Math 229 Lecture Notes: Product and Quotient Rules

Math 229 Lecture Notes: Product and Quotient Rules Professor Richard Blecksmith [email protected]

1. Time Out for Notation Update It is awkward to say “the derivative of xn is nxn−1 ” Using
′ the prime notation for derivatives, we could write xn = nxn−1 ” But most mathematicians prefer to use an alternative notation due to the Leibnitz, one of the co-inventors of calculus, along with • Issac Newton, who favored the prime notation

• • • •

2. Leibnitz Notation • In the Leibnitz notation, if y = f (x), then the derivative of f (x) is dy written dx d y • or sometimes dx d • or even f (x) dx • Thus the power rule can be expressed as d n
x = nxn−1 dx 3. Which Notation should you use • As a rule of thumb, the prime notation is best when working with generic functions: f , g, etc. • The Leibnitz notation is best when working with letters which represent physical quantities. • For example, in the formula A = πr2 – A stands for area (of a circle) – r stands for radius – π is the number 3.1415926 . . . 1

2

dA • The formula = 2π r tells us dr • the area is increasing by a factor of 2π times the radius

4. Negative and Fractional Powers Many important function in algebra can be written as powers if we allow negative exponents or fractional exponents. For example 1 = x−1 x 1 = x−2 x2 √ x = x1/2

5. New Uses for the Power Rule The good news is that the power rule still holds true when the exponent is negative or a fraction. Thus, d −1 d 1 x = (−1)x−2 = dx x dx d1 d −2 = x = (−2)x−3 2 dx x dx 1 d √
d 1/2 x = x−1/2 x = dx dx 2 6. Summary of Rules d n x = nxn−1 dx d c=0 • Constant Rule: dx d • Constant Times Rule: c · f (x) = c · f ′ (x) dx

• Power Rule:

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• Addition Rule: i d h f (x) + g(x) = f ′ (x) + g ′ (x) dx i d h • Subtraction Rule: f (x) − g(x) = f ′ (x) − g ′ (x) dx

7. Polynomials Together these rules enable us to differentiate all polynomials. d 3 7x − 5x2 + 8x − 4 dx = 21x2 −10x +8

d 11x5 + 13x4 − 6x3 + 12x2 + x − 17 dx

= 55x4 +52x3 −18x2 +24x +1

8. Picture of FOIL = ac + ad + bc + bd

(a + b) · (c + d)

a

ac First

ad Outer

b

bc Inner

bd Last

c

d

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9. Impressing your Friends You can use FOIL to calculate products in your head, thereby impressing friends and strangers. Multiply By FOIL this product is

32 × 53

(30 + 2) × (50 + 3) = 30 × 50 +3 × 30 +2 × 50 +2 × 3 = 1500 +90 +100 +6 = 1696

10. Picture of Product Rule g(x + h)

∆g

g(x)

f (x) ∆g

∆f ∆g

f (x)g(x)

∆f g(x)

∆f

f (x)

f (x + h)

∆f = f (x + h) − f (x) ∆g = g(x + h) − g(x)

11. Numerical Example Suppose f (x) = 1000 and g(x) = 700 and we increase f (x) by ∆f = 2 and g(x) by ∆f = 3:

5

703 700

3

3 × 1000 = 3000

2×3=6

f (x)g(x) = 700, 000

2 × 700 = 1400

1000

2

1002

The area increases by 2 × 700 + 3 × 1000 + 6 = 1400 + 3000 + 6 = 4406

12. Proof of Product Rule     f (x + h)g(x + h) = f (x) + ∆f · g(x) + ∆g = f (x)g(x)+ ∆f g(x)+ f (x)∆g+ ∆f ∆g f (x + h)g(x + h) − f (x)g(x) = ∆f g(x) + f (x)∆g + ∆f ∆g f (x + h)g(x + h) − f (x)g(x) ∆f g(x) + f (x) ∆g + ∆f ∆g ∆f g(x) = = + h h h ∆f ∆g ∆f ∆g f (x) ∆g ∆f ∆g h = + g(x) + f (x) + ·h h h h h h h h

13. Conclusion of Proof ∆f f (x + h) − f (x) = lim = f ′ (x) h→0 h h→0 h ∆g g(x + h) − g(x) • lim = lim = g ′ (x) h→0 h h→0 h d f (x + h)g(x + h) − f (x)g(x) [f (x)g(x)] = lim • h→0 dx h   ∆g ∆f ∆g ∆f g(x) + f (x) + h = [lim h→0 h h h h • lim

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   ∆f ∆g = lim g(x) + f (x) lim +0 h→0 h h→0 h = f ′ (x)g(x) + f (x)g ′ (x)

14. The Product Rule

(f · g)′ = f ′ · g + f · g ′

In words this says: the derivative of a prdouct is the sum of the derivative of the first function times the second plus the first function times the derivative of the second. Notice that the constant times rule is a special case of the product rule and the fact that the derivative of a constant function is zero: (cf )′ = (c)′ f + c(f ′ ) = 0 · f + c · f ′ = c · f ′ 15. Example • • • • • • • •

f (x) = x5 − 6x2 g(x) = x3 + 10x f ′ (x) = 5x4 − 12x g ′ (x) = 3x2 + 10 (f g)′ (x) = f ′ (x)g(x) + f (x)g ′ (x) = (5x4 − 12x)(x3 + 10x) + (x5 − 6x2 )(3x2 + 10) = 5x7 + 50x5 − 12x4 − 120x2 +3x7 + 10x5 − 18x4 − 60x2 = 8x7 + 60x5 − 30x4 − 180x2 16. Muliply First

• • • • • • •

f (x) = x5 − 6x2 g(x) = x3 + 10x By the product rule we got (f g)′ (x) = 8x7 + 60x5 − 30x4 − 180x2 p(x) = f (x)g(x) = (x5 − 6x2 )(x3 + 10x) = x8 + 10x6 − 6x5 − 60x3 p′ (x) = 8x7 + 60x5 − 30x4 − 180x2 the same answer

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17. Warning • The product rule does NOT say that the derivative of a product is the product of the derivatives. • For example, what is wrong with the following derivation: d d d 2 x = x· x= 1·1 =1 ? dx dx dx • Answer: You are not using the product rule correctly. • The correct use of the product rule gives d  d 2 d  x = x ·x+x· x = x · 1 + 1 · x = 2x dx dx dx • the correct answer. 18. The Quotient Rule Suppose that f (x) and g(x) are two functions whose derivatives we know, and that we want to find the derivative of the quotient r(x) =

f (x) g(x)

Dropping the “(x)” we have that f =r·g Using the product rule, f ′ = r′ · g + r · g ′ Now solve for r′ r′ · g = f ′ − r · g ′ = f ′ − fg · g ′ 19. The Quotient Rule Continued r′ · g = f ′ −

f ′ ·g g

Multiply through by 1/g: 1  ′ f ′ f ′ f g′ r′ = − 2 f − ·g = g g g g

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=

f ′ g f g′ f ′ g − f g′ − = g2 g2 g2

This last formula is the quotient rule:  f ′ g

=

f ′ g − f g′ g2

20. Product vs Quotient Rule Product Rule: (f · g)′ = f ′ g + f g ′  f ′ f ′ g − f g ′ Quotient Rule: = g g2 Note that we can obtain the quotient rule from the product rule by • changing the + to a − in the numerator • dividing by g 2 Warning: It doesn’t matter if you reverse the terms in the product rule, but it does matter in the quotient rule.

21. Old Problem Revisited Compute the derivative of y = dy = dx =

d (1) dx

· (x2 ) − 1 · (x2 )2

d (x2 ) dx

1 using the quotient rule. x2

=

0 · (x2 ) − 1 · (2x) x4

−2 −2x = 3 4 x x

Compare this answer with the result from using the power rule.

22. Quotient Rule Example Compute the derivative of y =

x2 − 2 x3 + 1

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By the quotient rule,
d 2 x − 2 · (x3 + 1) − (x2 − 2) · dy = dx dx (x3 + 1)2
2x · (x3 + 1) − (x2 − 2) · (3x2 ) = (x3 + 1)2

d (x3 dx

+ 1)

2x4 + 2x − (3x4 − 6x2 ) = (x3 + 1)2 =

−x4 + 6x2 + 2x (x3 + 1)2

Aren’t you glad you don’t have to use the limit definition?