Chapter 3. Differentiation 3.3 Differentiation Rules

Transcription

1 3.3 Dfferetato Rules 1 Capter 3. Dfferetato 3.3 Dfferetato Rules Dervatve of a Costat Fucto. If f as te costat value f(x) = c, te f x = [c] = 0. x Proof. From te efto: f (x) f(x + ) f(x) o c c 0 = 0. QED Power Rule for Postve Itegers If s a postve teger, te x [x ] = x 1. Note. Before we preset te proof of te Power Rule, we trouce te Bomal Teorem.

2 3.3 Dfferetato Rules 2 Teorem. Bomal Teorem Let a a b be real umbers a let be a postve teger. Te (a + b) = a + a 1 b + = =0 a b ( 1) a 2 b ab 1 + b 2 were =! ( )!! a! = ()( 1)( 2) (3)(2)(1). Note. We ca prove te Bomal Teorem usg Matematcal Iucto. Proof of te Power Rule. By efto, f f(x + ) f(x) (x) (x + ) x =0 x + =1 x x x x

3 3.3 Dfferetato Rules 3 =1 =1 =1 x x 1 + =2 = x 1. x 1 x 1 x 1 QED Note. See page 136 for a proof of te Power Rule tat oes t (explctly) use te Bomal Teorem. Power Rule (Geeral Verso) If s ay real umber, te x [x ] = x 1, for all x were te powers x a x 1 are efe.

4 3.3 Dfferetato Rules 4 Note. Te state of te Geeral Verso of te Power Rule s a bt premature. I fact, eve efg wat t meas to ave a rratoal expoet requres te use of expoetal fuctos. Noe-te-less, te text states t ow! Dervatve Costat Multple Rule If u s a fferetable fucto of x, a c s a costat, te [cu] = cu x x. Dervatve Sum Rule If u a v are fferetable fuctos of x, te ter sum u + v s fferetable at every pot were u a v are bot fferetable. At suc pots, u [u + v] = x x + v x. Note. Te proofs of te Dervatve Costat Multple Rule a te Dervatve Sum Rule follow from te correspog rules for lmts (amely, te Costat Multple Rule a te Sum Rule, respectvely).

5 3.3 Dfferetato Rules 5 Corollary. (Page 144 umber 71) If P(x) = a x + a 1 x a 2 x 2 +a 1 x+a 0, te P (x) = a x 1 +( 1)a 1 x a 2 x+a 1. Example. Page 143 umbers 4, 12, a 34. Note. We ow fferetate a expoetal fucto f(x) = a x were a > 0. By efto, f (x) f(x + ) f(x) a x+ a x a x a a x a xa 1 = a x lm a 1. a 1 We clam wtout justfcato tat lm exsts a s some umber L a epeet o a. (For a clea scusso of ts result, see sectos 7.2 a 7.3 of Tomas Calculus, Staar 11t Eto otes are avalable ole at ttp://faculty.etsu.eu/garerr/1920/12/ otes12.tm. Tere s a verso ts text secto 7.1.) Wt x = 0, we ave f (0) = a 0 a 1 a 1 lm = L a. We wll see te precse

6 3.3 Dfferetato Rules 6 value of L a secto 3.7. Now f (0) s te slope of te grap of y = a x at x = 0. Motvate by Fgure 3.11, we see tat tere s a value of a somewere betwee 2 a 3 suc tat ts slope s 0. Fgure 3.12, page 139 We efe e to be te umber for wc te slope of te le taget to y = e x e 1 s m = 1 at x = 0. Tat s, we efe e suc tat lm = 1. Oe ca eterme umercally (for a tecque, see pages 183 a 184) tat e Wat s atural about te atural expoetal fucto e x s a calculus property a fferetato property. Teorem. Dervatve of te Natural Expoetal Fucto. x [ex ] = e x.

7 3.3 Dfferetato Rules 7 Example. Dfferetate f(x) = x + 5e x. Dervatve Prouct Rule If u a v are fferetable at x, te so s ter prouct uv, a u [uv] = x x v + uv x = [u ]v + u[v ]. Proof. By efto we ave: u(x + )v(x + ) u(x)v(x) [uv] x u(x + )v(x + ) u(x + )v(x) + u(x + )v(x) u(x)v(x) v(x + ) v(x) u(x + ) u(x) u(x + ) + v(x) v(x + ) v(x) u(x + ) lm = u(x)[v (x)] + [u (x)]v(x). + v(x) lm u(x + ) u(x) were lm u(x + ) = u(x) sce u s cotuous at x by Teorem 1 of secto 2.1. QED Example. Dfferetate f(x) = (4x 3 5x 2 + 4)(7x 2 x).

8 3.3 Dfferetato Rules 8 Dervatve Quotet Rule If u a v are fferetable at x a f v(x) 0, te te quotet u/v s fferetable at x, a x [ ] u v = u x v uv x v 2 = [u ]v u[v ] v 2. Proof. By efto we ave: x [ ] u v u(x+) v(x+) u(x) v(x) v(x)u(x + ) u(x)v(x + ) v(x + )v(x) v(x)u(x + ) v(x)u(x) + v(x)u(x) u(x)v(x + ) v(x + )v(x) v(x) u(x+) u(x) u(x) v(x+) v(x) x 0 v(x + )v(x) v(x) u(x+) u(x) = v(x) lm u(x+) u(x) = v(x)u (x) u(x)v (x). v 2 (x) lm u(x) v(x+) v(x) lm v(x + )v(x) u(x) lm v(x+) v(x) v(x) lm v(x + ) QED Example. Page 143 umber 20. Example. Page 145 umber 78, page 143 umber 48.

9 3.3 Dfferetato Rules 9 Note. We wll follow my square brackets otato as escrbe te aout. Note. We ca also calculate ger orer ervatves: y = x [y ], y = x [y ], y (4) = x [y ],..., y () = x [y( 1) ]. Example. Page 143 umber 42.