1. Section 2.3 Day 1 Product & Quotient Rules & Higher order Derivatives
AP Calculus AB

2. Learning Targets Define & apply the Product Rule
Define & apply the Quotient Rule
Apply more derivatives of trigonometric functions
Determine higher order derivatives
Recognize & apply the relationship between position, velocity, and acceleration functions

3. “One D-two plus two D-one“
Product Rule
𝑑 𝑑𝑥 𝑓 𝑥 𝑔 𝑥 =𝑓 𝑥 𝑔 ′ 𝑥 +𝑔 𝑥 𝑓′(𝑥)
 “One D-two plus two D-one“
*Note: Because of the commutative property, this could also be solved as: 𝑔 𝑥 𝑓′(𝑥) + 𝑓 𝑥 𝑔 ′ 𝑥

4. Example 1: Product Rule Find the derivative of ℎ 𝑥 = 3𝑥− 2𝑥 2 5+4𝑥
Let 𝑓 𝑥 =3− 𝑥 2 and 𝑔 𝑥 =5+4𝑥
ℎ ′ 𝑥 =𝑓 𝑥 𝑔 ′ 𝑥 +𝑔 𝑥 𝑓′(𝑥)
= 3𝑥− 𝑥 (5+4𝑥)(3−4𝑥)
=−24 𝑥 2 +4𝑥+15

5. Example 2: Product Rule Find the derivative of 𝑦=3 𝑥 2 sin 𝑥
Let 𝑓 𝑥 =3 𝑥 2 and 𝑔 𝑥 = sin 𝑥
𝑦 ′ =𝑓 𝑥 𝑔 ′ 𝑥 +𝑔 𝑥 𝑓′(𝑥)
=3 𝑥 2 cos 𝑥 + sin 𝑥 (6𝑥)
=3 𝑥 2 cos 𝑥 +6𝑥 sin 𝑥

6. Example 3: Product Rule Find the derivative of 𝑦=2𝑥 cos 𝑥 −2 sin 𝑥
Let 𝑓 𝑥 =2𝑥, 𝑔 𝑥 = cos 𝑥 , and ℎ 𝑥 =−2 sin 𝑥
𝑦 ′ =𝑓 𝑥 𝑔 ′ 𝑥 +𝑔 𝑥 𝑓 ′ 𝑥 +ℎ′(𝑥)
=2𝑥 − sin 𝑥 + cos 𝑥 2 −2 cos 𝑥
=−2𝑥 sin 𝑥 +2 cos 𝑥 −2 cos 𝑥
=−2𝑥 sin 𝑥

7. Product Rule NON Example!!
1. 𝑑 𝑑𝑥 𝑥 3 ∙ 𝑥 4 = 𝑑 𝑑𝑥 [ 𝑥 7 ]
2. =7 𝑥 6
Non-Example: 𝑑 𝑑𝑥 [ 𝑥 3 ∙ 𝑥 4 ]
1. 𝑑 𝑑𝑥 𝑥 3 ∙ 𝑥 4
= 𝑑 𝑑𝑥 [ 𝑥 3 ]∙ 𝑑 𝑑𝑥 [ 𝑥 4 ]
2. =3 𝑥 2 ∙4 𝑥 3 =12 𝑥 5

8. Quotient Rule 𝑑 𝑑𝑥 𝑓 𝑥 𝑔 𝑥 = 𝑔 𝑥 𝑓 ′ 𝑥 −𝑓 𝑥 𝑔 ′ 𝑥 𝑔 𝑥 2
𝑑 𝑑𝑥 𝑓 𝑥 𝑔 𝑥 = 𝑔 𝑥 𝑓 ′ 𝑥 −𝑓 𝑥 𝑔 ′ 𝑥 𝑔 𝑥 2
“Low D-high minus high D-low, over the square of what’s below”

9. Example 4: Quotient Rule
Find the derivative of 𝑦= 5𝑥−2 𝑥 2 +1
Let 𝑓 𝑥 =5𝑥−2 and 𝑔 𝑥 = 𝑥 2 +1
𝑦 ′ = 𝑔 𝑥 𝑓 ′ 𝑥 −𝑓 𝑥 𝑔 ′ 𝑥 𝑔 𝑥 2 = 𝑥 − 5𝑥−2 2𝑥 𝑥
= 5 𝑥 2 +5 −10 𝑥 2 +4𝑥 𝑥 = −5 𝑥 2 +4𝑥+5 𝑥

10. Example 6: Quotient Rule
Find the derivative of 𝑦= sin 𝑥 cos 𝑥
Let 𝑓 𝑥 = sin 𝑥 and 𝑔 𝑥 = cos 𝑥
𝑦 ′ = 𝑔 𝑥 𝑓 ′ 𝑥 −𝑓 𝑥 𝑔 ′ 𝑥 𝑔 𝑥 2 = cos 𝑥 cos 𝑥 − sin 𝑥 − sin 𝑥 cos 𝑥 2
= cos 2 𝑥 + sin 2 𝑥 cos 2 𝑥 = 1 cos 2 𝑥 = sec 2 𝑥

11. Example 7: Quotient Rule
Find the derivative of 𝑦= 1− cos 𝑥 sin 𝑥
Let 𝑓 𝑥 =1− cos 𝑥 and 𝑔 𝑥 = sin 𝑥
𝑦 ′ = 𝑔 𝑥 𝑓 ′ 𝑥 −𝑓 𝑥 𝑔 ′ 𝑥 𝑔 𝑥 2 = sin 𝑥 sin 𝑥 − 1− cos 𝑥 cos 𝑥 sin 2 𝑥
= sin 2 𝑥 − cos 𝑥 + cos 2 𝑥 sin 2 𝑥 = 1−cos x sin 2 x

12. Example 8: Re-writing Find the derivative of 𝑦= 9 5 𝑥 2
Re-write into 𝑦= 9 5 ( 𝑥 −2 ) (Key: the coefficient in the denominator does not come up)
Use Power Rule: 𝑦 ′ =− 𝑥 −3 =− 18 5 𝑥 3

13. Example 9: Re-writing Find the derivative of 𝑦= 2𝑥+1 𝑥 2
You can use quotient rule or power rule. Let’s use power rule here.
Re-write into 𝑦= 2𝑥+1 𝑥 −2
Let 𝑓 𝑥 =2𝑥+1 and 𝑔 𝑥 = 𝑥 −2
𝑦 ′ =𝑓 𝑥 𝑔 ′ 𝑥 +𝑔 𝑥 𝑓′(𝑥)
= 2𝑥+1 −2 𝑥 −3 + 𝑥 −2 2
= −4𝑥−2 𝑥 𝑥 2

14. Quotient Rule Check Find the derivative of 𝑦= 5 𝑥 4 +1 𝑥
**Try solving using both the quotient rule and the product rule

15. Derivatives of Trigonometric Functions
𝑑 𝑑𝑥 sin 𝑥 =𝑐𝑜𝑠𝑥
𝑑 𝑑𝑥 cos 𝑥 =−𝑠𝑖𝑛𝑥
𝑑 𝑑𝑥 tan 𝑥 = sec 2 𝑥
𝑑 𝑑𝑥 cot 𝑥 = −csc 2 𝑥
𝑑 𝑑𝑥 sec 𝑥 = sec 𝑥 tan 𝑥
𝑑 𝑑𝑥 csc 𝑥 =− csc 𝑥 cot 𝑥

16. Derivatives of Trigonometric Functions
tan 𝑥
cot 𝑥
−c𝑠𝑐 𝑥
c𝑠𝑐 𝑥
sec 𝑥
sec 𝑥

17. Example 11: Derivatives of Trig Functions
Find the derivative of the following:
1.)
2.)

18. Example 11: Derivatives of Trig Functions
Find a partner and try problems 3 and 4 from your notes outline:
3. Write equations of tangent and normal lines to 𝑓 𝑥 =𝑠𝑖𝑛𝑥𝑠𝑒𝑐𝑥 at the point 𝑥= 3𝜋 4 . Show the analysis that leads to your answer.
1.) 𝑓 3𝜋 4 =𝑠𝑖𝑛 3𝜋 4 𝑠𝑒𝑐 3𝜋 4 =−1
2.) 𝑓 ′ 𝑥 =𝑐𝑜𝑠𝑥𝑠𝑒𝑐𝑥+𝑠𝑖𝑛𝑥𝑠𝑒𝑐𝑥𝑡𝑎𝑛𝑥
3.) 𝑓 ′ 3𝜋 4 =𝑐𝑜𝑠 3𝜋 4 𝑠𝑒𝑐 3𝜋 4 +𝑠𝑖𝑛 3𝜋 4 𝑠𝑒𝑐 3𝜋 4 𝑡𝑎𝑛 3𝜋 4
= − − − − −1 =1+1=2
4.) Tangent Line: 𝑦+1=2(𝑥− 3𝜋 4 ) Normal Line: 𝑦+1=− 1 2 (𝑥− 3𝜋 4 )
4. Determine 𝑑 2 𝑑 𝑏 2 sec⁡(𝑏)
𝑑 𝑑𝑏 sec⁡(𝑏) = sec⁡(𝑏)tan⁡(𝑏) 𝑑 2 𝑑 𝑏 2 sec⁡(𝑏) = 𝑑 𝑑𝑏 sec 𝑏 tan 𝑏 = sec 𝑏 𝑡𝑎𝑛 2 𝑏 + 𝑠𝑒𝑐 3 𝑏

19. Higher Order Derivatives
-You can take the 3rd, 4th, 5th, etc. derivative.
-This just means you have to use the derivative rules multiple times.
Notations:
2nd Derivative: 𝑦′′or 𝑑 2 𝑦 𝑑 𝑥 2 ,
3rd Derivative: 𝑦′′′ or 𝑑 3 𝑦 𝑑 𝑥 3
4th Derivative: 𝑓 (4) (𝑥) or 𝑑 4 𝑦 𝑑 𝑥 4 …and so on…

20. Example 10: Higher Order Derivatives
Find 𝑓′′′(𝑥) of 𝑓 𝑥 =4 𝑥 4 −9 𝑥 3 +2 𝑥 2 −9𝑥+5
𝑓 ′ 𝑥 =16 𝑥 3 −27 𝑥 2 +4𝑥−9
𝑓 ′′ 𝑥 =48 𝑥 2 −54𝑥+4
𝑓 ′′′ 𝑥 =96𝑥−54

21. Position, Velocity, Acceleration
Recall, that the velocity was the slope of the position function.
Let’s take a look at the velocity graph.
Notice that the slope of velocity is acceleration.

22. Speed Increasing vs Decreasing
Speed is increasing if velocity and acceleration have the SAME SIGN
Speed is decreasing if velocity and acceleration have DIFFERENT SIGNS

23. Speed Increasing vs Decreasing
Examples
1. Positive Velocity and Negative Acceleration
- Driving from 0-60pmh, then breaking
2. Negative Velocity and Positive Acceleration
- Ball thrown up in the air, but when it starts to fall down the velocity is negative, but its increasing in acceleration
3. Positive Velocity and Positive Acceleration
- Driving a car forward and accelerating in the same direction
4. Negative Velocity and Negative Acceleration
- Driving backwards and breaking

24. Exit Ticket for Feedback
1. Find 𝑓′(𝑥) of 𝑓 𝑥 = 𝑥 2 cos 𝑥
2. Find 𝑓′(𝑥) of 𝑓 𝑥 = 𝑥+1 𝑥−2
3. The position function of a particle is given by 𝑓 𝑥 = 𝑥 4 − 𝑥 2 +9𝑥. What is the acceleration of the particle at 𝑥=2?