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

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

“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

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 2 4 +(5+4𝑥)(3−4𝑥) =−24 𝑥 2 +4𝑥+15

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

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

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

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”

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 2 +1 5 − 5𝑥−2 2𝑥 𝑥 2 +1 2 = 5 𝑥 2 +5 −10 𝑥 2 +4𝑥 𝑥 2 +1 2 = −5 𝑥 2 +4𝑥+5 𝑥 2 +1 2

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

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

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 ′ =− 18 5 𝑥 −3 =− 18 5 𝑥 3

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 3 + 2 𝑥 2

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

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

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

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

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 = − 2 2 − 2 2 + − 2 2 − 2 2 −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

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…

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

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.

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

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

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?