of differentiation Mean Value Theorem and Taylor's Theorem Second and higher derivatives 1 Figure: f (x) = x sin x MATH1010 University Mathematics Differentiability of functions Differentiation Rules of differentiation Mean Value Theorem and Taylor's Theorem Second and higher derivatives Example Let 8 1 <x 2 sin ; if x 6= 0 f (x) = x : :0; if x = 0 1 Find f 0(x). 2 Determine whether f 0(x) is continuous at x = 0.