CHAIN RULE: DAY WITH TRIG FUNCTIONS Section.4A Calculus AP/Dual, Revised 018 viet.dang@humbleisd.net 7/30/018 1:44 AM.4A: Chain Rule Day 1
THE CHAIN RULE A. d dx f g x = f g x g x B. If f(x) is a differentiable function and g(x) is a differentiable function, then y = f(g(x)) is a differentiable function. 7/30/018 1:44 AM.4A: Chain Rule Day
TRIG FUNCTION DERIVATIVES C. Derivatives of Trig Functions (MEMORIZE THEM) 1.. d dx d dx sin x = cos x cos x = sin x 3. d dx tan x = sec x 4. 5. d dx d dx csc x = csc x cot x sec x = sec x tan x 6. d dx cot x = csc x 7/30/018 1:44 AM.4A: Chain Rule Day 3
TRIG DERIVATIVES So, when you have to ask your neighbor a question PSST... Positive SEC SEC TAN d dx d dx sec x = sec x tan x tan x = sec x 7/30/018 1:44 AM.4A: Chain Rule Day 4
TRIG DERIVATIVES Think opposite of PSST is to CHANGE Positive SEC SEC TAN CHANGE d CSC dx CSC d COT dx Negative( CHANGE) 7/30/018 1:44 AM.4A: Chain Rule Day 5 CSC CSC COT csc x = csc x cot x cot x = csc x
REVIEW Basic Functions sin x e x ln x Composite Functions sin ( x + 3) x e ln ( 3x 4 ) 7/30/018 1:44 AM.4A: Chain Rule Day 6
COMPOSITE FUNCTIONS sin ( x + 3) ( sin x) ( x + 3 )) 7/30/018 1:44 AM.4A: Chain Rule Day 7
Solve y for y = sin x + 3 EXAMPLE 1 ( ( )) ( ) y ' = f ' g x g ' x Original Derivative sin ( x + 3) cos( x + 3) x + 3 y y ( x ) ' = cos + 3 ( x ) ' = cos + 3 7/30/018 1:44 AM.4A: Chain Rule Day 8
EXAMPLE Solve y for y = cos x ( ( )) ( ) y ' = f ' g x g ' x cos( x) sin ( x) x sin ( x)( ) y ( x) ' = sin 7/30/018 1:44 AM.4A: Chain Rule Day 9
EXAMPLE 3 Solve y for y = cos x ( ( )) ( ) y ' = f ' g x g ' x ( cos x) ( cos x) 1 cos x sin x x 1 ( )( )( ) y ' = cos x sin x 1 ( )( ) y ' = cos x sin x 7/30/018 1:44 AM.4A: Chain Rule Day 10
EXAMPLE 4 Solve y for y = csc x 3 3 3 y ' = 3x csc x cot x 7/30/018 1:44 AM.4A: Chain Rule Day 11
EXAMPLE 5 Solve y for y = cos (3x) ( ) y ' = 18x sin 9x 7/30/018 1:44 AM.4A: Chain Rule Day 1
YOUR TURN Solve y for y = tan x y ' = tan xsec x 7/30/018 1:44 AM.4A: Chain Rule Day 13
WHEN YOU BUILD A CHAIN It can keep going and going 7/30/018 1:44 AM.4A: Chain Rule Day 14
REVISIT ORDER A. Wrapper B. Shell C. Gum 7/30/018 1:44 AM.4A: Chain Rule Day 15
WITH TRIG CHAIN RULE FUNCTIONS A. Make sure the exponent and negative are put in the appropriate spots B. Use parenthesis 7/30/018 1:44 AM.4A: Chain Rule Day 16
Solve dy for y = dx sin3 4x EXAMPLE 6 ( ( )) ( ) y ' = f ' g x g ' x ( sin ( 4x) ) 3 3( sin ( 4x) ) sin ( 4x) cos( 4x) 4x 4 ( ) ( ) ( )( ) y ' = 3 sin x cos 4x 4 ( ) ( ) ( ) f ' x = 1sin 4x cos 4x 7/30/018 1:44 AM.4A: Chain Rule Day 17
Solve dy dx for f x = 3cos3 x 3 + 1 EXAMPLE 7 f ' x 7x cos x 1 sin x 1 ( ) ( 3 ) ( 3 = + + ) 7/30/018 1:44 AM.4A: Chain Rule Day 18
Solve dy dx for f x = cot3 (x 5x) EXAMPLE 8 f ' x = 3cot x 5x csc x 5x x 5 ( ) ( ) ( )( ) 7/30/018 1:44 AM.4A: Chain Rule Day 19
Solve dy dx for f x = 4sec3 x + 1 YOUR TURN f ' x = 4sec x + 1 tan x + 1 ( ) 3 ( ) ( ) 7/30/018 1:44 AM.4A: Chain Rule Day 0
Solve f x for f x = x 4 sin x EXAMPLE 9 y = x 4 sin x 3 4 y ' = 4x sin x + x cos x y '' = fg ' + gf ' ( ) ( ) ( ) 3 3 4 4 cos + sin 1 + 4 cos + sin x x x x x x x x f '' x = 1x sin x + 8x cos x x sin x ( ) 3 4 7/30/018 1:44 AM.4A: Chain Rule Day 1
EXAMPLE 10 Solve f x for f x = 1 x 3 cos 5x 4 f ' ( x) = d d cos5x 1 x 1 x cos5x dx dx ( ) ( ) ( ) ( 4 cos5x ) ( ) ( ) 4 3 3 4 d d dx dx ( ) ( ) ( ) ( 4 3 ( ) )( 4 ) ( 4 cos5x 3 1 x 1 1 x sin 5x 5x ) ( 4 cos5x ) 7/30/018 1:44 AM.4A: Chain Rule Day
Solve f x for f x = 1 x 3 cos 5x 4 EXAMPLE 10 ( ) ( ) ( 4 3 ( ) )( 4 )( 3 cos5x 3 1 x 1 x sin 5x 0x ) ( 4 cos5x ) f ' ( ) 3 + ( ) ( ) ( ) ( ) ( x) cos5x 3 1 x 0x 1 x sin 5x = 4 3 4 cos 5x 4 ( ) ( ) 3 x x 4 x 3 ( x) ( x 4 ) 3 1 cos5 + 0 1 sin 5 cos 5x 4 7/30/018 1:44 AM.4A: Chain Rule Day 3
YOUR TURN Solve f x for f x = x 5 cos x 3 f ' ( x) = ( 3 3 3 3 ) 3 ( cos x ) x cosx + 6x sin x 15xsin x 7/30/018 1:44 AM.4A: Chain Rule Day 4
EXAMPLE 11 Determine the point(s) in the interval of 0, π at which the graph of cos x f x = has a horizontal tangent. +sin x sin x 1 f '( x) = + sin x ( ) sin x 1 = 0 sin x = 1 sin 1 x = 7 3 11 3,,, 6 3 6 3 7/30/018 1:44 AM.4A: Chain Rule Day 5
EXAMPLE 1 If y = tan u, u = v 1 dy, and v = ln x what is the value of v dx dy dy du dv = dx du dv dx dy dy du 1 dv ( tan u) = v ( ln x) dx du dv v dx dy ( ( )) 1 1 = sec 0 1+ 1 dx e e at x = e? 7/30/018 1:44 AM.4A: Chain Rule Day 6 x = e v= ln e= 1 1 u = 1 = 1 1 = 0 1
EXAMPLE 13 Determine the equation of the tangent line using the equation, y = 1 x + cos x at π, 1. 1 y = x + x ( cos ) 1/ 1 1/ y ' = x + cos x sin x 1 sin x y ' = x cos x ( ) ( ) 1 y 1 = x ( ) 1 y ' = 7/30/018 1:44 AM.4A: Chain Rule Day 7 y ' y ' ( ) sin cos 1 0 = 1 ( ) 1 ( ) = ( ) ( )
AP MULTIPLE CHOICE PRACTICE QUESTION 1 (NON-CALCULATOR) If f x = sin x π, then f 3π = (A) 1 (B) 3π (C) cos 3π (D) 3π 7/30/018 1:44 AM.4A: Chain Rule Day 8
AP MULTIPLE CHOICE PRACTICE QUESTION 1 (NON-CALCULATOR) If f x = sin x π, then f 3π = Vocabulary Connections and Process Answer and Justifications Derivative Chain Rule Trigonometric Derivative ( ) ( = sin ) f x x ( ) ( = )( ) f ' x cos x x ( ) ( ) ( ) ( ) ( ) f ' x = cos 3 3 ( ) = ( )( ) '( ) = cos( )( 3 ) f '( x) = ( 1)( 3 ) f ' x cos 3 3 f x D 7/30/018 1:44 AM.4A: Chain Rule Day 9
AP MULTIPLE CHOICE PRACTICE QUESTION (NON-CALCULATOR) Solve the derivative of f θ = sin θ (A) cos θ sin θ cos θ (B) sin θ (C) cos θ (D) sec θ 7/30/018 1:44 AM.4A: Chain Rule Day 30
AP MULTIPLE CHOICE PRACTICE QUESTION (NON-CALCULATOR) Solve the derivative of f θ = sin θ Vocabulary Connections and Process Answer and Justifications Derivative Chain Rule Trigonometric Derivative d d d sin = cos( ) d 1 sin = sin cos ( ) 1/ 1/ ( ) ( ) ( ) d d ( sin ) 1/ cos = sin A 7/30/018 1:44 AM.4A: Chain Rule Day 31
ASSIGNMENT Page 136 43-71 odd, 84, 87, 93, 10 7/30/018 1:44 AM.4A: Chain Rule Day 3