Do NowFind the tangents to the curve at the points wherethe slope is 4. What is the smallest slope of the curve? At whatvalue of x does the curve have this slope?

3y x x

{

Product and Quotient Rules

Section 3.3b

Do NowFind the tangents to the curve at the points wherethe slope is 4. What is the smallest slope of the curve? At whatvalue of x does the curve have this slope?

3y x x

The derivative: 23 1y x x Find where the slope is 4:

23 1 4x 23 3 0x

3 1 1 0x x 1x

Slope 4, points (1,2) and (–1,–2).

Tangent lines: 4 2y x 4 2y x

For smallest slope, minimize

23 1y x x The smallest slope is 1,and occurs at x = 0.Graphical support???

As we learned last class, the derivative of the sum of twofunctions is the sum of their derivatives (and the same holdstrue for differences of functions). Is there a similar rule forthe product of two functions?

2f x xLet

2 2d dx x x x

dx dx The derivative:

11 1d dx x

dx dx However,

We need to derive a new rule for products…

f x u x v x Let

0

limh

u x h v x h u x v xduv

dx h

The derivative:

0

limh

u x h v x h u x h v x u x h v x u x v x

h

Subtract and add u(x + h)v(x) in the numerator:

0

limh

v x h v x u x h u xu x h v x

h h

0 0 0

lim lim limh h h

v x h v x u x h u xu x h v x

h h

f x u x v x Let

The derivative:

0 0 0

lim lim limh h h

v x h v x u x h u xu x h v x

h h

d du x v x v x u x

dx dx

Rule 5: The Product Rule

d dv duuv u v

dx dx dx

The product of two differentiable functions u and v isdifferentiable, and

To find the derivative of a product of twofunctions: “The first times the derivative of thesecond plus the second times the derivative ofthe first.”

u x

f xv x

How about when we have a quotient?...

0limh

u x h u x

v x h v xd u

dx v h

The derivative:

Subtract and add v(x)u(x) in the numerator:

0

limh

v x u x h u x v x h

hv x h v x

0

limh

v x u x h v x u x v x u x u x v x h

hv x h v x

u x

f xv x

How about when we have a quotient?...

The derivative:

0

limh

v x u x h v x u x v x u x u x v x h

hv x h v x

0limh

u x h u x v x h v xv x u x

h hv x h v x

2d d

v x u x u x v xdx dx

v x

Rule 6: The Quotient Rule

2

du dvv ud u dx dx

dx v v

At a point where , the quotient of twodifferentiable functions is differentiable, and

To find the derivative of a quotient of twofunctions: “The bottom times the derivative ofthe top minus the top times the derivative ofthe bottom, all divided by the bottom squared.”

0v y u v

Practice Problems

2 1u x

Find if f x 2 31 3f x x x

3 3v x Let’s use the product rule with

and

2 31 3d

f x x xdx

2 2 31 3 3 2x x x x 4 2 43 3 2 6x x x x 4 25 3 6x x x Any other method for

finding this answer?

Practice Problems

2 1u x

Differentiate 2

2

1

1

xf x

x

2 1v x Use the quotient rule with and

2 2

22

1 2 1 2

1

x x x xf x

x

3 3

22

2 2 2 2

1

x x x x

x

22

4

1

x

x

Graphical support: 1y f x 2 NDERy f x

Practice Problems

Let be the product of the functions u and v.y uvFind if 2y

2 3u 2 4u 2 1v 2 2v

From the Product Rule: y uv uv vu At our particular point:

3 2 1 4

2 2 2 2 2y u v v u

2

Practice Problems

Suppose u and v are functions of x that are differentiable at x = 2.Also suppose that

2 3u 2 4u 2 1v 2 2v Find the values of the following derivatives at x = 2.

(a)d u

dx v

2

2 2 2 2

2

v u u v

v

2

1 4 3 2

1

10

Practice Problems

Suppose u and v are functions of x that are differentiable at x = 2.Also suppose that

2 3u 2 4u 2 1v 2 2v Find the values of the following derivatives at x = 2.

(b)d v

dx u

2

2 2 2 2

2

u v v u

u

2

3 2 1 4

3

10

9

Practice Problems

Suppose u and v are functions of x that are differentiable at x = 2.Also suppose that

2 3u 2 4u 2 1v 2 2v Find the values of the following derivatives at x = 2.

(c) 3 2 2d

u v uvdx

3 2 2 2 2d

u v uvdx

3 4 2 2 2 2 12