Mat 229 Lecture Notes: Prouct an Quotient Rules Professor Ricar Blecksmit ricar@mat.niu.eu 1. Time Out for Notation Upate It is awkwar to say te erivative of x n is nx n 1 Using te prime notation for erivatives, we coul write ( x n) nx n 1 But most matematicians prefer to use an alternative notation ue to te Leibnitz, one of te co-inventors of calculus, along wit Issac Newton, wo favore te prime notation 2. Leibnitz Notation In te Leibnitz notation, if y f(x), ten te erivative of f(x) is y written x or sometimes x y or even x f(x) Tus te power rule can be expresse as ( ) x n nx n 1 x 3. Wic Notation soul you use As a rule of tumb, te prime notation is best wen working wit generic functions: f, g, etc. Te Leibnitz notation is best wen working wit letters wic represent pysical quantities. For example, in te formula A πr 2 A stans for area (of a circle) r stans for raius π is te number 3.1415926... 1
2 A Te formula 2πr tells us r te area is increasing by a factor of 2π times te raius 4. Negative an Fractional Powers Many important function in algebra can be written as powers if we allow negative exponents or fractional exponents. For example 1 x x 1 1 x 2 x 2 x x 1/2 5. New Uses for te Power Rule Te goo news is tat te power rule still ols true wen te exponent is negative or a fraction. Tus, 1 ) x( x x x 1 ( 1)x 2 ( 1 ) x x 2 x x 2 ( 2)x 3 ( ) x x x x1/2 1 2 x 1/2 6. Summary of Rules Power Rule: x xn nx n 1 Constant Rule: x c 0 Constant Times Rule: x c f(x) c f (x)
3 Aition Rule: [ ] f(x)+g(x) f (x)+g (x) x [ ] Subtraction Rule: f(x) g(x) f (x) g (x) x 7. Polynomials Togeter tese rules enable us to ifferentiate all polynomials. x 7x3 5x 2 +8x 4 21x 2 10x +8 x 11x5 +13x 4 6x 3 +12x 2 +x 17 55x 4 +52x 3 18x 2 +24x +1 8. Picture of FOIL (a+b) (c+) ac +a +bc +b a ac First a Outer b bc Inner b Last c
4 9. Impressing your Friens You can use FOIL to calculate proucts in your ea, tereby impressing friens an strangers. Multiply By FOIL tis prouct is 32 53 (30+2) (50+3) 30 50 +3 30 +2 50 +2 3 1500 +90 +100 +6 1696 g(x+) g(x) g 10. Picture of Prouct Rule f(x) g f g f(x)g(x) f g(x) f(x) f f(x+) f f(x+) f(x) g g(x+) g(x) 11. Numerical Example Suppose f(x) 1000 an g(x) 700 an we increase f(x) by f 2 an g(x) by f 3:
5 703 3 700 3 1000 3000 2 3 6 f(x)g(x) 700,000 2 700 1400 1000 2 1002 Te area increases by 2 700+3 1000+6 1400+3000+6 4406 12. Proof of Prouct Rule f(x+)g(x+) [ f(x)+ f ] [g(x)+ g ] f(x)g(x)+ f g(x)+ f(x) g+ f g f(x+)g(x+) f(x)g(x) f g(x)+f(x) g + f g f(x+)g(x+) f(x)g(x) f g(x)+f(x) g + f g f(x) g + f g f g(x)+f(x) g + f g f g(x) + 13. Conclusion of Proof f lim 0 lim f(x+) f(x) 0 g(x+) g(x) g lim 0 lim 0 x [lim 0 f (x) g (x) f(x+)g(x+) f(x)g(x) [f(x)g(x)] lim [ 0 f g(x)+f(x) g + f ] g
6 [ f lim 0 f (x)g(x)+f(x)g (x) ] [ g(x)+f(x) lim 0 ] g +0 14. Te Prouct Rule (f g) f g +f g In wors tis says: te erivative of a prouct is te sum of te erivative of te first function times te secon plus te first function times te erivative of te secon. Notice tat te constant times rule is a special case of te prouct rule an te fact tat te erivative of a constant function is zero: (cf) (c) f +c(f ) 0 f +c f c f 15. Example f(x) x 5 6x 2 g(x) x 3 +10x f (x) 5x 4 12x g (x) 3x 2 +10 (fg) (x) f (x)g(x)+f(x)g (x) (5x 4 12x)(x 3 +10x)+(x 5 6x 2 )(3x 2 +10) 5x 7 +50x 5 12x 4 120x 2 +3x 7 +10x 5 18x 4 60x 2 8x 7 +60x 5 30x 4 180x 2 16. Muliply First f(x) x 5 6x 2 g(x) x 3 +10x By te prouct rule we got (fg) (x) 8x 7 +60x 5 30x 4 180x 2 p(x) f(x)g(x) (x 5 6x 2 )(x 3 +10x) x 8 +10x 6 6x 5 60x 3 p (x) 8x 7 +60x 5 30x 4 180x 2 te same answer
7 17. Warning Te prouct rule oes NOT say tat te erivative of a prouct is te prouct of te erivatives. For example, wat is wrong wit te following erivation: x x2 x x x x 1 1 1? Answer: You are not using te prouct rule correctly. Te correct use of te prouct rule gives ( ) x x2 x x x+x te correct answer. ( x x ) x 1+1 x 2x 18. Te Quotient Rule Suppose tat f(x) an g(x) are two functions wose erivatives we know, an tat we want to fin te erivative of te quotient r(x) f(x) g(x) Dropping te (x) we ave tat f r g Using te prouct rule, f r g +r g Now solve for r r g f r g f f g g 19. Te Quotient Rule Continue r g f f g g Multiply troug by 1/g: r 1 (f f ) g g g f g f g g 2
8 f g g f g f g f g 2 g 2 g 2 Tis last formula is te quotient rule: ( f g) f g f g g 2 Prouct Rule: (f g) f g +fg Quotient Rule: 20. Prouct vs Quotient Rule ( f g) f g f g g 2 Note tat we can obtain te quotient rule from te prouct rule by canging te + to a in te numerator iviing by g 2 Warning: It oesn t matter if you reverse te terms in te prouct rule, but it oes matter in te quotient rule. 21. Ol Problem Revisite Compute te erivative of y 1 using te quotient rule. x2 y x x (1) (x2 ) 1 x (x2 ) 0 (x2 ) 1 (2x) (x 2 ) 2 x 4 2x 2 x 4 x 3 Compare tis answer wit te result from using te power rule. 22. Quotient Rule Example Compute te erivative of y x2 2 x 3 +1
9 By te quotient rule, ( y x x x 2 2 ) (x3 +1) (x 2 2) x (x3 +1) (x 3 +1) 2 ( 2x ) (x 3 +1) (x 2 2) (3x 2 ) (x 3 +1) 2 2x4 +2x (3x 4 6x 2 ) (x 3 +1) 2 x4 +6x 2 +2x (x 3 +1) 2 Aren t you gla you on t ave to use te limit efinition?