lesson 16: inverse trigonometric functions (section 021 slides)
DESCRIPTION
We cover the inverses to the trigonometric functions sine, cosine, tangent, cotangent, secant, cosecant, and their derivatives. The remarkable fact is that although these functions and their inverses are transcendental (complicated) functions, the derivatives are algebraic functions. Also, we meet my all-time favorite function: arctan.TRANSCRIPT
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Section 3.5Inverse Trigonometric
FunctionsV63.0121.021, Calculus I
New York University
November 2, 2010
Announcements
I Midterm grades have been submittedI Quiz 3 this week in recitation on Section 2.6, 2.8, 3.1, 3.2I Thank you for the evaluations
. . . . . .
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. . . . . .
Announcements
I Midterm grades have beensubmitted
I Quiz 3 this week inrecitation on Section 2.6,2.8, 3.1, 3.2
I Thank you for theevaluations
V63.0121.021, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 2, 2010 2 / 40
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. . . . . .
Evaluations: The good
I “Exceptional competence and effectively articulate. (Do not firehim)”
I “Good guy”I “He’s the clear man”I “Love the juices”
V63.0121.021, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 2, 2010 3 / 40
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. . . . . .
Evaluations: The good
I “Exceptional competence and effectively articulate. (Do not firehim)”
I “Good guy”
I “He’s the clear man”I “Love the juices”
V63.0121.021, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 2, 2010 3 / 40
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. . . . . .
Evaluations: The good
I “Exceptional competence and effectively articulate. (Do not firehim)”
I “Good guy”I “He’s the clear man”
I “Love the juices”
V63.0121.021, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 2, 2010 3 / 40
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. . . . . .
Evaluations: The good
I “Exceptional competence and effectively articulate. (Do not firehim)”
I “Good guy”I “He’s the clear man”I “Love the juices”
V63.0121.021, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 2, 2010 3 / 40
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. . . . . .
Evaluations: The bad
I Too fast, not enough examples
I Not enough time to do everythingI Lecture is not the only learning time (recitation and independent
study)I I try to balance concept and procedure
I Too many proofsI In this course we care about conceptsI There will be conceptual problems on the examI Concepts are the keys to overcoming templated problems
V63.0121.021, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 2, 2010 4 / 40
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. . . . . .
Evaluations: The bad
I Too fast, not enough examplesI Not enough time to do everythingI Lecture is not the only learning time (recitation and independent
study)I I try to balance concept and procedure
I Too many proofsI In this course we care about conceptsI There will be conceptual problems on the examI Concepts are the keys to overcoming templated problems
V63.0121.021, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 2, 2010 4 / 40
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. . . . . .
Evaluations: The bad
I Too fast, not enough examplesI Not enough time to do everythingI Lecture is not the only learning time (recitation and independent
study)I I try to balance concept and procedure
I Too many proofs
I In this course we care about conceptsI There will be conceptual problems on the examI Concepts are the keys to overcoming templated problems
V63.0121.021, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 2, 2010 4 / 40
![Page 10: Lesson 16: Inverse Trigonometric Functions (Section 021 slides)](https://reader034.vdocuments.site/reader034/viewer/2022052307/5559c86ad8b42a93208b45f5/html5/thumbnails/10.jpg)
. . . . . .
Evaluations: The bad
I Too fast, not enough examplesI Not enough time to do everythingI Lecture is not the only learning time (recitation and independent
study)I I try to balance concept and procedure
I Too many proofsI In this course we care about conceptsI There will be conceptual problems on the examI Concepts are the keys to overcoming templated problems
V63.0121.021, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 2, 2010 4 / 40
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. . . . . .
Evaluations: technological comments
I Smart board issuesI laser pointer visibilityI slides sometimes move fast
V63.0121.021, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 2, 2010 5 / 40
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. . . . . .
Evaluations: The ugly
I “If class was even remotely interesting this class would beawesome.”
I “Sometimes condescending/rude.”I “Can’t pick his nose without checking his notes, and he still gets it
wrong the first time.”I “If I were chained to a desk and forced to see this guy teach, I
would chew my arm off in order to get free.”
V63.0121.021, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 2, 2010 6 / 40
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. . . . . .
Evaluations: The ugly
I “If class was even remotely interesting this class would beawesome.”
I “Sometimes condescending/rude.”
I “Can’t pick his nose without checking his notes, and he still gets itwrong the first time.”
I “If I were chained to a desk and forced to see this guy teach, Iwould chew my arm off in order to get free.”
V63.0121.021, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 2, 2010 6 / 40
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. . . . . .
Evaluations: The ugly
I “If class was even remotely interesting this class would beawesome.”
I “Sometimes condescending/rude.”I “Can’t pick his nose without checking his notes, and he still gets it
wrong the first time.”
I “If I were chained to a desk and forced to see this guy teach, Iwould chew my arm off in order to get free.”
V63.0121.021, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 2, 2010 6 / 40
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. . . . . .
Evaluations: The ugly
I “If class was even remotely interesting this class would beawesome.”
I “Sometimes condescending/rude.”I “Can’t pick his nose without checking his notes, and he still gets it
wrong the first time.”I “If I were chained to a desk and forced to see this guy teach, I
would chew my arm off in order to get free.”
V63.0121.021, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 2, 2010 6 / 40
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. . . . . .
A slide on slides
I ProI “Powerpoints explain topics carefully step-by-step”I “Powerpoint and lesson flow smoothly”I “Can visualize material well”I “I like that you post slides beforehand”
I ConI “I would like to have him use the chalkboard more.”I “It’s so unnatural to learn math via powerpoint.”I “I hate powerpoint.”
I Why I like themI Board handwriting not an issueI Easy to put online; notetaking is more than transcription
I What we can doI if you have suggestions for details to put in, I’m listeningI Feel free to ask me to fill in something on the board
V63.0121.021, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 2, 2010 7 / 40
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. . . . . .
A slide on slides
I ProI “Powerpoints explain topics carefully step-by-step”I “Powerpoint and lesson flow smoothly”I “Can visualize material well”I “I like that you post slides beforehand”
I ConI “I would like to have him use the chalkboard more.”I “It’s so unnatural to learn math via powerpoint.”I “I hate powerpoint.”
I Why I like themI Board handwriting not an issueI Easy to put online; notetaking is more than transcription
I What we can doI if you have suggestions for details to put in, I’m listeningI Feel free to ask me to fill in something on the board
V63.0121.021, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 2, 2010 7 / 40
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. . . . . .
A slide on slides
I ProI “Powerpoints explain topics carefully step-by-step”I “Powerpoint and lesson flow smoothly”I “Can visualize material well”I “I like that you post slides beforehand”
I ConI “I would like to have him use the chalkboard more.”I “It’s so unnatural to learn math via powerpoint.”I “I hate powerpoint.”
I Why I like themI Board handwriting not an issueI Easy to put online; notetaking is more than transcription
I What we can doI if you have suggestions for details to put in, I’m listeningI Feel free to ask me to fill in something on the board
V63.0121.021, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 2, 2010 7 / 40
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. . . . . .
My handwriting
V63.0121.021, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 2, 2010 8 / 40
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. . . . . .
A slide on slides
I ProI “Powerpoints explain topics carefully step-by-step”I “Powerpoint and lesson flow smoothly”I “Can visualize material well”I “I like that you post slides beforehand”
I ConI “I would like to have him use the chalkboard more.”I “It’s so unnatural to learn math via powerpoint.”I “I hate powerpoint.”
I Why I like themI Board handwriting not an issueI Easy to put online; notetaking is more than transcription
I What we can doI if you have suggestions for details to put in, I’m listeningI Feel free to ask me to fill in something on the board
V63.0121.021, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 2, 2010 9 / 40
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. . . . . .
Objectives
I Know the definitions,domains, ranges, andother properties of theinverse trignometricfunctions: arcsin, arccos,arctan, arcsec, arccsc,arccot.
I Know the derivatives of theinverse trignometricfunctions.
V63.0121.021, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 2, 2010 10 / 40
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. . . . . .
Outline
Inverse Trigonometric Functions
Derivatives of Inverse Trigonometric FunctionsArcsineArccosineArctangentArcsecant
Applications
V63.0121.021, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 2, 2010 11 / 40
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. . . . . .
What is an inverse function?
DefinitionLet f be a function with domain D and range E. The inverse of f is thefunction f−1 defined by:
f−1(b) = a,
where a is chosen so that f(a) = b.
Sof−1(f(x)) = x, f(f−1(x)) = x
V63.0121.021, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 2, 2010 12 / 40
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. . . . . .
What is an inverse function?
DefinitionLet f be a function with domain D and range E. The inverse of f is thefunction f−1 defined by:
f−1(b) = a,
where a is chosen so that f(a) = b.
Sof−1(f(x)) = x, f(f−1(x)) = x
V63.0121.021, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 2, 2010 12 / 40
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. . . . . .
What functions are invertible?
In order for f−1 to be a function, there must be only one a in Dcorresponding to each b in E.
I Such a function is called one-to-oneI The graph of such a function passes the horizontal line test: any
horizontal line intersects the graph in exactly one point if at all.I If f is continuous, then f−1 is continuous.
V63.0121.021, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 2, 2010 13 / 40
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. . . . . .
Graphing the inverse function
I If b = f(a), then f−1(b) = a.
I So if (a,b) is on the graphof f, then (b,a) is on thegraph of f−1.
I On the xy-plane, the point(b,a) is the reflection of(a,b) in the line y = x.
I Therefore:
..x
.y
. .(a,b)
..(b,a)
.y = x
FactThe graph of f−1 is the reflection of the graph of f in the line y = x.
V63.0121.021, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 2, 2010 14 / 40
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. . . . . .
Graphing the inverse function
I If b = f(a), then f−1(b) = a.I So if (a,b) is on the graph
of f, then (b,a) is on thegraph of f−1.
I On the xy-plane, the point(b,a) is the reflection of(a,b) in the line y = x.
I Therefore:
..x
.y
. .(a,b)
..(b,a)
.y = x
FactThe graph of f−1 is the reflection of the graph of f in the line y = x.
V63.0121.021, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 2, 2010 14 / 40
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. . . . . .
Graphing the inverse function
I If b = f(a), then f−1(b) = a.I So if (a,b) is on the graph
of f, then (b,a) is on thegraph of f−1.
I On the xy-plane, the point(b,a) is the reflection of(a,b) in the line y = x.
I Therefore:
..x
.y
. .(a,b)
..(b,a)
.y = x
FactThe graph of f−1 is the reflection of the graph of f in the line y = x.
V63.0121.021, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 2, 2010 14 / 40
![Page 29: Lesson 16: Inverse Trigonometric Functions (Section 021 slides)](https://reader034.vdocuments.site/reader034/viewer/2022052307/5559c86ad8b42a93208b45f5/html5/thumbnails/29.jpg)
. . . . . .
Graphing the inverse function
I If b = f(a), then f−1(b) = a.I So if (a,b) is on the graph
of f, then (b,a) is on thegraph of f−1.
I On the xy-plane, the point(b,a) is the reflection of(a,b) in the line y = x.
I Therefore:
..x
.y
. .(a,b)
..(b,a)
.y = x
FactThe graph of f−1 is the reflection of the graph of f in the line y = x.
V63.0121.021, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 2, 2010 14 / 40
![Page 30: Lesson 16: Inverse Trigonometric Functions (Section 021 slides)](https://reader034.vdocuments.site/reader034/viewer/2022052307/5559c86ad8b42a93208b45f5/html5/thumbnails/30.jpg)
. . . . . .
Graphing the inverse function
I If b = f(a), then f−1(b) = a.I So if (a,b) is on the graph
of f, then (b,a) is on thegraph of f−1.
I On the xy-plane, the point(b,a) is the reflection of(a,b) in the line y = x.
I Therefore:
..x
.y
. .(a,b)
..(b,a)
.y = x
FactThe graph of f−1 is the reflection of the graph of f in the line y = x.
V63.0121.021, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 2, 2010 14 / 40
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. . . . . .
Graphing the inverse function
I If b = f(a), then f−1(b) = a.I So if (a,b) is on the graph
of f, then (b,a) is on thegraph of f−1.
I On the xy-plane, the point(b,a) is the reflection of(a,b) in the line y = x.
I Therefore:.
.x
.y
. .(a,b)
..(b,a)
.y = x
FactThe graph of f−1 is the reflection of the graph of f in the line y = x.
V63.0121.021, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 2, 2010 14 / 40
![Page 32: Lesson 16: Inverse Trigonometric Functions (Section 021 slides)](https://reader034.vdocuments.site/reader034/viewer/2022052307/5559c86ad8b42a93208b45f5/html5/thumbnails/32.jpg)
. . . . . .
Graphing the inverse function
I If b = f(a), then f−1(b) = a.I So if (a,b) is on the graph
of f, then (b,a) is on thegraph of f−1.
I On the xy-plane, the point(b,a) is the reflection of(a,b) in the line y = x.
I Therefore:.
.x
.y
. .(a,b)
..(b,a)
.y = x
FactThe graph of f−1 is the reflection of the graph of f in the line y = x.
V63.0121.021, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 2, 2010 14 / 40
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. . . . . .
arcsin
Arcsin is the inverse of the sine function after restriction to [−π/2, π/2].
. .x
.y
.sin.
.−π
2
.
.π
2
.y = x
.
. .arcsin
I The domain of arcsin is [−1,1]I The range of arcsin is
[−π
2,π
2
]
V63.0121.021, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 2, 2010 15 / 40
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. . . . . .
arcsin
Arcsin is the inverse of the sine function after restriction to [−π/2, π/2].
. .x
.y
.sin.
.
.
.−π
2
.
.π
2
.y = x
.
. .arcsin
I The domain of arcsin is [−1,1]I The range of arcsin is
[−π
2,π
2
]
V63.0121.021, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 2, 2010 15 / 40
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. . . . . .
arcsin
Arcsin is the inverse of the sine function after restriction to [−π/2, π/2].
. .x
.y
.sin.
.
.
.−π
2
.
.π
2
.y = x
.
. .arcsin
I The domain of arcsin is [−1,1]I The range of arcsin is
[−π
2,π
2
]
V63.0121.021, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 2, 2010 15 / 40
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. . . . . .
arcsin
Arcsin is the inverse of the sine function after restriction to [−π/2, π/2].
. .x
.y
.sin.
.
.
.−π
2
.
.π
2
.y = x
.
. .arcsin
I The domain of arcsin is [−1,1]I The range of arcsin is
[−π
2,π
2
]
V63.0121.021, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 2, 2010 15 / 40
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. . . . . .
arccos
Arccos is the inverse of the cosine function after restriction to [0, π]
. .x
.y
.cos..0
..π
.y = x
.
. .arccos
I The domain of arccos is [−1,1]I The range of arccos is [0, π]
V63.0121.021, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 2, 2010 16 / 40
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. . . . . .
arccos
Arccos is the inverse of the cosine function after restriction to [0, π]
. .x
.y
.cos.
.
..0
..π
.y = x
.
. .arccos
I The domain of arccos is [−1,1]I The range of arccos is [0, π]
V63.0121.021, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 2, 2010 16 / 40
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. . . . . .
arccos
Arccos is the inverse of the cosine function after restriction to [0, π]
. .x
.y
.cos.
.
..0
..π
.y = x
.
. .arccos
I The domain of arccos is [−1,1]I The range of arccos is [0, π]
V63.0121.021, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 2, 2010 16 / 40
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. . . . . .
arccos
Arccos is the inverse of the cosine function after restriction to [0, π]
. .x
.y
.cos.
.
..0
..π
.y = x
.
. .arccos
I The domain of arccos is [−1,1]I The range of arccos is [0, π]
V63.0121.021, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 2, 2010 16 / 40
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. . . . . .
arctan
Arctan is the inverse of the tangent function after restriction to[−π/2, π/2].
. .x
.y
.tan
.−3π2
.−π
2.π
2 .3π2
.y = x
.arctan
.−π
2
.π
2
I The domain of arctan is (−∞,∞)
I The range of arctan is(−π
2,π
2
)I lim
x→∞arctan x =
π
2, limx→−∞
arctan x = −π
2
V63.0121.021, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 2, 2010 17 / 40
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. . . . . .
arctan
Arctan is the inverse of the tangent function after restriction to[−π/2, π/2].
. .x
.y
.tan
.−3π2
.−π
2.π
2 .3π2
.y = x
.arctan
.−π
2
.π
2
I The domain of arctan is (−∞,∞)
I The range of arctan is(−π
2,π
2
)I lim
x→∞arctan x =
π
2, limx→−∞
arctan x = −π
2
V63.0121.021, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 2, 2010 17 / 40
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. . . . . .
arctan
Arctan is the inverse of the tangent function after restriction to[−π/2, π/2].
. .x
.y
.tan
.−3π2
.−π
2.π
2 .3π2
.y = x
.arctan
.−π
2
.π
2
I The domain of arctan is (−∞,∞)
I The range of arctan is(−π
2,π
2
)I lim
x→∞arctan x =
π
2, limx→−∞
arctan x = −π
2
V63.0121.021, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 2, 2010 17 / 40
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. . . . . .
arctan
Arctan is the inverse of the tangent function after restriction to[−π/2, π/2].
. .x
.y
.tan
.−3π2
.−π
2.π
2 .3π2
.y = x
.arctan
.−π
2
.π
2
I The domain of arctan is (−∞,∞)
I The range of arctan is(−π
2,π
2
)I lim
x→∞arctan x =
π
2, limx→−∞
arctan x = −π
2
V63.0121.021, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 2, 2010 17 / 40
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. . . . . .
arcsec
Arcsecant is the inverse of secant after restriction to[0, π/2) ∪ (π,3π/2].
. .x
.y
.sec
.−3π2
.−π
2.π
2 .3π2
.y = x
.
.
.π
2
.3π2
I The domain of arcsec is (−∞,−1] ∪ [1,∞)
I The range of arcsec is[0,
π
2
)∪(π2, π]
I limx→∞
arcsec x =π
2, limx→−∞
arcsec x =3π2
V63.0121.021, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 2, 2010 18 / 40
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. . . . . .
arcsec
Arcsecant is the inverse of secant after restriction to[0, π/2) ∪ (π,3π/2].
. .x
.y
.sec
.−3π2
.−π
2.π
2 .3π2
.
.
.y = x
.
.
.π
2
.3π2
I The domain of arcsec is (−∞,−1] ∪ [1,∞)
I The range of arcsec is[0,
π
2
)∪(π2, π]
I limx→∞
arcsec x =π
2, limx→−∞
arcsec x =3π2
V63.0121.021, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 2, 2010 18 / 40
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. . . . . .
arcsec
Arcsecant is the inverse of secant after restriction to[0, π/2) ∪ (π,3π/2].
. .x
.y
.sec
.−3π2
.−π
2.π
2 .3π2
.
.
.y = x
.
.
.π
2
.3π2
I The domain of arcsec is (−∞,−1] ∪ [1,∞)
I The range of arcsec is[0,
π
2
)∪(π2, π]
I limx→∞
arcsec x =π
2, limx→−∞
arcsec x =3π2
V63.0121.021, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 2, 2010 18 / 40
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. . . . . .
arcsec
Arcsecant is the inverse of secant after restriction to[0, π/2) ∪ (π,3π/2].
. .x
.y
.sec
.−3π2
.−π
2.π
2 .3π2
.
.
.y = x
.
.
.π
2
.3π2
I The domain of arcsec is (−∞,−1] ∪ [1,∞)
I The range of arcsec is[0,
π
2
)∪(π2, π]
I limx→∞
arcsec x =π
2, limx→−∞
arcsec x =3π2
V63.0121.021, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 2, 2010 18 / 40
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. . . . . .
Values of Trigonometric Functions
x 0π
6π
4π
3π
2
sin x 0 12
√22
√32
1
cos x 1√32
√22
12
0
tan x 01√3
1√3 undef
cot x undef√3 1
1√3
0
sec x 12√3
2√2
2 undef
csc x undef 22√2
2√3
1
V63.0121.021, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 2, 2010 19 / 40
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. . . . . .
Check: Values of inverse trigonometric functions
Example
FindI arcsin(1/2)I arctan(−1)
I arccos
(−√22
)
Solution
Iπ
6I −π
4I
3π4
V63.0121.021, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 2, 2010 20 / 40
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. . . . . .
Check: Values of inverse trigonometric functions
Example
FindI arcsin(1/2)I arctan(−1)
I arccos
(−√22
)
Solution
Iπ
6
I −π
4I
3π4
V63.0121.021, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 2, 2010 20 / 40
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. . . . . .
What is arctan(−1)?
. .
.
..3π/4
..−π/4
.sin(3π/4) =√22
.cos(3π/4) = −√22.sin(π/4) = −
√22
.cos(π/4) =√22
I Yes, tan(3π4
)= −1
I But, the range of arctan is(−π
2,π
2
)I Another angle whose
tangent is −1 is −π
4, and
this is in the right range.I So arctan(−1) = −π
4
V63.0121.021, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 2, 2010 21 / 40
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. . . . . .
What is arctan(−1)?
. .
.
..3π/4
..−π/4
.sin(3π/4) =√22
.cos(3π/4) = −√22
.sin(π/4) = −√22
.cos(π/4) =√22
I Yes, tan(3π4
)= −1
I But, the range of arctan is(−π
2,π
2
)I Another angle whose
tangent is −1 is −π
4, and
this is in the right range.I So arctan(−1) = −π
4
V63.0121.021, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 2, 2010 21 / 40
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. . . . . .
What is arctan(−1)?
. .
.
..3π/4
..−π/4
.sin(3π/4) =√22
.cos(3π/4) = −√22
.sin(π/4) = −√22
.cos(π/4) =√22
I Yes, tan(3π4
)= −1
I But, the range of arctan is(−π
2,π
2
)
I Another angle whosetangent is −1 is −π
4, and
this is in the right range.I So arctan(−1) = −π
4
V63.0121.021, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 2, 2010 21 / 40
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. . . . . .
What is arctan(−1)?
. .
.
..3π/4
..−π/4
.sin(3π/4) =√22
.cos(3π/4) = −√22
.sin(π/4) = −√22
.cos(π/4) =√22
I Yes, tan(3π4
)= −1
I But, the range of arctan is(−π
2,π
2
)I Another angle whose
tangent is −1 is −π
4, and
this is in the right range.
I So arctan(−1) = −π
4
V63.0121.021, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 2, 2010 21 / 40
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. . . . . .
What is arctan(−1)?
. .
.
..3π/4
..−π/4
.sin(3π/4) =√22
.cos(3π/4) = −√22
.sin(π/4) = −√22
.cos(π/4) =√22
I Yes, tan(3π4
)= −1
I But, the range of arctan is(−π
2,π
2
)I Another angle whose
tangent is −1 is −π
4, and
this is in the right range.I So arctan(−1) = −π
4
V63.0121.021, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 2, 2010 21 / 40
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. . . . . .
Check: Values of inverse trigonometric functions
Example
FindI arcsin(1/2)I arctan(−1)
I arccos
(−√22
)
Solution
Iπ
6I −π
4
I3π4
V63.0121.021, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 2, 2010 22 / 40
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. . . . . .
Check: Values of inverse trigonometric functions
Example
FindI arcsin(1/2)I arctan(−1)
I arccos
(−√22
)
Solution
Iπ
6I −π
4I
3π4
V63.0121.021, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 2, 2010 22 / 40
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. . . . . .
Caution: Notational ambiguity
..sin2 x = (sin x)2 .sin−1 x = (sin x)−1
I sinn x means the nth power of sin x, except when n = −1!I The book uses sin−1 x for the inverse of sin x, and never for
(sin x)−1.
I I use csc x for1
sin xand arcsin x for the inverse of sin x.
V63.0121.021, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 2, 2010 23 / 40
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. . . . . .
Outline
Inverse Trigonometric Functions
Derivatives of Inverse Trigonometric FunctionsArcsineArccosineArctangentArcsecant
Applications
V63.0121.021, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 2, 2010 24 / 40
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. . . . . .
The Inverse Function Theorem
Theorem (The Inverse Function Theorem)
Let f be differentiable at a, and f′(a) ̸= 0. Then f−1 is defined in anopen interval containing b = f(a), and
(f−1)′(b) =1
f′(f−1(b))
In Leibniz notation we have
dxdy
=1
dy/dx
Upshot: Many times the derivative of f−1(x) can be found by implicitdifferentiation and the derivative of f:
V63.0121.021, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 2, 2010 25 / 40
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. . . . . .
The Inverse Function Theorem
Theorem (The Inverse Function Theorem)
Let f be differentiable at a, and f′(a) ̸= 0. Then f−1 is defined in anopen interval containing b = f(a), and
(f−1)′(b) =1
f′(f−1(b))
In Leibniz notation we have
dxdy
=1
dy/dx
Upshot: Many times the derivative of f−1(x) can be found by implicitdifferentiation and the derivative of f:
V63.0121.021, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 2, 2010 25 / 40
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. . . . . .
Illustrating the Inverse Function Theorem.
.
Example
Use the inverse function theorem to find the derivative of the square rootfunction.
Solution (Newtonian notation)
Let f(x) = x2 so that f−1(y) =√y. Then f′(u) = 2u so for any b > 0 we have
(f−1)′(b) =1
2√b
Solution (Leibniz notation)
If the original function is y = x2, then the inverse function is defined by x = y2.Differentiate implicitly:
1 = 2ydydx
=⇒ dydx
=1
2√x
V63.0121.021, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 2, 2010 26 / 40
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. . . . . .
Illustrating the Inverse Function Theorem.
.
Example
Use the inverse function theorem to find the derivative of the square rootfunction.
Solution (Newtonian notation)
Let f(x) = x2 so that f−1(y) =√y. Then f′(u) = 2u so for any b > 0 we have
(f−1)′(b) =1
2√b
Solution (Leibniz notation)
If the original function is y = x2, then the inverse function is defined by x = y2.Differentiate implicitly:
1 = 2ydydx
=⇒ dydx
=1
2√x
V63.0121.021, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 2, 2010 26 / 40
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. . . . . .
Illustrating the Inverse Function Theorem.
.
Example
Use the inverse function theorem to find the derivative of the square rootfunction.
Solution (Newtonian notation)
Let f(x) = x2 so that f−1(y) =√y. Then f′(u) = 2u so for any b > 0 we have
(f−1)′(b) =1
2√b
Solution (Leibniz notation)
If the original function is y = x2, then the inverse function is defined by x = y2.Differentiate implicitly:
1 = 2ydydx
=⇒ dydx
=1
2√x
V63.0121.021, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 2, 2010 26 / 40
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. . . . . .
Derivation: The derivative of arcsin
Let y = arcsin x, so x = sin y. Then
cos ydydx
= 1 =⇒ dydx
=1
cos y=
1cos(arcsin x)
To simplify, look at a righttriangle:
cos(arcsin x) =√1− x2
So
Fact
ddx
arcsin(x) =1√
1− x2.
. .
. .y = arcsin x
.1 .x
.√1− x2
V63.0121.021, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 2, 2010 27 / 40
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. . . . . .
Derivation: The derivative of arcsin
Let y = arcsin x, so x = sin y. Then
cos ydydx
= 1 =⇒ dydx
=1
cos y=
1cos(arcsin x)
To simplify, look at a righttriangle:
cos(arcsin x) =√1− x2
So
Fact
ddx
arcsin(x) =1√
1− x2
.
. .
. .y = arcsin x
.1 .x
.√1− x2
V63.0121.021, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 2, 2010 27 / 40
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. . . . . .
Derivation: The derivative of arcsin
Let y = arcsin x, so x = sin y. Then
cos ydydx
= 1 =⇒ dydx
=1
cos y=
1cos(arcsin x)
To simplify, look at a righttriangle:
cos(arcsin x) =√1− x2
So
Fact
ddx
arcsin(x) =1√
1− x2
.
. .
. .y = arcsin x
.1 .x
.√1− x2
V63.0121.021, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 2, 2010 27 / 40
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. . . . . .
Derivation: The derivative of arcsin
Let y = arcsin x, so x = sin y. Then
cos ydydx
= 1 =⇒ dydx
=1
cos y=
1cos(arcsin x)
To simplify, look at a righttriangle:
cos(arcsin x) =√1− x2
So
Fact
ddx
arcsin(x) =1√
1− x2
.
. .
. .y = arcsin x
.1 .x
.√1− x2
V63.0121.021, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 2, 2010 27 / 40
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. . . . . .
Derivation: The derivative of arcsin
Let y = arcsin x, so x = sin y. Then
cos ydydx
= 1 =⇒ dydx
=1
cos y=
1cos(arcsin x)
To simplify, look at a righttriangle:
cos(arcsin x) =√1− x2
So
Fact
ddx
arcsin(x) =1√
1− x2
.
. .
. .y = arcsin x
.1 .x
.√1− x2
V63.0121.021, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 2, 2010 27 / 40
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. . . . . .
Derivation: The derivative of arcsin
Let y = arcsin x, so x = sin y. Then
cos ydydx
= 1 =⇒ dydx
=1
cos y=
1cos(arcsin x)
To simplify, look at a righttriangle:
cos(arcsin x) =√1− x2
So
Fact
ddx
arcsin(x) =1√
1− x2
.
. .
. .y = arcsin x
.1 .x
.√1− x2
V63.0121.021, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 2, 2010 27 / 40
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. . . . . .
Derivation: The derivative of arcsin
Let y = arcsin x, so x = sin y. Then
cos ydydx
= 1 =⇒ dydx
=1
cos y=
1cos(arcsin x)
To simplify, look at a righttriangle:
cos(arcsin x) =√1− x2
So
Fact
ddx
arcsin(x) =1√
1− x2.
. .
. .y = arcsin x
.1 .x
.√1− x2
V63.0121.021, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 2, 2010 27 / 40
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. . . . . .
Graphing arcsin and its derivative
I The domain of f is [−1,1],but the domain of f′ is(−1,1)
I limx→1−
f′(x) = +∞
I limx→−1+
f′(x) = +∞ ..|.−1
.|.1
.
. .arcsin
.1√
1− x2
V63.0121.021, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 2, 2010 28 / 40
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. . . . . .
Composing with arcsin
Example
Let f(x) = arcsin(x3 + 1). Find f′(x).
SolutionWe have
ddx
arcsin(x3 + 1) =1√
1− (x3 + 1)2ddx
(x3 + 1)
=3x2√
−x6 − 2x3
V63.0121.021, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 2, 2010 29 / 40
![Page 75: Lesson 16: Inverse Trigonometric Functions (Section 021 slides)](https://reader034.vdocuments.site/reader034/viewer/2022052307/5559c86ad8b42a93208b45f5/html5/thumbnails/75.jpg)
. . . . . .
Composing with arcsin
Example
Let f(x) = arcsin(x3 + 1). Find f′(x).
SolutionWe have
ddx
arcsin(x3 + 1) =1√
1− (x3 + 1)2ddx
(x3 + 1)
=3x2√
−x6 − 2x3
V63.0121.021, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 2, 2010 29 / 40
![Page 76: Lesson 16: Inverse Trigonometric Functions (Section 021 slides)](https://reader034.vdocuments.site/reader034/viewer/2022052307/5559c86ad8b42a93208b45f5/html5/thumbnails/76.jpg)
. . . . . .
Derivation: The derivative of arccos
Let y = arccos x, so x = cos y. Then
− sin ydydx
= 1 =⇒ dydx
=1
− sin y=
1− sin(arccos x)
To simplify, look at a righttriangle:
sin(arccos x) =√1− x2
So
Fact
ddx
arccos(x) = − 1√1− x2
.
.1.√1− x2
.x. .y = arccos x
V63.0121.021, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 2, 2010 30 / 40
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. . . . . .
Derivation: The derivative of arccos
Let y = arccos x, so x = cos y. Then
− sin ydydx
= 1 =⇒ dydx
=1
− sin y=
1− sin(arccos x)
To simplify, look at a righttriangle:
sin(arccos x) =√1− x2
So
Fact
ddx
arccos(x) = − 1√1− x2
.
.1.√1− x2
.x. .y = arccos x
V63.0121.021, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 2, 2010 30 / 40
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. . . . . .
Graphing arcsin and arccos
..|.−1
.|.1
.
. .arcsin
.
. .arccos
Note
cos θ = sin(π2− θ)
=⇒ arccos x =π
2− arcsin x
So it’s not a surprise that theirderivatives are opposites.
V63.0121.021, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 2, 2010 31 / 40
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. . . . . .
Graphing arcsin and arccos
..|.−1
.|.1
.
. .arcsin
.
. .arccosNote
cos θ = sin(π2− θ)
=⇒ arccos x =π
2− arcsin x
So it’s not a surprise that theirderivatives are opposites.
V63.0121.021, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 2, 2010 31 / 40
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. . . . . .
Derivation: The derivative of arctan
Let y = arctan x, so x = tan y. Then
sec2 ydydx
= 1 =⇒ dydx
=1
sec2 y= cos2(arctan x)
To simplify, look at a righttriangle:
cos(arctan x) =1√
1+ x2
So
Fact
ddx
arctan(x) =1
1+ x2.
.
.. .y = arctan x
.x
.1
.√1+ x2
V63.0121.021, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 2, 2010 32 / 40
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. . . . . .
Derivation: The derivative of arctan
Let y = arctan x, so x = tan y. Then
sec2 ydydx
= 1 =⇒ dydx
=1
sec2 y= cos2(arctan x)
To simplify, look at a righttriangle:
cos(arctan x) =1√
1+ x2
So
Fact
ddx
arctan(x) =1
1+ x2
.
.
.. .y = arctan x
.x
.1
.√1+ x2
V63.0121.021, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 2, 2010 32 / 40
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. . . . . .
Derivation: The derivative of arctan
Let y = arctan x, so x = tan y. Then
sec2 ydydx
= 1 =⇒ dydx
=1
sec2 y= cos2(arctan x)
To simplify, look at a righttriangle:
cos(arctan x) =1√
1+ x2
So
Fact
ddx
arctan(x) =1
1+ x2
.
.
.. .y = arctan x
.x
.1
.√1+ x2
V63.0121.021, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 2, 2010 32 / 40
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. . . . . .
Derivation: The derivative of arctan
Let y = arctan x, so x = tan y. Then
sec2 ydydx
= 1 =⇒ dydx
=1
sec2 y= cos2(arctan x)
To simplify, look at a righttriangle:
cos(arctan x) =1√
1+ x2
So
Fact
ddx
arctan(x) =1
1+ x2
.
.
.. .y = arctan x
.x
.1
.√1+ x2
V63.0121.021, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 2, 2010 32 / 40
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. . . . . .
Derivation: The derivative of arctan
Let y = arctan x, so x = tan y. Then
sec2 ydydx
= 1 =⇒ dydx
=1
sec2 y= cos2(arctan x)
To simplify, look at a righttriangle:
cos(arctan x) =1√
1+ x2
So
Fact
ddx
arctan(x) =1
1+ x2
.
.
.. .y = arctan x
.x
.1
.√1+ x2
V63.0121.021, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 2, 2010 32 / 40
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. . . . . .
Derivation: The derivative of arctan
Let y = arctan x, so x = tan y. Then
sec2 ydydx
= 1 =⇒ dydx
=1
sec2 y= cos2(arctan x)
To simplify, look at a righttriangle:
cos(arctan x) =1√
1+ x2
So
Fact
ddx
arctan(x) =1
1+ x2
.
.
.. .y = arctan x
.x
.1
.√1+ x2
V63.0121.021, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 2, 2010 32 / 40
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. . . . . .
Derivation: The derivative of arctan
Let y = arctan x, so x = tan y. Then
sec2 ydydx
= 1 =⇒ dydx
=1
sec2 y= cos2(arctan x)
To simplify, look at a righttriangle:
cos(arctan x) =1√
1+ x2
So
Fact
ddx
arctan(x) =1
1+ x2.
.
.. .y = arctan x
.x
.1
.√1+ x2
V63.0121.021, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 2, 2010 32 / 40
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. . . . . .
Graphing arctan and its derivative
. .x
.y
.arctan
.1
1+ x2
.π/2
.−π/2
I The domain of f and f′ are both (−∞,∞)
I Because of the horizontal asymptotes, limx→±∞
f′(x) = 0
V63.0121.021, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 2, 2010 33 / 40
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. . . . . .
Composing with arctan
Example
Let f(x) = arctan√x. Find f′(x).
Solution
ddx
arctan√x =
11+
(√x)2 d
dx√x =
11+ x
· 12√x
=1
2√x+ 2x
√x
V63.0121.021, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 2, 2010 34 / 40
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. . . . . .
Composing with arctan
Example
Let f(x) = arctan√x. Find f′(x).
Solution
ddx
arctan√x =
11+
(√x)2 d
dx√x =
11+ x
· 12√x
=1
2√x+ 2x
√x
V63.0121.021, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 2, 2010 34 / 40
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. . . . . .
Derivation: The derivative of arcsec
Try this first.
Let y = arcsec x, so x = sec y. Then
sec y tan ydydx
= 1 =⇒ dydx
=1
sec y tan y=
1x tan(arcsec(x))
To simplify, look at a righttriangle:
tan(arcsec x) =√x2 − 11
So
Fact
ddx
arcsec(x) =1
x√x2 − 1
.
.x
.1. .y = arcsec x
.√
x2 − 1
V63.0121.021, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 2, 2010 35 / 40
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. . . . . .
Derivation: The derivative of arcsec
Try this first. Let y = arcsec x, so x = sec y. Then
sec y tan ydydx
= 1 =⇒ dydx
=1
sec y tan y=
1x tan(arcsec(x))
To simplify, look at a righttriangle:
tan(arcsec x) =√x2 − 11
So
Fact
ddx
arcsec(x) =1
x√x2 − 1
.
.x
.1. .y = arcsec x
.√
x2 − 1
V63.0121.021, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 2, 2010 35 / 40
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. . . . . .
Derivation: The derivative of arcsec
Try this first. Let y = arcsec x, so x = sec y. Then
sec y tan ydydx
= 1 =⇒ dydx
=1
sec y tan y=
1x tan(arcsec(x))
To simplify, look at a righttriangle:
tan(arcsec x) =√x2 − 11
So
Fact
ddx
arcsec(x) =1
x√x2 − 1
.
.x
.1. .y = arcsec x
.√
x2 − 1
V63.0121.021, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 2, 2010 35 / 40
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. . . . . .
Derivation: The derivative of arcsec
Try this first. Let y = arcsec x, so x = sec y. Then
sec y tan ydydx
= 1 =⇒ dydx
=1
sec y tan y=
1x tan(arcsec(x))
To simplify, look at a righttriangle:
tan(arcsec x) =√x2 − 11
So
Fact
ddx
arcsec(x) =1
x√x2 − 1
.
.x
.1. .y = arcsec x
.√
x2 − 1
V63.0121.021, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 2, 2010 35 / 40
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. . . . . .
Derivation: The derivative of arcsec
Try this first. Let y = arcsec x, so x = sec y. Then
sec y tan ydydx
= 1 =⇒ dydx
=1
sec y tan y=
1x tan(arcsec(x))
To simplify, look at a righttriangle:
tan(arcsec x) =√x2 − 11
So
Fact
ddx
arcsec(x) =1
x√x2 − 1
.
.x
.1
. .y = arcsec x
.√
x2 − 1
V63.0121.021, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 2, 2010 35 / 40
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. . . . . .
Derivation: The derivative of arcsec
Try this first. Let y = arcsec x, so x = sec y. Then
sec y tan ydydx
= 1 =⇒ dydx
=1
sec y tan y=
1x tan(arcsec(x))
To simplify, look at a righttriangle:
tan(arcsec x) =√x2 − 11
So
Fact
ddx
arcsec(x) =1
x√x2 − 1
.
.x
.1. .y = arcsec x
.√
x2 − 1
V63.0121.021, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 2, 2010 35 / 40
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. . . . . .
Derivation: The derivative of arcsec
Try this first. Let y = arcsec x, so x = sec y. Then
sec y tan ydydx
= 1 =⇒ dydx
=1
sec y tan y=
1x tan(arcsec(x))
To simplify, look at a righttriangle:
tan(arcsec x) =√x2 − 11
So
Fact
ddx
arcsec(x) =1
x√x2 − 1
.
.x
.1. .y = arcsec x
.√
x2 − 1
V63.0121.021, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 2, 2010 35 / 40
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. . . . . .
Derivation: The derivative of arcsec
Try this first. Let y = arcsec x, so x = sec y. Then
sec y tan ydydx
= 1 =⇒ dydx
=1
sec y tan y=
1x tan(arcsec(x))
To simplify, look at a righttriangle:
tan(arcsec x) =√x2 − 11
So
Fact
ddx
arcsec(x) =1
x√x2 − 1
.
.x
.1. .y = arcsec x
.√
x2 − 1
V63.0121.021, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 2, 2010 35 / 40
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. . . . . .
Another Example
Example
Let f(x) = earcsec 3x. Find f′(x).
Solution
f′(x) = earcsec 3x · 13x√
(3x)2 − 1· 3
=3earcsec 3x
3x√9x2 − 1
V63.0121.021, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 2, 2010 36 / 40
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. . . . . .
Another Example
Example
Let f(x) = earcsec 3x. Find f′(x).
Solution
f′(x) = earcsec 3x · 13x√
(3x)2 − 1· 3
=3earcsec 3x
3x√9x2 − 1
V63.0121.021, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 2, 2010 36 / 40
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. . . . . .
Outline
Inverse Trigonometric Functions
Derivatives of Inverse Trigonometric FunctionsArcsineArccosineArctangentArcsecant
Applications
V63.0121.021, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 2, 2010 37 / 40
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. . . . . .
Application
ExampleOne of the guiding principles ofmost sports is to “keep youreye on the ball.” In baseball, abatter stands 2 ft away fromhome plate as a pitch is thrownwith a velocity of 130 ft/sec(about 90mph). At what ratedoes the batter’s angle of gazeneed to change to follow theball as it crosses home plate?
SolutionLet y(t) be the distance from the ball to home plate, and θ the angle thebatter’s eyes make with home plate while following the ball. We knowy′ = −130 and we want θ′ at the moment that y = 0.
V63.0121.021, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 2, 2010 38 / 40
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. . . . . .
Application
ExampleOne of the guiding principles ofmost sports is to “keep youreye on the ball.” In baseball, abatter stands 2 ft away fromhome plate as a pitch is thrownwith a velocity of 130 ft/sec(about 90mph). At what ratedoes the batter’s angle of gazeneed to change to follow theball as it crosses home plate?
SolutionLet y(t) be the distance from the ball to home plate, and θ the angle thebatter’s eyes make with home plate while following the ball. We knowy′ = −130 and we want θ′ at the moment that y = 0.V63.0121.021, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 2, 2010 38 / 40
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. . . . . .
SolutionLet y(t) be the distance from the ball to home plate, and θ the angle thebatter’s eyes make with home plate while following the ball. We knowy′ = −130 and we want θ′ at the moment that y = 0.
We have θ = arctan(y/2). Thus
dθdt
=1
1+ (y/2)2· 12dydt
When y = 0 and y′ = −130,then
dθdt
∣∣∣∣y=0
=1
1+ 0·12(−130) = −65 rad/sec
The human eye can only trackat 3 rad/sec!
..2 ft
.y
.130 ft/sec
.
.θ
V63.0121.021, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 2, 2010 39 / 40
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. . . . . .
SolutionLet y(t) be the distance from the ball to home plate, and θ the angle thebatter’s eyes make with home plate while following the ball. We knowy′ = −130 and we want θ′ at the moment that y = 0.We have θ = arctan(y/2). Thus
dθdt
=1
1+ (y/2)2· 12dydt
When y = 0 and y′ = −130,then
dθdt
∣∣∣∣y=0
=1
1+ 0·12(−130) = −65 rad/sec
The human eye can only trackat 3 rad/sec!
..2 ft
.y
.130 ft/sec
.
.θ
V63.0121.021, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 2, 2010 39 / 40
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. . . . . .
SolutionLet y(t) be the distance from the ball to home plate, and θ the angle thebatter’s eyes make with home plate while following the ball. We knowy′ = −130 and we want θ′ at the moment that y = 0.We have θ = arctan(y/2). Thus
dθdt
=1
1+ (y/2)2· 12dydt
When y = 0 and y′ = −130,then
dθdt
∣∣∣∣y=0
=1
1+ 0·12(−130) = −65 rad/sec
The human eye can only trackat 3 rad/sec!
..2 ft
.y
.130 ft/sec
.
.θ
V63.0121.021, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 2, 2010 39 / 40
![Page 106: Lesson 16: Inverse Trigonometric Functions (Section 021 slides)](https://reader034.vdocuments.site/reader034/viewer/2022052307/5559c86ad8b42a93208b45f5/html5/thumbnails/106.jpg)
. . . . . .
SolutionLet y(t) be the distance from the ball to home plate, and θ the angle thebatter’s eyes make with home plate while following the ball. We knowy′ = −130 and we want θ′ at the moment that y = 0.We have θ = arctan(y/2). Thus
dθdt
=1
1+ (y/2)2· 12dydt
When y = 0 and y′ = −130,then
dθdt
∣∣∣∣y=0
=1
1+ 0·12(−130) = −65 rad/sec
The human eye can only trackat 3 rad/sec!
..2 ft
.y
.130 ft/sec
.
.θ
V63.0121.021, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 2, 2010 39 / 40
![Page 107: Lesson 16: Inverse Trigonometric Functions (Section 021 slides)](https://reader034.vdocuments.site/reader034/viewer/2022052307/5559c86ad8b42a93208b45f5/html5/thumbnails/107.jpg)
. . . . . .
Summary
y y′
arcsin x1√
1− x2
arccos x − 1√1− x2
arctan x1
1+ x2
arccot x − 11+ x2
arcsec x1
x√x2 − 1
arccsc x − 1x√x2 − 1
I Remarkable that thederivatives of thesetranscendental functionsare algebraic (or evenrational!)
V63.0121.021, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 2, 2010 40 / 40