MATH 12002 - CALCULUS I 2.4: Product and Quotient Rules Professor - - PowerPoint PPT Presentation

MATH 12002 - CALCULUS I §2.4: Product and Quotient Rules

Professor Donald L. White

Department of Mathematical Sciences Kent State University

D.L. White (Kent State University) 1 / 10

The Product Rule

Product Rule

If f (x) and g(x) are differentiable functions, then d dx (f (x) · g(x)) = d dx f (x)

• · g(x) + f (x) ·

d dx g(x)

• =

f ′(x)g(x) + f (x)g′(x). This can be written informally as (fg)′ = f ′g + fg′. Since both the addition and multiplication of functions are commutative, there are several equivalent ways to write this rule: (fg)′ = f ′g + fg′ = fg′ + f ′g = f ′g + g′f = g′f + f ′g = gf ′ + fg′ = fg′ + gf ′ = gf ′ + g′f = g′f + gf ′. They are all the same, so which one you use is up to you.

D.L. White (Kent State University) 2 / 10

Product Rule Examples

1 Find the derivative of F(x) = (x2 + 3x + 5)(6x3 − 7).

By the product rule (with f (x) = x2 + 3x + 5 and g(x) = 6x3 − 7), F ′(x) = d dx (x2 + 3x + 5)

• (6x3 − 7) + (x2 + 3x + 5)

d dx (6x3 − 7)

• =

(2x + 3)(6x3 − 7) + (x2 + 3x + 5)(18x2).

D.L. White (Kent State University) 3 / 10

Product Rule Examples

2 Find the derivative of F(x) = 2√x(3x 4 3 + 9x 7 9 ).

By the product rule (with f (x) = 2√x = 2x

1 2 and g(x) = 3x 4 3 + 9x 7 9 ),

F ′(x) = d dx (2√x) 3x

4 3 + 9x 7 9

• + (2√x)

d dx

• 3x

4 3 + 9x 7 9

• =
• 1

2 · 2x− 1

2

3x

4 3 + 9x 7 9

• + (2√x)
• 4

3 · 3x

1 3 + 7

9 · 9x− 2

9

• =
• x− 1

2

3x

4 3 + 9x 7 9

• + (2√x)
• 4x

1 3 + 7x− 2 9

• .

D.L. White (Kent State University) 4 / 10

Product Rule Examples

3 Find the derivative of F(x) = 3

√x(x2 + 5)(x5 + 2). We can compute this derivative using the product rule with f (x) =

3

√x(x2 + 5) and g(x) = x5 + 2, so that F ′(x) = d dx 3 √x(x2 + 5)

• (x5 + 2) +

3 √x(x2 + 5) d dx (x5 + 2)

• and then use the product rule to evaluate

d dx 3 √x(x2 + 5)

• .

But it is worth looking at the general situation of a product of 3 (or more) factors.

D.L. White (Kent State University) 5 / 10

Product Rule Examples

Let’s consider a product of three functions f (x)g(x)h(x). Using the product rule with f (x)g(x) as the first function and h(x) as the second, we have (fgh)′ = [(fg) · h]′ = (fg)′h + (fg)h′ = [f ′g + fg′]h + (fg)h′ = (f ′g)h + (fg′)h + (fg)h′ = f ′gh + fg′h + fgh′. So the derivative of a product of three functions is just a sum of three terms, each of which is a product of the derivative of one factor times the

• ther two factors.

Similarly, the derivative of a product f (x)g(x)h(x)k(x) of four factors is (fghk)′ = f ′ghk + fg′hk + fgh′k + fghk′. This pattern holds for any number of factors.

D.L. White (Kent State University) 6 / 10

Product Rule Examples

Hence the derivative of F(x) =

3

√x(x2 + 5)(x5 + 2) = x1/3(x2 + 5)(x5 + 2) is F ′(x) = d

dx ( 3

√x)

• (x2 + 5)(x5 + 2) + ( 3

√x) d

dx (x2 + 5)

• (x5 + 2)

+ ( 3 √x)(x2 + 5) d

dx (x5 + 2)

• =

1 3x− 2

3 (x2 + 5)(x5 + 2) + 3

√x(2x)(x5 + 2) +

3

√x(x2 + 5)(5x4).

D.L. White (Kent State University) 7 / 10

The Quotient Rule

Quotient Rule

If f (x) and g(x) are differentiable functions, then d dx f (x) g(x)

• =

d

dx f (x)

• · g(x) − f (x) ·

d

dx g(x)

• g(x)2

= f ′(x)g(x) − f (x)g′(x) g(x)2 . This can be written informally as

• f

g

′ = f ′g−fg′

g2

. Note that because of the subtraction in the numerator, we do not have as much freedom with the quotient rule as we did with the product rule. The first term can be written as f ′g or gf ′ and the second as fg′ or g′f , but the derivative f ′ of the numerator must be in the first term and the derivative g′ of the denominator must be in the second term.

D.L. White (Kent State University) 8 / 10

Quotient Rule Examples

1 Find the derivative of F(x) =

x3 + 5 2x7 − 5x−2 . By the quotient rule (with f (x) = x3 + 5 and g(x) = 2x7 − 5x−2), F ′(x) = d

dx (x3 + 5)

• (2x7 − 5x−2) − (x3 + 5)

d

dx (2x7 − 5x−2)

• (2x7 − 5x−2)2

= (3x2)(2x7 − 5x−2) − (x3 + 5)(14x6 + 10x−3) (2x7 − 5x−2)2 .

D.L. White (Kent State University) 9 / 10

Quotient Rule Examples

2 Find the derivative of F(x) =

√x 4x3 + 7. By the quotient rule (with f (x) = √x = x1/2 and g(x) = 4x3 + 7), F ′(x) = d

dx (√x)

• (4x3 + 7) − (√x)

d

dx (4x3 + 7)

• (4x3 + 7)2

= (1

2x−1/2)(4x3 + 7) − (√x)(12x2)

(4x3 + 7)2 .

D.L. White (Kent State University) 10 / 10