MAT01A1: Differentiation of Polynomials & Exponential Functions + te Prouct & Quotient Rules Dr Craig 22 Marc 2017
Semester Test 1 Scripts will be available for collection from Tursay morning. For marking queries: Q2 Q8: Dr Craig Q9 Q14: Ms Ricarson
Announcements: Ceck Blackboar at least twice a week. Please ceck your stuent email aress regularly, or set up forwaring to anoter email account. 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. Saturay class tis week: 10am to 1pm. Venue will be announce on Blackboar.
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
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) = 3x2 x f(x) = x 4 f(x) = 4x3 x
Te power rule: if n is a positive integer ten x (xn ) = nx n 1 Proof: We will use a formula for x n a n. We will also use te fact tat f (a) x a f(x) f(a) x a (see equation 5, Section 2.7, page 146)
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 2 +... + a.a n 2 + a n 1 = n.a n 1
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 +.
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 ) = 1 2. 1 x = 1 2 x
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
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) ]
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
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).
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 2 5 2 x + 3
Example: fin te points on te curve y = x 4 6x 2 + 4 were te tangent line is orizontal.
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
Te erivative of an exponential function: Note tat an ence a 1 lim 0 = f (0) x ax = a x a 1 lim = a x.f (0) 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.
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
Te erivative of e x is: x (ex ) = e x
Te proofs of bot of te ifferentiation rules tat follow (te Prouct Rule an te Quotient Rule) will be in te scope of bot te secon Semester Test an te final exam. PDFs of tese proofs ave been poste on Blackboar. We believe tese proofs to be easier to unerstan from our limit-base approac to erivatives tan te proofs in te textbook.
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
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) ]
Proof of te prouct rule continue... [ 0 f(x + ) g(x+) g(x) ] [ + lim 0 [ ] f(x + ) lim g(x+) g(x) 0 0 + lim 0 g(x) lim 0 = f(x).g (x) + g(x).f (x) g(x) f(x+) f(x) [ f(x + ) f(x) ] ]
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)
Fin f (x) if f(x) = x.e x Fin f (t) if f(t) = t(a + bt) If f(x) = g(x) x, wit g(4) = 2 an g (4) = 3, fin f (4).
Wat if we ave 3 factors? Let (x) = 2x 2.3x 4 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 We coul also write (x) as 6x 13/2 an get ( 2x 2.3x 4 x ) = ) (6x 13/2 = 39x 11/2 x x
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 + )
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+)
0 [ ] [ g(x) f(x+) f(x) + f(x) g(x).g(x + ) ] g(x) g(x+) 0 [ ] [ g(x) f(x+) f(x) f(x) + lim 0 lim 0 g(x).g(x + ) g(x) g(x+) ] = g(x) lim 0 [ ] f(x+) f(x) [ + f(x) lim 0 lim 0 g(x).g(x + ) ] g(x) g(x+) = g(x).f (x) f(x).g (x) [ g(x)] 2
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