The quotient rule for differentation E. Kim 1 Warm up Use the - - PowerPoint PPT Presentation

The quotient rule for differentation

• E. Kim

1

Warm up

Use the product rule1 to find the derivative of h(x) = (5x3 + 9x)(xex + 7x2)

1In words, the derivative of a product is the derivative of the first factor

times the second, plus the first factor times the derivative of the second.

2

The Quotient Rule

Question

If k(x) = f(x)

g(x), what is k′(x)?

3

The Quotient Rule

Question

If k(x) = f(x)

g(x), what is k′(x)? ◮ Mult by g(x) on both sides:

3

The Quotient Rule

Question

If k(x) = f(x)

g(x), what is k′(x)? ◮ Mult by g(x) on both sides: k(x) g(x) = f(x)

3

The Quotient Rule

Question

If k(x) = f(x)

g(x), what is k′(x)? ◮ Mult by g(x) on both sides: k(x) g(x) = f(x) ◮ Differentiate each side:

3

The Quotient Rule

Question

If k(x) = f(x)

g(x), what is k′(x)? ◮ Mult by g(x) on both sides: k(x) g(x) = f(x) ◮ Differentiate each side: k′(x) g(x) + k(x) g′(x) = f′(x)

3

The Quotient Rule

Question

If k(x) = f(x)

g(x), what is k′(x)? ◮ Mult by g(x) on both sides: k(x) g(x) = f(x) ◮ Differentiate each side: k′(x) g(x) + k(x) g′(x) = f′(x) ◮ Replace k(x) as above:

3

The Quotient Rule

Question

If k(x) = f(x)

g(x), what is k′(x)? ◮ Mult by g(x) on both sides: k(x) g(x) = f(x) ◮ Differentiate each side: k′(x) g(x) + k(x) g′(x) = f′(x) ◮ Replace k(x) as above: k′(x) g(x) + f(x) g(x) g′(x) = f′(x)

3

The Quotient Rule

Question

If k(x) = f(x)

g(x), what is k′(x)? ◮ Mult by g(x) on both sides: k(x) g(x) = f(x) ◮ Differentiate each side: k′(x) g(x) + k(x) g′(x) = f′(x) ◮ Replace k(x) as above: k′(x) g(x) + f(x) g(x) g′(x) = f′(x) ◮ Rewrite fraction:

3

The Quotient Rule

Question

If k(x) = f(x)

g(x), what is k′(x)? ◮ Mult by g(x) on both sides: k(x) g(x) = f(x) ◮ Differentiate each side: k′(x) g(x) + k(x) g′(x) = f′(x) ◮ Replace k(x) as above: k′(x) g(x) + f(x) g(x) g′(x) = f′(x) ◮ Rewrite fraction: k′(x) g(x) + f(x) g′(x) g(x)

= g(x) f′(x)

g(x)

3

The Quotient Rule

Question

If k(x) = f(x)

g(x), what is k′(x)? ◮ Mult by g(x) on both sides: k(x) g(x) = f(x) ◮ Differentiate each side: k′(x) g(x) + k(x) g′(x) = f′(x) ◮ Replace k(x) as above: k′(x) g(x) + f(x) g(x) g′(x) = f′(x) ◮ Rewrite fraction: k′(x) g(x) + f(x) g′(x) g(x)

= g(x) f′(x)

g(x) ◮ Subtract on both sides:

3

The Quotient Rule

Question

If k(x) = f(x)

g(x), what is k′(x)? ◮ Mult by g(x) on both sides: k(x) g(x) = f(x) ◮ Differentiate each side: k′(x) g(x) + k(x) g′(x) = f′(x) ◮ Replace k(x) as above: k′(x) g(x) + f(x) g(x) g′(x) = f′(x) ◮ Rewrite fraction: k′(x) g(x) + f(x) g′(x) g(x)

= g(x) f′(x)

g(x) ◮ Subtract on both sides: k′(x) g(x) = g(x) f′(x)−f(x) g′(x) g(x)

3

The Quotient Rule

Question

If k(x) = f(x)

g(x), what is k′(x)? ◮ Mult by g(x) on both sides: k(x) g(x) = f(x) ◮ Differentiate each side: k′(x) g(x) + k(x) g′(x) = f′(x) ◮ Replace k(x) as above: k′(x) g(x) + f(x) g(x) g′(x) = f′(x) ◮ Rewrite fraction: k′(x) g(x) + f(x) g′(x) g(x)

= g(x) f′(x)

g(x) ◮ Subtract on both sides: k′(x) g(x) = g(x) f′(x)−f(x) g′(x) g(x) ◮ Divide on both sides:

3

The Quotient Rule

Question

If k(x) = f(x)

g(x), what is k′(x)? ◮ Mult by g(x) on both sides: k(x) g(x) = f(x) ◮ Differentiate each side: k′(x) g(x) + k(x) g′(x) = f′(x) ◮ Replace k(x) as above: k′(x) g(x) + f(x) g(x) g′(x) = f′(x) ◮ Rewrite fraction: k′(x) g(x) + f(x) g′(x) g(x)

= g(x) f′(x)

g(x) ◮ Subtract on both sides: k′(x) g(x) = g(x) f′(x)−f(x) g′(x) g(x) ◮ Divide on both sides: k′(x) = g(x) f′(x)−f(x) g′(x) [g(x)]2

3

The Quotient Rule

Question

If k(x) = f(x)

g(x), what is k′(x)? ◮ Mult by g(x) on both sides: k(x) g(x) = f(x) ◮ Differentiate each side: k′(x) g(x) + k(x) g′(x) = f′(x) ◮ Replace k(x) as above: k′(x) g(x) + f(x) g(x) g′(x) = f′(x) ◮ Rewrite fraction: k′(x) g(x) + f(x) g′(x) g(x)

= g(x) f′(x)

g(x) ◮ Subtract on both sides: k′(x) g(x) = g(x) f′(x)−f(x) g′(x) g(x) ◮ Divide on both sides: k′(x) = g(x) f′(x)−f(x) g′(x) [g(x)]2

The Quotient Rule

f(x) g(x) ′ = g(x) f′(x) − f(x) g′(x) [g(x)]2

3

The Quotient Rule

... in Newton notation...

f(x) g(x) ′ = g(x) f′(x) − f(x) g′(x) [g(x)]2

... in Leibnitz notation...

d dx f(x) g(x)

• = g(x) d

dx[f(x)] − f(x) d dx[g(x)]

[g(x)]2 In the upcoming examples, I’ve tried to label the steps which are just algebra with the text “alg”.

4

Example [B]

Exercise: If k(x) = x3 − 1 x8 + 1, find k′(x).

5

Example [B]

Exercise: If k(x) = x3 − 1 x8 + 1, find k′(x). Solution: f(x) = x3 − 1 and g(x) = x8 + 1 k′(x) = (x8 + 1)(x3 − 1)′ − (x3 − 1)(x8 + 1)′ (x8 + 1)2 = (x8 + 1)(3x2 − 0) − (x3 − 1)(8x7 + 0) (x8 + 1)2

alg

= (x8 + 1)(3x2) − (x3 − 1)(8x7) (x8 + 1)2

alg

= (3x10 + 3x2) − (8x10 − 8x7) x16 + 2x8 + 1

alg

= −5x10 + 3x2 + 8x7 x16 + 2x8 + 1

5

Example [B]

Exercise: If z(x) = e−x, find dz dx.

6

Example [B]

Exercise: If z(x) = e−x, find dz dx. Solution: z(x) = 1 ex , so use f(x) = 1 and g(x) = ex. dz dx = ex d

dx[1] − 1 d dx[ex]

(ex)2 = ex · 0 − 1 · ex (ex)2

alg

= −ex (ex)2

alg

= −1 ex

In the future...

We’ll gain a NEW way to find the derivative of z(x) = e−x

6

Two Mnemonics

f(x) g(x) ′ = g(x) f′(x) − f(x) g′(x) [g(x)]2

7

Two Mnemonics

f(x) g(x) ′ = g(x) f′(x) − f(x) g′(x) [g(x)]2

Hi over Lo

hi lo ′

7

Two Mnemonics

f(x) g(x) ′ = g(x) f′(x) − f(x) g′(x) [g(x)]2

Hi over Lo

hi lo ′ = lo d(hi) − hi d(lo) lo lo

7

Two Mnemonics

f(x) g(x) ′ = g(x) f′(x) − f(x) g′(x) [g(x)]2

Hi over Lo

hi lo ′ = lo d(hi) − hi d(lo) lo lo

Hi over Ho

hi ho ′

7

Two Mnemonics

f(x) g(x) ′ = g(x) f′(x) − f(x) g′(x) [g(x)]2

Hi over Lo

hi lo ′ = lo d(hi) − hi d(lo) lo lo

Hi over Ho

hi ho ′ = ho d(hi) − hi d(ho) ho ho

7

Example [B]

Exercise: If f(x) = ex + x5 1 + 2x3 , find f′(x).

8

Example [B]

Exercise: If f(x) = ex + x5 1 + 2x3 , find f′(x). Solution: f′(x) = (1 + 2x3) · (ex + x5)′ − (ex + x5) · (1 + 2x3)′ (1 + 2x3)2 = (1 + 2x3) · (ex + 5x4) − (ex + x5) · (0 + 6x2) (1 + 2x3)2

alg

= (1 + 2x3) · (ex + 5x4) − (ex + x5) · (6x2) (1 + 2x3)2

alg

= (ex + 5x4 + 2x3ex + 10x7) − (6x2ex + 6x7) 1 + 4x3 + 4x6

8

Example, on your own... [B]

Exercise: If y = (x − 1)(x2 − 2x) x4 , compute dy dx.

9

Example, on your own... [B]

Exercise: If y = (x − 1)(x2 − 2x) x4 , compute dy dx. Solution: dy dx = [x4] · d

dx[(x − 1)(x2 − 2x)] − [(x − 1)(x2 − 2x)] · d dx[x4]

[x4]2

PR

= [x4][(1)(x2 − 2x) + (x − 1)(2x − 2)] − [(x − 1)(x2 − 2x)][4x3] [x4]2

alg

= [x4][(x2 − 2x) + (2x2 − 4x + 2)] − [x3 − 3x2 + 2x][4x3] [x4]2

alg

= [x4][3x2 − 6x + 2] − [x3 − 3x2 + 2x][4x3] [x4]2

alg

= [3x6 − 6x5 + 2x4] − [4x6 − 12x5 + 8x4] x8

alg

= −x6 + 6x5 − 6x4 x8

alg

= −x−2 + 6x−3 − 6x−4

9

Revisit the same example

Exercise: If y = (x − 1)(x2 − 2x) x4 , compute dy dx.

2no calculus yet... 10

Revisit the same example

Exercise: If y = (x − 1)(x2 − 2x) x4 , compute dy dx. Second Solution:

◮ First FOIL out numerator: y = x3 − 3x2 + 2x

x4

◮ More algebra2: y = x−1 − 3x−2 + 2x−3. ◮ Use Sum/Difference Rule and Power Rule to take derivative:

dy dx = (−1)x−2 − 3 · (−2)x−3 + 2 · (−3)x−4 which simplifies to

alg

= −x−2 + 6x−3 − 6x−4

Same example, two methods

Avoiding the Quotient Rule was MUCH FASTER!!!

2no calculus yet... 10

11

Example [B]

Exercise: If f(x) = x3 + 2x 4 , find f′(x).

12

Example [B]

Exercise: If f(x) = x3 + 2x 4 , find f′(x). Slow Solution: f′(x) = 4 · (x3 + 2x)′ − (x3 + 2x)(4)′ (4)2 = 4(3x2 + 2) − (x3 + 2x)(0) 16 Faster Solution: Use the Product Rule on f(x) = 1

4(x3 + 2x) ◮ f′(x) = 1 4(x3 +2x)′ +( 1 4)′(x3 +2x) = 1 4(3x2 +2)+0(x3 +2x)

Fastest Solution: Constant Multiple Rule is faster than using the Quotient Rule (or even the Product Rule) f′(x) = 1 4(x3 + 2x)′ = 1 4(3x2 + 2)

12

Example [B]

Exercise: If f(x) = x3 + 2x 4 , find f′(x). Slow Solution: f′(x) = 4 · (x3 + 2x)′ − (x3 + 2x)(4)′ (4)2 = 4(3x2 + 2) − (x3 + 2x)(0) 16 Faster Solution: Use the Product Rule on f(x) = 1

4(x3 + 2x) ◮ f′(x) = 1 4(x3 +2x)′ +( 1 4)′(x3 +2x) = 1 4(3x2 +2)+0(x3 +2x)

Fastest Solution: Constant Multiple Rule is faster than using the Quotient Rule (or even the Product Rule) f′(x) = 1 4(x3 + 2x)′ = 1 4(3x2 + 2)

12

Example [B]

Exercise: If f(x) = x2/5 + ex 10x6 − 4ex , compute d f dx.

13

Example [B]

Exercise: If f(x) = x2/5 + ex 10x6 − 4ex , compute d f dx. Solution: For some problems, though, the Quotient Rule really is the best strategy. d f dx = [10x6 − 4ex] d

dx[x2/5 + ex] − [x2/5 + ex] d dx[10x6 − 4ex]

[10x6 − 4ex]2 = [10x6 − 4ex][ 2

5x−3/5 + ex] − [x2/5 + ex][60x5 − 4ex]

[10x6 − 4ex]2

13