IMPLICIT DIFFERENTIATION Section.5 Calculus AP/Dual, Revised 017 viet.dang@humbleisd.net 7/30/018 1:47 AM.5: Implicit Differentiation 1
REVIEW Solve or d of 4 + 3 = 6 4 3 6 4 3 6 4 3 4 ' 3 3 7/30/018 1:47 AM.5: Implicit Differentiation
REVIEW 4 ' 3 3 This form is written in eplicit form, as it means it represents what equals (or where the variable is on one side) 7/30/018 1:47 AM.5: Implicit Differentiation 3
REVIEW Solve d of 7 d 7 7 d 7 14 7/30/018 1:47 AM.5: Implicit Differentiation 4
GENERAL PRACTICE Take the derivative in respects to or known as d 1) ) 3 3) 4 5 4) cos () 1 11 1 6 6 0 0 4 4 sin sin 7/30/018 1:47 AM.5: Implicit Differentiation 5
GENERAL PRACTICE Take the derivative in respects to or known as d 1) ) 3 d 1 11 1 6 d 3) 4 5 0 d 4 4) cos () sin d 7/30/018 1:47 AM.5: Implicit Differentiation 6
DEFINITIONS A. Implicit Differentiation takes the derivative with both sides with respects to in a relationship between,, and d ; essentiall an application of the CHAIN RULE B. An IMPLICIT function is written to which one has to complete additional steps to isolate the variable. Functions written are not solved for as it is not clearl stated but it is implied. C. Eamples include: 1. = whereas it can be written implicitl as = 1. 3 + 4 = can be written with implicit differentiation 7/30/018 1:47 AM.5: Implicit Differentiation 7
STEPS A. Differentiate both sides of the equation with respect to. MUST Include a d or ever time to differentiate with B. Collect all d on LEFT side and all other stuff to the right side C. Factor out d D. Divide both sides b the etra stuff 7/30/018 1:47 AM.5: Implicit Differentiation 8
EXAMPLE 1 Solve d of + = 5 Do we have an terms or an terms we took the derivative in respects to? 5 d d d 0 d 0 5 d d d 7/30/018 1:47 AM.5: Implicit Differentiation 9
EXAMPLE 1 Solve d of + = 5 d 0 d 0 d d 7/30/018 1:47 AM.5: Implicit Differentiation 10
EXAMPLE Solve d of 3 = + 5 d 3 7/30/018 1:47 AM.5: Implicit Differentiation 11
EXAMPLE 3 Solve d of 33 + + 4 = 7 d 8 9 7/30/018 1:47 AM.5: Implicit Differentiation 1
YOUR TURN Solve d of 3 + 5 = 4 d 3 5 7/30/018 1:47 AM.5: Implicit Differentiation 13
EXAMPLE 4 Solve d of 3 3 = 3 3 3 3 is being multiplied: Use the PRODUCT RULE d d d 3 3 d d 3 3 1 1 3 3 d d 3 1 1 3 3 3 7/30/018 1:47 AM.5: Implicit Differentiation 14
Solve d of 3 3 = EXAMPLE 4 d d 3 1 1 3 3 3 d 3 1 1 3 3 3 d 1 3 3 3 3 1 7/30/018 1:47 AM.5: Implicit Differentiation 15
EXAMPLE 5 Solve d of sin cos = 5 d 5 sin sin cos cos 7/30/018 1:47 AM.5: Implicit Differentiation 16
EXAMPLE 6 Solve d of sin = e in 0, π in terms of. sin 7/30/018 1:47 AM.5: Implicit Differentiation 17 e d cos e d e cos e e d cos e
EXAMPLE 6 Solve d of sin = e in 0, π in terms of. d cos e d cos sin d cot 7/30/018 1:47 AM.5: Implicit Differentiation 18
YOUR TURN Solve d of sin cos = 1 d cos cos sin sin 7/30/018 1:47 AM.5: Implicit Differentiation 19
EXAMPLE 7 Solve the second derivative of + = 16 16 0 d 0 d 7/30/018 1:47 AM.5: Implicit Differentiation 0
EXAMPLE 7 Solve the second derivative of + = 16 d d 1 d 7/30/018 1:47 AM.5: Implicit Differentiation 1 d d
EXAMPLE 7 Solve the second derivative of + = 16 d d 7/30/018 1:47 AM.5: Implicit Differentiation d
EXAMPLE 7 Solve the second derivative of + = 16 d d 3 d 3 7/30/018 1:47 AM.5: Implicit Differentiation 3
EXAMPLE 7 Solve the second derivative of + = 16 d 3 d d 16 3 7/30/018 1:47 AM.5: Implicit Differentiation 4 16 3
EXAMPLE 8 Solve the equation of the tangent line to the graph of + = 5 at the point 3, 4. 7/30/018 1:47 AM.5: Implicit Differentiation 5 5 0 d 0 d
EXAMPLE 8 Solve the equation of the tangent line to the graph of + = 5 at the point 3, 4. d 3 4 4 3 4 3 4 3 4 3 m 1 1 7/30/018 1:47 AM.5: Implicit Differentiation 6
EXAMPLE 8 Solve the equation of the tangent line to the graph of + = 5 at the point 3, 4. 3 4 3 4 7/30/018 1:47 AM.5: Implicit Differentiation 7
EXAMPLE 9 Find the equation of the tangent line to the graph of 7 + + 7 = 15 at the point 1, 1. 7 7 15 14 d d 14 0 d d 14 14 d 14 14 7/30/018 1:47 AM.5: Implicit Differentiation 8
EXAMPLE 9 Find the equation of the tangent line to the graph of 7 + + 7 = 15 at the point 1, 1. d 14 14 d 14 14 d 141 1 1 141 d 15 15 1 1 1 7/30/018 1:47 AM.5: Implicit Differentiation 9
YOUR TURN Solve the equation of the tangent line to the graph of + 4 = 4 at the point, 1. 1 1 7/30/018 1:47 AM.5: Implicit Differentiation 30
AP MULTIPLE CHOICE PRACTICE QUESTION 1 (NON-CALCULATOR) If 5 + 7 = 0, then solve d (A) 5+7 (B) 5 7 (C) 5 7 (D) 10 + 14 7/30/018 1:47 AM.5: Implicit Differentiation 31
AP MULTIPLE CHOICE PRACTICE QUESTION 1 (NON-CALCULATOR) If 5 + 7 = 0, then solve d 14 10 C d 14 10 7/30/018 1:47 AM.5: Implicit Differentiation 3 Vocabular Connections and Process Answer and Justifications Derivative Implicit Diff. Power Rule Product Rule d d 5 10 7 14 d 1 d 10 d d 14 0 d d d d 14 10 14 14 5 7 5 7
AP MULTIPLE CHOICE PRACTICE QUESTION If tan + =, then d = (A) tan + (B) sec + (C) sin + 1 (D) cos + 1 (NON-CALCULATOR) 7/30/018 1:47 AM.5: Implicit Differentiation 33
AP MULTIPLE CHOICE PRACTICE QUESTION If tan + =, then d = (NON-CALCULATOR) Vocabular Connections and Process Answer and Justifications Derivative Implicit Diff. Trig Derivative Power Rule d tan d sec 1 1 d sec 1 1 sec sec 1 d 1 sec D 7/30/018 1:47 AM.5: Implicit Differentiation 34 d 1 cos d cos 1
ASSIGNMENT Worksheet 7/30/018 1:47 AM.5: Implicit Differentiation 35