MAT1A01: Differentiation of Polynomials & Exponential Functions + the Product & Quotient Rules

Transcription

1 MAT1A01: Differentiation of Polynomials & Exponential Functions + te Prouct & Quotient Rules Dr Craig 22 Marc 2016

2 Semester Test 1 Results ave been publise on Blackboar uner My Graes. Scripts will be available for collection from Wenesay morning. Collect your script before te vacation so tat you can learn from your mistakes an improve your unerstaning. For marking queries: Q2 Q6: Dr Craig Q7 Q11: Ms Ricarson

3 Sick Test B-Les 102, Bring your stuent car.

4 Announcements: Ceck Blackboar at least twice a week Please ceck your stuent aress regularly, or set up forwaring to anoter account Remember: 8 10 ours of mats per week (outsie of lecture time) is wat is require if you in t acieve your goal in te Semester Test.

5 Announcements: As state in te Learning Guie, everyone wo score < 40% for Semester Test 1 will be require to atten compulsory Saturay morning classes. Anyone else wo wants to atten tese classes is welcome to join. Te compulsory classes will start from Saturay 09 April.

6 Derivatives of polynomials: Te simplest polynomial is te constant polynomial f(x) = c (c R) Using te limit efinition of te erivative: f(x + ) f(x) f(x) x 0 c c 0 0 = 0 0

7 Derivatives of power functions: Te slope of te function y = x is 1 at every value of x. Hence for f(x) = x we ave xx = 1. Using te efinition of te erivative of a function, one can calculate te following erivatives: f(x) = x 2 f(x) = 2x x f(x) = x 3 f(x) = 3x2 x f(x) = x 4 f(x) = 4x3 x

8 Te power rule: If n is a positive integer: x (xn ) = nx n 1 Proof: We will use a formula for x n a n. We will also use te fact tat f f(x) f(a) (a) x a x a (see equation 5, Section 2.7, page 146)

9 Proof continue... (x n a n ) = (x a)(x n 1 + x n 2 a xa n 2 + a n 1 ) f (a) x a f(x) f(a) x a x a x n a n x a x a (x n 1 + x n 2 a xa n 2 + a n 1 ) = a n 1 + a.a n a.a n 2 + a n 1 = n.a n 1

10 Tere is also a proof of x xn = n.x n 1 tat uses te Binomial Teorem (see page 175). Te Binomial Teorem allows us to easily expan (x + ) n for any n Z +.

11 Later, in Section 3.6, we will prove te power rule for n any real number: x (xn ) = nx n 1 Example: f(x) = x x f(x) = x = x x (x0.5 ) = (0.5)(x 0.5 ) = x = 1 2 x

12 Te constant multiple rule: If c is a constant an f is a ifferentiable function, ten: x [cf(x)] = c x f(x) Example: f(x) = 3x 4 x (f(x)) = x (3x4 ) = 3 x (x4 ) = 3(4x 3 ) = 12x 3

13 Te sum rule: If f an g are bot ifferentiable, ten: [f(x) + g(x)] = x x f(x) + x g(x) Proof: (f + g)(x + ) (f + g)(x) [f(x)+g(x)] x 0 [ ] [ ] f(x + ) + g(x + ) f(x) + g(x) 0 [ f(x + ) f(x) + 0 [ f(x + ) f(x) ] 0 g(x + ) g(x) ] + lim 0 [ g(x + ) g(x) ]

14 Example of te Sum Rule: Let f(x) = x 4 + x 2. Ten x f(x) = x (x4 + x 2 ) = x (x4 ) + x (x2 ) = 4x 3 + 2x

15 Te ifference rule: If f an g are bot ifferentiable, ten: [f(x) g(x)] = x x f(x) x g(x) Proof: Tis can be prove by combining f(x) g(x) = f(x) + ( 1)(g(x)) wit te Sum Rule an Constant Multiple Rule. Alternatively, it can be prove using a similar meto to tat use to prove te Sum Rule (omework).

16 Te Power Rule, Constant Rule, Sum Rule, an Difference Rule can be combine to fin te erivative of functions like te following: f(x) = 2x 4 x 3 5 x + 3x f (x) = 8x 3 3x x + 3

17 Example: fin te points on te curve y = x 4 6x were te tangent line is orizontal.

18 Te erivative of an exponential function: Let f(x) = a x. Ten f(x + ) f(x) f(x) x 0 a x+ a x 0 0 a x a a x 0 a x (a 1) = a x lim 0 a 1

19 Te erivative of an exponential function: Note tat x ax = a x a 1 lim 0 = a x.f (0) Q: Wat oes tis mean? A: Te rate of cange of te function at any point is proportional to te value of te function at tat point.

20 Definition of te number e: Earlier we saw tat for f(x) = a x, f (0) 0 a 1 Tere is a number a between 2 an 3 suc tat f (0) = 1. We use tis to efine te number e as follows: e is te number suc tat lim 0 e 1 = 1

21 Te erivative of e x is: x (ex ) = e x

22 Te prouct rule: If f an g are bot ifferentiable, ten: [f(x).g(x)] = f(x) g(x) + g(x) x x x f(x) Example: observe tat x 3 = x(x 2 ) [ x(x 2 ) ] = x x x (x2 ) + x 2 x (x) = x(2x) + (x 2 )(1) = 2x 2 + x 2 = 3x 2

23 Proof of te prouct rule: f(x + ).g(x + ) f(x).g(x) [f(x).g(x)] x 0 f(x+).g(x+) f(x+).g(x)+f(x+).g(x) f(x).g(x) 0 ( ) ( ) g(x+) g(x) +g(x) f(x+) f(x) 0 f(x+) ( ) g(x+) g(x) + g(x) ] [ f(x + ) g(x+) g(x) + lim 0 [ f(x + ) 0 [ 0 ( f(x+) f(x) )] g(x) f(x+) f(x) ]

24 Proof of te prouct rule continue... [ 0 f(x + ) g(x+) g(x) ] [ + lim 0 [ ] f(x + ) lim g(x+) g(x) lim 0 g(x) lim 0 = f(x).g (x) + g(x).f (x) g(x) f(x+) f(x) [ f(x + ) f(x) ] ]

25 Anoter example of te prouct rule: Suppose (x) = x 2 e x. Let f(x) = x 2 an g(x) = e x. Ten x (x) = x [f(x).g(x)] = x [x2 e x ] = x 2.e x + e x.2x = e x (x 2 + 2x)

26 Fin f (x) if f(x) = x.e x Fin f (t) if f(t) = t(a + bt) If f(x) = xg(x), wit g(4) = 2 an g (4) = 3, fin f (4).

27 Wat if we ave 3 factors? Let (x) = 2x 2.3x 4 x. We can write (x) as 6x 13/2 an get ( 2x 2.3x 4 x ) = ) (6x 13/2 = 39x 11/2 x x Write (x) = 2x 2 (3x 4 x) an ten: x ((x)) = 4x(3x4 ( x) + 2x 2 x 3x 4 x ) = 4x(3x 4 ( x) + 2x 2 12x 3 ) x + 3x4 2 x = 12x 5 x + 24x 5 x + 3x6 x = 12x 11/2 + 24x 11/2 + 3x 11/2 = 39x 11/2

28 Te quotient rule: If f an g are bot ifferentiable, ten: [ ] f(x) = g(x) x [f(x)] f(x) x [g(x)] x g(x) [g(x)] 2 Proof: x [ ] f(x) g(x) 0 f(x+) g(x+) f(x) g(x) 0 g(x).f(x + ) f(x).g(x + ).g(x).g(x + )

29 0 g(x).f(x + ) f(x).g(x + ).g(x).g(x + ) 0 g(x).f(x + ) f(x).g(x) + f(x).g(x) f(x).g(x + ).g(x).g(x + ) [ ] [ ] g(x) f(x + ) f(x) + f(x) g(x) g(x + ) 0.g(x).g(x + ) 0 [ ] [ g(x) f(x+) f(x) + f(x) g(x).g(x + ) ] g(x) g(x+)

30 0 [ ] [ g(x) f(x+) f(x) + f(x) g(x).g(x + ) ] g(x) g(x+) 0 = g(x) lim 0 [ ] [ g(x) f(x+) f(x) f(x) + lim 0 lim 0 g(x).g(x + ) [ ] f(x+) f(x) [ + f(x) lim 0 lim 0 g(x).g(x + ) = g(x).f (x) f(x).g (x) [ g(x)] 2 g(x) g(x+) ] ] g(x) g(x+)

31 Example: y = ex x 2 y = (x2 ) x (ex ) (e x ) x (x2 ) [x 2 ] 2 = (x2 )(e x ) (e x )(2x) x 4 = (ex )(x 2 2x) x 4