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Derivatives of Inverse Trigonometric Functions

Chapter 4.3

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Derivatives of Inverse Functions

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Derivatives of Inverse Functions

How are the slopes of the tangent lines from the previous slide related? The slope of the tangent line at ( 𝑥 0 , 𝑦 0 ) for 𝑓 is 𝑦= 𝑓 ′ 𝑥 0 𝑥− 𝑥 0 +𝑓( 𝑥 0 ) Since the tangent line at ( 𝑦 0 , 𝑥 0 ) for 𝑔 is a reflection of the above tangent line, we can find its equation in terms of 𝑓′ by swapping 𝑥 and 𝑦, then solving for 𝑦 𝑥= 𝑓 ′ 𝑥 0 𝑦− 𝑥 0 +𝑓 𝑥 0 𝑦= 1 𝑓 ′ 𝑥 0 𝑥+ 𝑥 0 𝑓 ′ 𝑥 0 −𝑓 𝑥 0 𝑓 ′ 𝑥 0

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Derivatives of Inverse Functions

𝑦= 1 𝑓 ′ 𝑥 0 𝑥+ 𝑥 0 𝑓 ′ 𝑥 0 −𝑓 𝑥 0 𝑓 ′ 𝑥 0 The slopes are reciprocals This gives us a hint about the relationship between the derivative of a function and its inverse (when an inverse exists) The next theorem asserts the existence of the derivative of inverse function (without proof) In addition, the relationship is given and can be proven using the Chain Rule

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Derivatives of Inverse Functions

THEOREM: If 𝑓 is differentiable at every point of an interval 𝐼 and 𝑑𝑓 𝑑𝑥 is never zero on 𝐼, then 𝑓 has an inverse and 𝑔 𝑥 = 𝑓 −1 (𝑥) is differentiable at every point of the interval 𝑓(𝐼). Furthermore, 𝑑 𝑑𝑥 𝑓 −1 𝑥 = 𝑔 ′ 𝑥 = 1 𝑓 ′ 𝑔 𝑥

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Derivatives of Inverse Functions

THEOREM: PROOF: We can easily prove the last part of this theorem using the Chain Rule. If 𝑔 𝑥 = 𝑓 −1 (𝑥), then 𝑓 𝑔 𝑥 =𝑥 Now, 𝑔 is an inner function. So we apply the Chain Rule to get 𝑔 ′ 𝑥 ⋅ 𝑓 ′ 𝑔 𝑥 =1 𝑔 ′ 𝑥 = 1 𝑓 ′ 𝑔 𝑥

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Derivative of Arcsine The function 𝑥= sin 𝑦 is differentiable on the open interval − 𝜋 2 < 𝑦< 𝜋 2 Its derivative cos 𝑦 is positive (i.e., non-zero) in the same interval By the previous theorem, there must exist a function 𝑦= arcsin 𝑥 that is differentiable on the interval We can use implicit differentiation to find the derivative of arcsine

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Derivative of Arcsine 𝑦= arcsin 𝑥 ⟹ sin 𝑦 =𝑥 Since 𝑦 is a function of 𝑥, then sin 𝑦 =𝑥 is an implicit function 𝑦 ′ ⋅ cos 𝑦 =1⟹ 𝑦 ′ = 1 cos 𝑦 This division presents no problems since the cosine is non-zero in the interval − 𝜋 2 <𝑦< 𝜋 2 Since cos 𝑦= 1− sin 𝑦 2 , then 𝑦 ′ = 1 1− sin 𝑦 2

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Derivative of Arcsine cos 𝑦= 1− sin 𝑦 2 ⟹ 𝑦 ′ = 1 1− sin 𝑦 2 But sin 𝑦 =𝑥, so 𝑦′= 1 1− 𝑥 2 More generally, if 𝑢 is a differentiable function of 𝑥, then 𝑑 𝑑𝑥 arcsin 𝑢 = 𝑢 ′ 1− 𝑢 2 , 𝑢 <1

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Example 1: Applying the Derivative

Differentiate 𝑦= arcsin 𝑥 2 .

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Example 1: Applying the Derivative

Differentiate 𝑦= arcsin 𝑥 2 . We have 𝑥 2 as an inner function, so if 𝑢= 𝑥 2 and 𝑦= arcsin 𝑢 , then 𝑑𝑦 𝑑𝑥 = 𝑑𝑦 𝑑𝑢 ⋅ 𝑑𝑢 𝑑𝑥 = 1 1− 𝑢 2 ⋅2𝑥= 2𝑥 1− 𝑥 = 2𝑥 1− 𝑥 4

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Derivative of Arctangent

The domain of arctangent is −∞,∞ It is differentiable for all real numbers We find the derivative of arctangent in the same way we found the derivative of arcsine 𝑦= arctan 𝑥 ⟹𝑥= tan 𝑦 Use implicit differentiation (since 𝑦 is a function of 𝑥) 1= sec 2 𝑦 ⋅ 𝑦 ′ ⟹ 𝑦 ′ = 1 sec 2 𝑦

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Derivative of Arctangent

1= sec 2 𝑦 ⋅ 𝑦 ′ ⟹ 𝑦 ′ = 1 sec 2 𝑦 Use the Pythagorean identity sec 2 𝑦 =1+ tan 2 𝑦 : 𝑦 ′ = 1 1+ tan 2 𝑦 But 𝑥= tan 𝑦 so we get 𝑦 ′ = 1 1+ 𝑥 2

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Derivative of Arctangent

𝑦 ′ = 1 1+ 𝑥 2 Note that this is defined for all 𝑥 More generally, if 𝑢 is a function of 𝑥, then 𝑦= arctan 𝑢 ⟹ 𝑦 ′ = 𝑢 ′ 1+ 𝑢 2

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Example 2: A Moving Particle

A particle moves along the 𝑥-axis so that its position at any time 𝑡≥0 is given by 𝑥 𝑡 = arctan 𝑡 . What is the velocity of the particle when 𝑡=16?

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Example 2: A Moving Particle

A particle moves along the 𝑥-axis so that its position at any time 𝑡≥0 is given by 𝑥 𝑡 = arctan 𝑡 . What is the velocity of the particle when 𝑡=16? We have 𝑡 as an inner function, so if 𝑢= 𝑡 and 𝑥= arctan 𝑢 , then 𝑑𝑥 𝑑𝑡 = 𝑑𝑥 𝑑𝑢 ⋅ 𝑑𝑢 𝑑𝑡 = 1 1+ 𝑢 2 ⋅ 1 2 𝑡 = 1 2 𝑡 𝑡 = 1 2 𝑡 1+𝑡

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Example 2: A Moving Particle

A particle moves along the 𝑥-axis so that its position at any time 𝑡≥0 is given by 𝑥 𝑡 = arctan 𝑡 . What is the velocity of the particle when 𝑡=16? 𝑑𝑥 𝑑𝑡 = 1 2 𝑡 1+𝑡 At 𝑡=16 the velocity is 𝑑𝑥 𝑑𝑡 = = 1 8⋅17 = 1 136

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Derivative of Arcsecant

We must be careful to follow the requirements of the theorem on derivatives of inverses when determining the derivative of arcsecant These are sec 𝑥 must be differentiable over its domain, 0, 𝜋 2 ∪ 𝜋 2 ,𝜋 𝑑 𝑑𝑥 [ sec 𝑥 ] must not equal zero in this interval Note from the graph of 𝑦= sec 𝑥 that the derivatives at both −1 and 1 are horizontal lines So the interval we will choose will be 0, 𝜋 2 ∪ 𝜋 2 ,𝜋

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Derivative of Arcsecant

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Derivative of Arcsecant

Having set up the conditions required by the theorem, we now conclude that sec 𝑥 has an inverse that that the inverse is differentiable over 𝑓 𝐼 , or −∞,−1 ∪ 1,∞ Now, if 𝑦= arcsec 𝑥 , then sec 𝑦 =𝑥 Differentiate this implicitly sec 𝑦 =𝑥⟹ sec 𝑦 tan 𝑦 ⋅ 𝑦 ′ =1 𝑦 ′ = 1 sec 𝑦 tan 𝑦 Note that the denominator will not be zero because of the domain we have chosen

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Derivative of Arcsecant

𝑦 ′ = 1 sec 𝑦 tan 𝑦 It remains to rewrite this as a function of 𝑥 Since 𝑥= sec 𝑦 and sec 2 𝑦 =1+ tan 2 𝑦 ⟹ tan 𝑦 =± sec 2 𝑦 −1 , then 𝑦 ′ =± 1 𝑥 𝑥 2 −1 How do we deal with the ±?

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Derivative of Arcsecant

𝑦 ′ =± 1 𝑥 𝑥 2 −1 Recall that the domain of 𝑦′ is −∞,−1 ∪ 1,∞ , or 𝑥 >1 We can rewrite the derivative as a piecewise function 𝑦 ′ = + 1 𝑥 𝑥 2 −1 , 𝑥>1 − 1 𝑥 𝑥 2 −1 , 𝑥<−1

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Derivative of Arcsecant

𝑦 ′ = + 1 𝑥 𝑥 2 −1 , 𝑥>1 − 1 𝑥 𝑥 2 −1 , 𝑥<−1 This is the same as 𝑦 ′ = 1 𝑥 𝑥 2 −1 = 1 𝑥 𝑥 2 −1 = 1 𝑥 𝑥 2 −1

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Derivative of Arcsecant

Hence, 𝑑 𝑑𝑥 arcsec 𝑥 = 1 𝑥 𝑥 2 −1 If 𝑢 is a function of 𝑥, then 𝑑 𝑑𝑥 arcsec 𝑢 = 𝑢 ′ 𝑢 𝑢 2 −1

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Example 3: Finding the Derivative of Arcsecant

Differentiate 𝑓 𝑥 = arcsec 5 𝑥 4 .

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Example 3: Finding the Derivative of Arcsecant

Differentiate 𝑓 𝑥 = arcsec 5 𝑥 4 . 5 𝑥 4 is an inner function. Using the Chain Rule we have 𝑢=5 𝑥 4 and 𝑦= arcsec 𝑢 𝑑𝑢 𝑑𝑥 =20 𝑥 3 and 𝑑𝑦 𝑑𝑢 = 1 𝑢 𝑢 2 −1 Therefore, 𝑑𝑦 𝑑𝑥 = 𝑑𝑦 𝑑𝑢 ⋅ 𝑑𝑢 𝑑𝑥 =20 𝑥 3 ⋅ 1 5 𝑥 𝑥 −1 = 4 𝑥 25 𝑥 8 −1

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Derivatives of the Other Three Inverses

We could use the same procedure to find the derivatives of arccosine, arccotangent, and arccosecant But it is easier to use identities; the one below shows how arccosine and arcsine are related 𝑦= sin 𝑥 ⟹ arcsin 𝑦 =𝑥 𝑦= cos 𝜋 2 −𝑥 ⟹ arccos 𝑦= 𝜋 2 −𝑥 Therefore, arccos 𝑦 = 𝜋 2 − arcsin 𝑦

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Derivatives of the Other Three Inverses

The other two are similarly derived They are arccos 𝑥 = 𝜋 2 − arcsin 𝑥 arccot 𝑥 = 𝜋 2 − arctan 𝑥 arccsc 𝑥 = 𝜋 2 − arcsec 𝑥

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Derivatives of the Other Three Inverses

The derivative of arccosecant is 𝑑 𝑑𝑥 arccos 𝑥 = 𝑑 𝑑𝑥 𝜋 2 − arcsin 𝑥 =0− 𝑑 𝑑𝑥 arcsin 𝑥 = −1 1− 𝑥 2

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Derivatives of the Other Three Inverses

All six derivatives are given below 𝑑 𝑑𝑥 arcsin 𝑥 = 1 1− 𝑥 𝑑 𝑑𝑥 arccos 𝑥 = −1 1− 𝑥 2 𝑑 𝑑𝑥 arctan 𝑥 = 1 𝑥 𝑑 𝑑𝑥 arccot 𝑥 = −1 𝑥 2 +1 𝑑 𝑑𝑥 arcsec 𝑥 = 1 𝑥 𝑥 2 −1 𝑑 𝑑𝑥 arccsc 𝑥 = −1 𝑥 𝑥 2 −1

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Calculator Conversion Identities

Calculators have keys for arcsine, arccosine, and arctangent The following should be used to calculate arccosecant, arcsecant, and arccotangent arcsec 𝑥 = arccos 1 𝑥 arccsc 𝑥 = arcsin 1 𝑥 arccot 𝑥 = 𝜋 2 − arctan 𝑥

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Example 4: A Tangent Line to the Arccotangent Curve

Find an equation for the line tangent to the graph of 𝑦= arccot 𝑥 at 𝑥= −1.

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Example 4: A Tangent Line to the Arccotangent Curve

Find an equation for the line tangent to the graph of 𝑦= arccot 𝑥 at 𝑥=−1. The derivative of arccot 𝑥 is 𝑑 𝑑𝑥 arccot 𝑥 = −1 𝑥 2 +1 The slope of the tangent line at 𝑥=−1 is −1 − =− 1 2 The tangent line equation is 𝑦=− 1 𝑥 𝑥+1 + 3𝜋 4

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Exercise 4.3

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