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

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1 MAT1A01: Differentiation of Polynomials & Exponential Functions + te Prouct & Quotient Rules Dr Craig 17 April 2013

2 Reminer Mats Learning Centre: C-Ring 512 My office: C-Ring 533A (Stats Dept corrior) Consulting ours: Monay 12pm-1pm, Wenesay 12pm-1pm Tel: Or, just rop in an see if I am tere. If I am free, I will elp.

3 Yesteray s class test Drawing te grap of a function f (x) IS NOT enoug to sow tat it is continuous on (, ). If you are going to apply te Intermeiate Value Teorem, you nee to state tat you are using it. You cannot simply calculate f (a) an f (b) an ave tat as your answer.

4 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) = lim x 0 = lim 0 c c = lim 0 0 = 0

5 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 x x = 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) = x 4 2 f (x) = 3x x 3 f (x) = 4x x

6 Te power rule: If n is a positive integer: x (x n ) = nx n 1 Proof: We will use te formula x n a n = (x a)(x n 1 + x n 2 a xa n 2 + a n 1 ) We will also use te fact tat f (a) = lim x a f (x) f (a) x a (see equation 5, Section 2.7, page 146)

7 Proof continue... f (a) = lim x a f (x) f (a) x a = lim x a x n a n x a = lim 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

8 Later, in Section 3.6, we will prove te power rule for n any real number: x (x n ) = nx n 1

9 Later, in Section 3.6, we will prove te power rule for n any real number: Example: f (x) = x x (x n ) = nx n 1

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

11 Te constant multiple rule: If c is a constant an f is a ifferentiable function, ten: x [cf (x)] = c x f (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

13 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 (3x 4 ) = 3 x (x 4 ) = 3(4x 3 ) = 12x 3

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

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

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

17 Example of te Sum Rule: Let f (x) = x 4 + x 2. Ten

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

19 Te ifference rule: If f an g are bot ifferentiable, ten: x [f (x) g(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).

20 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

21 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

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

23 Te erivative of an exponential function: Note tat x ax = a x a 1 lim 0 Q: Wat oes tis mean? = a x.f (0)

24 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.

25 Definition of te number e: Earlier we saw tat for f (x) = a x, f (0) = lim 0 a 1 We use tis to efine te number e as follows: e is te number suc tat lim 0 e 1 = 1 Ten we see tat te erivative of e x is: x (ex ) = e x

26 Te prouct rule: If f an g are bot ifferentiable, ten: x [f (x).g(x)] = f (x) x g(x) + g(x) x f (x) Example: 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 [x 2 e x ] = x 2.e x + e x.2x = e x (x 2 + 2x)

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

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

29 Examples of te prouct rule: Note tat x 3 = x(x 2 )

30 Examples of te prouct rule: Note tat x 3 = x(x 2 ) x x 3 = x x (x 2 ) + x 2 x (x) = x(2x) + (x 2 )(1)

31 Examples of te prouct rule: Note tat x 3 = x(x 2 ) x x 3 = x x (x 2 ) + x 2 x (x) = x(2x) + (x 2 )(1) = 2x 2 + x 2

32 Examples of te prouct rule: Note tat x 3 = x(x 2 ) x x 3 = x x (x 2 ) + x 2 x (x) = x(2x) + (x 2 )(1) = 2x 2 + x 2 = 3x 2

33 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:

34 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) ] = lim 0 f (x+) g(x+) f (x) g(x) = lim 0 g(x).f (x + ) f (x).g(x + ).g(x).g(x + )

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

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

37 Using te quotient rule: Consier Now y = ex x 2 y = (x 2 ) x (ex ) (e x ) x (x 2 ) [x 2 ] 2 = (x 2 )(e x ) (e x )(2x) x 4 = (ex )(x 2 2x) x 4

38 Derivatives of Trig Functions Before calculating te erivatives of te trig functions, we nee to prove two important limit calculations: sin θ lim = 1 θ 0 θ

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