3.3 Rules for Differentiation Colorado National Monument Greg Kelly, Hanford High School, Richland,...

3.3 Rules for Differentiation

Colorado National MonumentGreg Kelly, Hanford High School, Richland, WashingtonPhoto by Vickie Kelly, 2003

The “Coke Ovens”, named because they resemble brick “beehive” ovens that were used to manufacture coke from coal for steel making.

If the derivative of a function is its slope, then for a constant function, the derivative must be zero.

0dc

dx

example: 3y

0y

The derivative of a constant is zero.

If we find derivatives with the difference quotient:

2 22

0limh

x h xdx

dx h

2 2 2

0

2limh

x xh h x

h

2x

3 33

0limh

x h xdx

dx h

3 2 2 3 3

0

3 3limh

x x h xh h x

h

23x

1

1 1

1 2 1

1 3 3 1

1 4 6 4 1

1 5 10 10 5 1

(Pascal’s Triangle)

2

4dx

dx

4 3 2 2 3 4 4

0

4 6 4limh

x x h x h xh h x

h

34x

2 3

We observe a pattern: 2x 23x 34x 45x 56x …

1n ndx nx

dx

examples:

4f x x

34f x x

8y x

78y x

power rule

We observe a pattern: 2x 23x 34x 45x 56x …

d ducu c

dx dx

examples:

1n ndcx cnx

dx

constant multiple rule:

5 4 47 7 5 35dx x x

dx

When we used the difference quotient, we observed that since the limit had no effect on a constant coefficient, that the constant could be factored to the outside.

(Each term is treated separately)

d ducu c

dx dx

constant multiple rule:

sum and difference rules:

d du dvu v

dx dx dx d du dv

u vdx dx dx

4 12y x x 34 12y x

4 22 2y x x

34 4dy

x xdx

This makes sense, because:

0limh

f x h g x h f x g x

h

0 0

lim limh h

f x h f x g x h g x

h h

Example:Find the horizontal tangents of: 4 22 2y x x

34 4dy

x xdx

Horizontal tangents occur when slope = zero.34 4 0x x

3 0x x

2 1 0x x

1 1 0x x x

0, 1, 1x

Plugging the x values into the original equation, we get:

2, 1, 1y y y

(The function is even, so we only get two horizontal tangents.)

4 22 2y x x

4 22 2y x x

2y

4 22 2y x x

2y

1y

4 22 2y x x

4 22 2y x x

First derivative (slope) is zero at:

0, 1, 1x

34 4dy

x xdx

Product Rule:

?duv

dx We can use the definition of derivative to find

a formula for the derivative of a product.

0

limh

f x h g x h f x g xdf x g x

dx h

We need to rewrite this as a limit that we can evaluate.

This would work:

0

limh

f x h f x

h

Product Rule:

?duv

dx We can use the definition of derivative to find

a formula for the derivative of a product.

0

limh

f x h g x h f x g xdf x g x

dx h

This would work:

0

limh

f x h f x

h

If we subtract

f x g x h we can factor

factor out . g x h

Product Rule:

?duv

dx We can use the definition of derivative to find

a formula for the derivative of a product.

0

limh

f x h g x h f x g xdf x g x

dx h

If we subtract

f x g x h we can factor

factor out . g x h

But if we subtract

f x g x h will need to

add it back in.

Product Rule:

?duv

dx We can use the definition of derivative to find

a formula for the derivative of a product.

0

limh

f x h g x h f x g xdf x g x

dx h

0

limh

f x g x hf x h g x h f x g x

h

f x g x h

0

limh

f x g x hf fx h g x h f x g x

h

x g x h

0 0

lim limh h

f x g x hf x h g x h f x g xx

h h

f x g h

We are going to subtract and add the same expression to the limit:

Many calculus books (including ours) give this formula with the terms in a different order.We are going to use this order to be consistent with the quotient rule (next) and with the derivative of cross products (next year.)

0 0

lim limh h

f x h g x h f x g x h f x g x h f x g x

h h

0 0

lim limh h

g x h g x hf x h f x g x h f x

h h

x gf x

0 0

lim limh h

f x h f x g x h g x

h

g x xh f

h

000 0

lim lli iim m mlh hhh

g x hf x h f x g x h g x

f xh h

df x g x f x g x x

df x g

x

d du dvuv v u

dx dx dx

Substituting u and v, we get a formula for the derivative of a product:

Evaluating the limits:

The limit of a product = the product of the limits:

Using the distributive property:

Product Rule:

d du dvuv v u

dx dx dx Notice that this is not just the

product of two derivatives.

This is sometimes memorized as: duv u v uv

dx

2 33 2 5d

x x xdx

5 3 32 5 6 15d

x x x xdx

5 32 11 15d

x x xdx

4 210 33 15x x

2 3x 26 5x 32 5x x

4 2 4 2 24 10 6 5 18 15x x x x x

4 210 33 15x x

2x

Quotient rule:

?d u

dx v

Once again we can use the definition of derivative to find a formula.

0limh

f x h f x

f x g x h g xd

dx g x h

Clearing the complex fraction:

g x h g x

g x h g x

0

limh

f x h g x f x g x h

h g x h g x

0

limh

f x g x ff x h g x f x g x h

h g x h g x

x g x

Again we are going to subtract and add the same expression:

0

limh

f x g x ff x h g x f x g x h

h g x h g x

x g x

0 0

1lim limh h

f x h f x g x g x h

h g x h

f xx

x

g

g

20 0

1lim limh h

f x h f x g x h g x

h hf x

g xg x

2

1f xdf x g x f x g x

dx g x g x

2

d u u v uv

dx v v

Substituting u and v, we get the formula for the derivative of a quotient:

Evaluating the limits:

The limit of a product = the product of the limits:

Factoring each side of the numerator and factoring the denominator:

(and evaluating this limit:)

0

limh

ff x h f x g xx f x g xg x g x h

h g x h g x

Quotient Rule:

2

du dvv ud u dx dx

dx v v

or 2

d u u v uv

dx v v

3

2

2 5

3

d x x

dx x

2 2 3

22

6 5 3 2 5 2

3

x x x x x

x

Higher Order Derivatives:

dyy

dx is the first derivative of y with respect to x.

2

2

dy d dy d yy

dx dx dx dx

is the second derivative.

(y double prime)

dyy

dx

is the third derivative.

4 dy y

dx is the fourth derivative.

We will learn later what these higher order derivatives are used for.