23 2 xxxf )(

hxxhxhx

xfh

)())()((lim)(

2323 22

0

hxxhxhxhx

h

)())((lim

232233 222

0

hxxhxhxhx

h

232233 222

0

lim

hhxhh

h

2

0

23

lim

hxh

230

lim x23 xxf 23 )(

3.3 –Differentiation RulesREVIEW: Use the Limit Definition to find the derivative of the given function.

𝑓 ′(𝑥 )=limh→0

𝑓 (𝑥+h )− 𝑓 (𝑥 )h

hhxh

h

)(lim

230

3.3 –Differentiation RulesWhat are the rules?

𝑓 (𝑥 )=3 𝑥2 𝑓 ′ (𝑥 )=6 𝑥

𝑓 (𝑥 )=5𝑥5−6 𝑥3+𝑥 𝑓 ′ (𝑥 )=25 𝑥4−18 𝑥2+1

𝑓 (𝑥 )=7 𝑥6+9𝑥− 4−12𝑥+5 𝑓 ′ (𝑥 )=42𝑥5−36 𝑥−5−12

𝑓 (𝑥 )=52 𝑓 ′ (𝑥 )=0

𝑓 (𝑥 )=5𝑥−8 𝑓 ′ (𝑥 )=5𝑓 (𝑥 )=−9 𝑥+10 𝑓 ′ (𝑥 )=−9

𝑓 (𝑥 )=2𝑥7−4 𝑥5+𝑥2 𝑓 ′ (𝑥 )=14 𝑥6−20 𝑥4+2𝑥

𝑓 (𝑥 )=4 𝑥2+𝑥 𝑓 ′ (𝑥 )=8𝑥+1

𝑓 (𝑥 )=6 𝑥− 3+8 𝑥3+2𝑥+7 𝑓 ′ (𝑥 )=−18𝑥− 4+24 𝑥3+2

3.3 –Differentiation Rules

𝑓 (𝑥 )=5 𝑓 ′ (𝑥 )=0 𝑓 (𝑥 )=−17 𝑓 ′ (𝑥 )=0

𝑓 (𝑥 )=𝑥2 𝑓 ′ (𝑥 )=2𝑥 𝑓 (𝑥 )=𝑥7 𝑓 ′ (𝑥 )=7 𝑥6

𝑓 (𝑥 )=𝑥−5𝑓 ′ (𝑥 )=−5𝑥− 6

𝑓 (𝑥 )=𝑥57 𝑓 ′ (𝑥 )=5

7𝑥− 27

𝑓 (𝑥 )=𝑥− 8𝑓 ′ (𝑥 )=−8 𝑥− 9

3.3 –Differentiation Rules

𝑓 (𝑥 )=4 𝑥2𝑓 ′ (𝑥 )=4 ∙2𝑥=8 𝑥

𝑓 (𝑥 )=−9 𝑥7 𝑓 ′ (𝑥 )=−9 ∙7 𝑥6=−63 𝑥6

𝑓 (𝑥 )=−3 𝑥−5 𝑓 ′ (𝑥 )=−3¿

𝑓 (𝑥 )=2𝑥57 𝑓 ′ (𝑥 )=2∙ 5

7𝑥− 27=10

7𝑥− 27

𝑓 (𝑥 )=6 𝑥− 8 𝑓 ′ (𝑥 )=6 ∙ (−8 ) 𝑥− 9=−48 𝑥−9

3.3 –Differentiation Rules

𝑓 (𝑥 )=4 𝑒𝑥 𝑓 ′ (𝑥 )=4𝑒𝑥

𝑓 (𝑥 )=−6 𝑥−5+6 𝑥−8 𝑓 ′ (𝑥 )=30𝑥−6−48 𝑥− 9

𝑓 (𝑥 )=3 𝑥4+2𝑥57 −8 𝑓 ′ (𝑥 )=12 𝑥3+ 10

7𝑥− 27

𝑓 (𝑥 )=4 𝑥2−9 𝑥7 𝑓 ′ (𝑥 )=8𝑥−63 𝑥6

𝑓 (𝑥 )=−6𝑒𝑥+5 𝑥7+2𝑓 ′ (𝑥 )=−6𝑒𝑥+35 𝑥6

3.3 –Differentiation Rules

𝑓 (𝑥 )=(4 𝑥+𝑥4 ) (3 𝑥+2𝑥2 )

𝑓 ′ (𝑥 )= (4 𝑥 +𝑥4 )

𝑢=(4 𝑥+𝑥4 )v=(3𝑥+2 𝑥2 )

(3+4 𝑥 )+¿(3 𝑥+2 𝑥2 )(4+4 𝑥3 )

𝑓 ′ (𝑥 )=12 𝑥+16𝑥2+3 𝑥4+4 𝑥5+12𝑥+12𝑥4+8 𝑥2+8𝑥5

𝑓 ′ (𝑥 )=12 𝑥5+15𝑥4+24 𝑥2+24 𝑥

3.3 –Differentiation Rules

𝑓 (𝑥 )=(1+𝑥2 ) (𝑥34 −𝑥−3)

𝑓 ′ (𝑥 )= (1+𝑥2 )

𝑢=(1+𝑥2)v=(𝑥34 −𝑥−3)

( 34 𝑥−14+3 𝑥− 4)+¿(𝑥 34−𝑥−3)(2 𝑥 )

𝑓 ′ (𝑥 )= 34𝑥− 14 +3 𝑥− 4+ 3

4𝑥74+3𝑥−2+2𝑥

74 −2 𝑥− 2

𝑓 ′ (𝑥 )=114𝑥74+ 34𝑥− 14+𝑥− 2+3 𝑥− 4

3.3 –Differentiation Rules

𝑔 (𝑥 )= 𝑥𝑒𝑥

𝑢=𝑥𝑣=𝑒𝑥

𝑑𝑑𝑥 ( 𝑥𝑒𝑥 )=¿𝑒

𝑥

❑(1 )− 𝑒𝑥𝑥(𝑒𝑥)2

𝑑𝑑𝑥 ( 𝑥𝑒𝑥 )=𝑒𝑥−𝑥𝑒𝑥

𝑒2𝑥

𝑑𝑑𝑥 ( 𝑥𝑒𝑥 )=𝑒𝑥 (1− 𝑥 )

𝑒2𝑥

𝑑𝑑𝑥 ( 𝑥𝑒𝑥 )= (1−𝑥 )

𝑒𝑥

3.3 –Differentiation Rules

𝑔 (𝑥 )= 3 𝑥3 𝑥2+2

𝑢=3 𝑥 𝑣=3 𝑥2+2

𝑑𝑑𝑥 ( 3 𝑥

3 𝑥2+2 )=¿(3 𝑥¿¿2+2)

¿(3 )− (6 𝑥)3 𝑥(3 𝑥2+2 )2

𝑑𝑑𝑥 ( 3 𝑥

3 𝑥2+2 )=9 𝑥2+6−18 𝑥2

(3𝑥2+2 )2𝑑𝑑𝑥 ( 3 𝑥

3 𝑥2+2 )=−9 𝑥2+6

(3 𝑥2+2 )2

3.4 –The Derivative as a Rate of ChangePosition, Displacement, Velocity, Speed, and Acceleration

Position: the location of an object as determined by a position function.

Displacement: the distance an object is away from its initial position after travelling for a certain interval of time.

5 feet

3 feet

1 foot

5−3+1

𝐷𝑖𝑠𝑝𝑙𝑎𝑐𝑒𝑚𝑒𝑛𝑡 : 3 𝑓𝑒𝑒𝑡 𝑓𝑟𝑜𝑚 h𝑡 𝑒 𝑖𝑛𝑖𝑡𝑖𝑎𝑙𝑝𝑜𝑖𝑛𝑡

𝑝𝑜𝑠𝑖𝑡𝑖𝑣𝑒3 𝑓𝑒𝑒𝑡→ h𝑟𝑖𝑔 𝑡 𝑜𝑓 𝑖𝑛𝑖𝑡𝑖𝑎𝑙𝑝𝑜𝑖𝑛𝑡

3.4 –The Derivative as a Rate of ChangePosition, Displacement, Velocity, Speed, and Acceleration

Position: the location of an object as determined by a position function.

Displacement: the distance an object is away from its initial position after travelling for a certain interval of time.

9 feet

5 feet

1 foot

−9+5−1

𝐷𝑖𝑠𝑝𝑙𝑎𝑐𝑒𝑚𝑒𝑛𝑡 :−5 𝑓𝑒𝑒𝑡 𝑓𝑟𝑜𝑚 h𝑡 𝑒 𝑖𝑛𝑖𝑡𝑖𝑎𝑙𝑝𝑜𝑖𝑛𝑡

𝑛𝑒𝑔𝑎𝑡𝑖𝑣𝑒5 𝑓𝑒𝑒𝑡→ 𝑙𝑒𝑓𝑡𝑜𝑓 𝑖𝑛𝑖𝑡𝑖𝑎𝑙𝑝𝑜𝑖𝑛𝑡

3.4 –The Derivative as a Rate of ChangePosition, Displacement, Velocity, Speed, and Acceleration

Position: the location of an object as determined by a position function.

Displacement: the distance an object is away from its initial position after travelling for a certain interval of time.

𝑚𝑖𝑙𝑒𝑚𝑎𝑟𝑘𝑒𝑟

50−30+20

𝐷𝑖𝑠𝑝𝑙𝑎𝑐𝑒𝑚𝑒𝑛𝑡 : 40𝑚𝑖𝑙𝑒𝑠 𝑓𝑟𝑜𝑚 h𝑡 𝑒 𝑖𝑛𝑖𝑡𝑖𝑎𝑙𝑝𝑜𝑖𝑛𝑡

𝑝𝑜𝑠𝑖𝑡𝑖𝑣𝑒 40𝑚𝑖𝑙𝑒𝑠→ h𝑟𝑖𝑔 𝑡 𝑜𝑓 𝑖𝑛𝑖𝑡𝑖𝑎𝑙𝑝𝑜𝑖𝑛𝑡

100 150 150−100=50120 120−150=−30

140 140−120=20

3.4 –The Derivative as a Rate of ChangePosition, Displacement, Velocity, Speed, and Acceleration

Velocity: the change in position with respect to a change in time. It is a rate of change with direction.

Speed: the absolute value of velocity. It is a rate of change without direction.

𝑣 (𝑡 )=𝑠 ′ (𝑡 )=𝑑𝑠𝑑𝑡

The velocity function, , is obtain by differentiating the position function with respect to time.

𝑠 (𝑡 )=4 𝑡2+𝑡

𝑣 (𝑡 )=𝑠 ′ (𝑡)=8 𝑡+1

𝑠 (𝑡 )=5 𝑡3−6 𝑡 2+6

𝑣 (𝑡 )=𝑠 ′ (𝑡)=15 𝑡2−12 𝑡

𝑠𝑝𝑒𝑒𝑑=|𝑣 (𝑡 )|=|𝑠 ′(𝑡 )|=|𝑑 𝑠𝑑𝑡 |

3.4 –The Derivative as a Rate of Change

Position, Displacement, Velocity, Speed, and Acceleration

Acceleration: the change in velocity with respect to a change in time. It is a rate of change with direction.

The acceleration function, , is obtain by differentiating the velocity function with respect to time. It is also the 2nd derivative of the position function.

𝑎 (𝑡 )=𝑣 ′ (𝑡 )=𝑑𝑣𝑑𝑡

=𝑠′ ′ (𝑡 )= 𝑑2𝑠𝑑 𝑡2

𝑠 (𝑡 )=4 𝑡2+𝑡

𝑣 (𝑡 )=𝑠 ′ (𝑡)=8 𝑡+1

𝑠 (𝑡 )=5 𝑡3−6 𝑡 2+6

𝑣 (𝑡 )=𝑠 ′ (𝑡)=15 𝑡2−12 𝑡

𝑎 (𝑡 )=𝑣 ′ (𝑡 )=𝑠′ ′ (𝑡 )=8 𝑎 (𝑡 )=𝑣 ′ (𝑡 )=𝑠 ′ ′ (𝑡)=30 𝑡−12

3.4 –The Derivative as a Rate of ChangeEconomics

Marginal cost of production: the rate of change of costs with respect to the level of production.

𝑐 (𝑥 )

𝑚𝑎𝑟𝑔𝑖𝑛𝑎𝑙𝑐𝑜𝑠𝑡=𝑐 ′ (𝑥 )

Cost of production: a function of the units produced (x) that generates the cost of producing those units .

Average cost of production: the cost of production function divided by the number of units produced at that cost.

𝑎𝑣𝑒𝑟𝑎𝑔𝑒𝑐𝑜𝑠𝑡=𝑐(𝑥 )𝑥

Marginal cost is an approximation of the cost to produce one more unit after producing x units

3.4 –The Derivative as a Rate of ChangeEconomics

𝑐 (𝑥 ) 𝑚𝑎𝑟𝑖𝑔𝑖𝑛𝑎𝑙𝑐𝑜𝑠𝑡=𝑐 ′ (𝑥)Cost of production: 𝑎𝑣𝑒𝑟𝑎𝑔𝑒𝑐𝑜𝑠𝑡=𝑐(𝑥 )𝑥

A cost function is given as follows:

Find the average cost in producing 50 units.

Find the marginal cost to produce the 51st unit.

Find the actual cost to produce the 51st unit.

𝑎𝑣𝑒𝑟𝑎𝑔𝑒𝑐𝑜𝑠𝑡=𝑐(𝑥 )𝑥

¿100+8 (50)+0.1 (50 )2  

50¿75050 ¿ $15

𝑐 ′ (𝑥 )=8+0.2 𝑥 𝑐 ′ (50 )=8+0.2 (50) ¿ $18

𝑐 (50 )=100+8 (50)+0.1 (50 )2¿750

𝑐 (51 )=100+8(51)+0.1 (51 )2¿768.1

768.1−750¿ $18.10

¿𝑐 (50)50

3.4 –The Derivative as a Rate of Change