MAT 1339-S14 Class 2

Transcription

1 MAT 1339-S14 Class 2 July 07, 2014 Contents 1 Rate of Cange Introduction to Derivatives Derivatives Derivative of Polynomial function Te Product Rule Te Cain Rule Quotient Rule Rate of Cange 1.5 Introduction to Derivatives Definition 1.1 (Definition of te Derivative). Te derivative of a function y f(x) is a new function denoted by Wic is defined by f (x) or y or dy dx or d dx f(x), f(x + ) f(x) f (x), if te it exists. Wic represents te slope of te tangent, or instantaneous rate of cange, at any point on te curve of y f(x). Differentiation is te process (operation) of finding a derivative. Te derivative of a function f(x) at x a is f(a + ) f(a) f (a), if te it exists. If te it exists, we say f is differentiable at x a. 1

2 Example 1.2. Use te definition of te derivative to find te derivative of te following functions: (a) f(x) x (b) f(x) x 2 (c) f(x) x 3 f (x) f(x + ) f(x) (x + ) 3 x 3 using f(x) x 3 x 3 + 3x 2 + 3x x 3 3x 2 + 3x x 2 + 3x + 2 divide by because 0 3x 2. (d) f(x) 1 x, connect te formula of f wit te grap of f? 2

3 Conjecture: If f(x) x n, ten f (x) nx n 1. See te Power Rules page 77 Example 1.3. Let y f(t) 4.9t t + 1 by using te definition of te derivative find: (a) Instantaneous rate of cange at t 1 and t 1.5 (b) Slope of te tangent line at t 1 and t 1.5 Solution: To solve te equestion we need to calculate f (t) in general. Using te definition of te derivative f (t) f(t + ) f(t) [ 4.9(t + ) (t + ) + 1] ( 4.9t t + 1) t + 10 simplify te numerator t + 10 divide by because 0 9.8t

4 Teorem 1.4 (Differentiability Implies Continuity). If f is differentiable at x a ten f is continuous at x a. Proof: Example 1.5 (Continuity Doesn t Imply Differentiability). { 1 x + 2, if x > 2 f(x) 2 x + 5, if x 2. Is f continuous at x 2? find f (2), if exists? explain. 4

5 2 Derivatives 2.1 Derivative of Polynomial function Te goal of tis capter is to calculate derivatives efficiently. First we want to tackle te derivative of a polynomial function. Teorem 2.1. Let f(x) a n x n + a n 1 x n a 1 x + a 0 be a polynomial function, ten te derivative of f(x) is f (x) na n x n 1 + (n 1)a n 1 x n a 1. To prove tis teorem, we need te following rules. Constant Rule If f(x) c, were c is a constant, ten f (x) 0. Proof. f (x) f(x + ) f(x) Example 2.2. If f(x) 100, ten f (x) If g(x) 78, ten g (x) If (x) 25, ten (x) 5

6 Power Rule If f(x) x n, ten f (x) nx n 1. Proof: See page 77 Example 2.3. If f(x) x 100, ten f (x) If g(x) x 78, ten g (x) If (x) 1 x 25, ten (x) If f(x) 5 x 2, ten f (x) Note: If you are not familiar wit te exponent laws, it s your please review Prerequisite Skills Questions 3 and 4 (p. 70). Sum Rule and Difference Rule If H(x) f(x) + g(x), ten H (x) f (x) + g (x). Proof. H (x) H(x + ) H(x) If H(x) f(x) g(x), ten H (x) f (x) g (x). Proof. Exercise (similar to te proof of te sum rule). Example 2.4. If f(x) x x 78 x 25, ten f (x) 100x x 77 ( 25x 26 ) 100x x x 26. 6

7 Constant Multiple Rule If f(x) cg(x), were c is a constant, ten f (x) cg (x). Proof. f (x) f(x + ) f(x) cg(x + ) cg(x) c(g(x + ) g(x)) g(x + ) g(x) c cg (x). Example 2.5. If f(x) x x 22, ten f (x) 2009x (9)(22)x x x 21. If f(x) (2x + 3)(1 x) 2x 2x x 2x 2 x + 3, ten 2.2 Te Product Rule Teorem 2.6. f (x) 2(2x) 1 4x 1. If P (x) f(x)g(x), ten P (x) f (x)g(x) + f(x)g (x). Proof. P (x) P (x + ) P (x) f(x + )g(x + ) f(x)g(x) f(x + )g(x + ) f(x)g(x + ) + f(x)g(x + ) f(x)g(x) (f(x + ) f(x))g(x + ) + f(x)(g(x + ) g(x)) f(x)(g(x + ) g(x)) (f(x + ) f(x))g(x + ) + f(x + ) f(x) f (x)g(x) + f(x)g (x). g(x + ) + f(x) 0 g(x + ) g(x) 7

8 Example 2.7. Find f (x) for te following functions using te product rule. (a) f(x) (2x + 3)(1 x), ten f (x) (2x + 3) (1 x) + (2x + 3)(1 x) 2(1 x) + (2x + 3)( 1) 2 2x 2x 3 4x 1. (b) f(x) (x 2 + 2)(4x 5). Example 2.8. If y (x + 1 x )( x 1 x 2 + 1), ten y ( x + 1 ) ( ) ( x 1 x x x + 1 ) ( x 1 2 x x + 1 ( ) ( ) ( x 1 x 2 x x + 1 ) ( 1 2 x 2 x + 2 ). x 3 ) 2.4 Te Cain Rule Te cain rule is used to calculate te derivative of a composite function. Let s first review wat are composite functions by examples. Example 2.9. Let f(x) x and g(x) 1 x 2. Ten f g(x) f(g(x)) ( ) (g(x)) x 2 (x 2) g f(x) g(f(x)) 1 f(x) 2 1 (x 3 + 1) 2 1 x 3 1. Teorem If C(x) f(g(x)), ten C (x) f (g(x))g (x). We first illustrate ow to use te cain rule troug examples, ten we will prove te teorem. Example Let y (x 2 + 1) Ten 8

9 Example Let y 5x + 3 (5x + 3) 1 2. Ten Example Let y (5x 3 + 6x 2 ) 4. Ten From tese tree examples, we can summarize tat wen using te cain rule tere are basically tree steps: 1- find te derivative of te outside function 2- Inside function left uncanged (Leave it alone) 3- multiply te derivative of te inside function Questions combining te cain rule and te product rule can become more difficult. Here are two example. Example Let y 3x(1 x) 2. Ten y (3x) (1 x) 2 + 3x((1 x) 2 ) product rule 3(1 x) 2 + 3x(2(1 x)(1 x) ) cain rule 3(1 x) 2 + 3x(2(1 x))( 1)). Example Let y x 2 (x 3 + 5x) 10. Ten y (x 2 ) (x 3 + 5x) 10 + x 2 ((x 3 + 5x) 10 ) product rule 2x(x 3 + 5x) 10 + x 2 (10(x 3 + 5x) 9 (x 3 + 5x) ) cain rule 2x(x 3 + 5x) 10 + x 2 (10(x 3 + 5x) 9 (3x 2 + 5)). 2.5 Quotient Rule Te quotient rule is used to calculate te derivative of a quotient of two functions f(x) g(x). It is actually very easy because it is a special case of combining te cain rule and te product rule. Teorem If Q(x) f(x) g(x), ten Q (x) f (x)g(x) f(x)g (x) (g(x)) 2. 9

10 Proof. We can write Q(x) as a product of two functions So Q(x) f(x) g(x) f(x)((g(x)) 1 ). Q (x) f (x)(g(x)) 1 + f(x)((g(x)) 1 ) product rule f (x)(g(x)) 1 + f(x)(( 1)(g(x)) 2 g (x)) cain rule f (x) g(x) f(x)g (x) (g(x)) 2 f (x)g(x) f(x)g (x) (g(x)) 2. Example Let ten y 2x5 + 6x 3 + 4, x 4 2x y (2x5 + 6x 3 + 4) (x 4 2x) (2x 5 + 6x 3 + 4)(x 4 2x) (x 4 2x) 2 (10x4 + 18x 2 )(x 4 2x) (2x 5 + 6x 3 + 4)(4x 3 2) (x 4 2x) 2 Of course, we can also combine te cain rule and te quotient rule wen f(x) or g(x) (or bot) is a composite function. Example Let y 2x5 + 6x (3x 2 + 1) 6, ten y (2x5 + 6x 3 + 4) ( 19 (3x 2 + 1) 6 ) (2x 5 + 6x 3 + 4)( 19 (3x 2 + 1) 6 ) ( 19 (3x 2 + 1) 6 ) 2 quotient rule (10x4 + 18x 2 )( 19 (3x 2 + 1) 6 ) (2x 5 + 6x 3 + 4)( 6 19 (3x2 + 1) (6x)) ( 19. cain rule (3x 2 + 1) 6 ) 2 Note: From tis example, we can see tat it is easier if we first calculate te derivative of 19 (3x 2 + 1) 6 separately (as we did in Example 1) and use it later in te quotient rule. Tis elps to avoid unnecessary mistakes. 10

11 Example (!) Were Functions Aren t, Continuous? wy? and were are not Differentiable? wy? 1, if x < 0 x x 2 f(x) (1 x), if 0 x 2 2 (x 3) 1 3, if x > 2. 11