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Warm Up: pg 164 #59-61, 63, 64 and pg 156 QQ #4 61) A
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Derivatives of Inverse Trig Functions
Ch 3.8 Derivatives of Inverse Trig Functions
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Review from PreCalc: sin(ΞΈ) = πππ βπ¦π
a ratio an angle 60Β° 2 1 EX: sin(30Β°) = 1 2 30Β° 3 But inverse trig asks the opposite question EX: Arcsin( 1 2 ) = find the angle whose ratio is 1 2 =30Β°
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The 3 main Inverse Trig Functions
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Remember the restricted ranges β¦
Sine and Tangent are in Quadrant I and IV Cosine is in Quadrant I and II
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The Reciprocal Function is NOT the inverse function
Inverse notation: Reciprocal Notation: arcsin(ratio) or π ππ β1 (πππ‘ππ) 1 sinβ‘(30Β°) = csc(30Β°) This does not mean reciprocal
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What is the derivative of the arcsine?
** YOUR FRQ on the next TEST will include deriving one of the 3 main trig functions We find the derivative of y = arcsin(x) as follows: Start with y = sin(x) Inverse: x = sin(y) Take d/dx = cos(y)yβ Isolate yβ cosβ‘(π¦) =π¦β² Replace cos(y) With info from β π₯ 2 =π¦β² Triangle picture 1 x y 1β π₯ 2 We know y = arcsin(x) only exists between - Ο 2 β€ y β€ Ο 2 and -1 < x < 1
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What is the derivative of the arctan?
** YOUR FRQ on the next TEST will include deriving one of the 3 main trig functions Start with y = tan(x) Inverse: x = tan(y) Take d/dx = secΒ²(y)yβ Isolate yβ secΒ²(π¦) =π¦β² Replace cos(y) With info from π₯Β² =π¦β² Triangle picture 1β π₯ 2 x 1
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You Try! y = arccos(x) Start with y = cos(x) Inverse: x = cos(y)
Take d/dx = -sin(y)yβ Isolate yβ sin(π¦) =π¦β² Replace cos(y) With info from β π₯ 2 =π¦β² Triangle picture 1 1β π₯ 2 x
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What are the formal definitions of the inverse trig functionsβ
If u is a differentiable function of x with |u| < 1, we apply the chain rule to get: π ππ₯ π ππ β1 π’= 1 1βπ’Β² ππ’ ππ₯ π ππ₯ πππ β1 π’=β 1 1βπ’Β² ππ’ ππ₯ π ππ₯ π‘ππ β1 π’= 1 1+π’Β² ππ’ ππ₯ π ππ₯ πππ‘ β1 π’=β 1 1+π’Β² ππ’ ππ₯ π ππ₯ π ππ β1 π’= 1 |π’| π’ 2 β1 ππ’ ππ₯ π ππ₯ ππ π β1 π’=β 1 |π’| π’ 2 β1 ππ’ ππ₯
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Find Full Assignment Online
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EX 1: Applying the Arcsine formula
a) Find π ππ₯ ( π ππ β1 ( 3 π₯Β² )) b) Find π ππ₯ ( 1 π ππ β1 (2π₯) )
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EX 2: Applying the Arctan formula
A particle moves along the x-axis so its position at any time t β₯ 0 is x(t) = π‘ππ β1 ( π‘ ). What is the velocity of the particle when t = 16?
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You try! π ππ₯ ( π ππ β1 (1βπ₯)) b) Find π ππ₯ ( ππ π β1 ( π₯ 2 )) c) π ππ₯ ( π ππ β1 (5 π₯ 4 )) d) A particle moves along the x-axis so its position at any time t β₯ is x(t) = π ππ β1 ( π‘ 4 ). What is the velocity of the particle when t = 4?
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EX 3: Derivatives of inverse functions
a) Find an equation for the line tangent to the graph of y = tan(x) at the point ( π 4 , 1) b) Find an equation for the line tangent to the graph of y = π‘ππ β1 (x) at the point (1, π 4 )
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What did you notice in the last EX?
*The Derivative of a function is the reciprocal of the derivative of its inverse*
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EX 4: Derivatives of inverse functions
Let f(x) = π₯ 5 +2 π₯ 3 +π₯β1. Find f(1) b) Find fβ(1) c) Find π β1 (3) d) Find π β1 (3)β² So for #28: F(1) = 3 fβ(1) =12 (3,1) fβ(3) = 1/12
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You Try! Let f(x) = cos π₯ +3π₯ Find f(0) b) Find fβ(0)
c) Find π β1 (1) d) Find π β1 (1)β²
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3.8: pg 170 QR #1-3, 6- 10, EX: #1- 19odd, 24, 26, 29 DIDNβT GET TO 3.8B
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