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Section 3.5Inverse Trigonometric
FunctionsV63.0121.041, Calculus I
New York University
November 1, 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
. . . . . .
. . . . . .
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
V63.0121.041, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 1, 2010 2 / 32
. . . . . .
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.
V63.0121.041, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 1, 2010 3 / 32
. . . . . .
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
V63.0121.041, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 1, 2010 4 / 32
. . . . . .
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
V63.0121.041, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 1, 2010 4 / 32
. . . . . .
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.
V63.0121.041, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 1, 2010 5 / 32
. . . . . .
Outline
Inverse Trigonometric Functions
Derivatives of Inverse Trigonometric FunctionsArcsineArccosineArctangentArcsecant
Applications
V63.0121.041, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 1, 2010 6 / 32
. . . . . .
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
]
V63.0121.041, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 1, 2010 7 / 32
. . . . . .
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
]
V63.0121.041, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 1, 2010 7 / 32
. . . . . .
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
]
V63.0121.041, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 1, 2010 7 / 32
. . . . . .
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
]
V63.0121.041, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 1, 2010 7 / 32
. . . . . .
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]
V63.0121.041, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 1, 2010 8 / 32
. . . . . .
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]
V63.0121.041, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 1, 2010 8 / 32
. . . . . .
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]
V63.0121.041, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 1, 2010 8 / 32
. . . . . .
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]
V63.0121.041, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 1, 2010 8 / 32
. . . . . .
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
V63.0121.041, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 1, 2010 9 / 32
. . . . . .
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
V63.0121.041, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 1, 2010 9 / 32
. . . . . .
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
V63.0121.041, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 1, 2010 9 / 32
. . . . . .
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
V63.0121.041, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 1, 2010 9 / 32
. . . . . .
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
V63.0121.041, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 1, 2010 10 / 32
. . . . . .
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
V63.0121.041, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 1, 2010 10 / 32
. . . . . .
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
V63.0121.041, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 1, 2010 10 / 32
. . . . . .
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
V63.0121.041, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 1, 2010 10 / 32
. . . . . .
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
V63.0121.041, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 1, 2010 11 / 32
. . . . . .
Check: Values of inverse trigonometric functions
Example
FindI arcsin(1/2)I arctan(−1)
I arccos
(−√22
)
Solution
Iπ
6I −π
4I
3π4
V63.0121.041, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 1, 2010 12 / 32
. . . . . .
Check: Values of inverse trigonometric functions
Example
FindI arcsin(1/2)I arctan(−1)
I arccos
(−√22
)
Solution
Iπ
6
I −π
4I
3π4
V63.0121.041, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 1, 2010 12 / 32
. . . . . .
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
V63.0121.041, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 1, 2010 13 / 32
. . . . . .
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
V63.0121.041, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 1, 2010 13 / 32
. . . . . .
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
V63.0121.041, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 1, 2010 13 / 32
. . . . . .
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
V63.0121.041, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 1, 2010 13 / 32
. . . . . .
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
V63.0121.041, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 1, 2010 13 / 32
. . . . . .
Check: Values of inverse trigonometric functions
Example
FindI arcsin(1/2)I arctan(−1)
I arccos
(−√22
)
Solution
Iπ
6I −π
4
I3π4
V63.0121.041, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 1, 2010 14 / 32
. . . . . .
Check: Values of inverse trigonometric functions
Example
FindI arcsin(1/2)I arctan(−1)
I arccos
(−√22
)
Solution
Iπ
6I −π
4I
3π4
V63.0121.041, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 1, 2010 14 / 32
. . . . . .
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.
V63.0121.041, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 1, 2010 15 / 32
. . . . . .
Outline
Inverse Trigonometric Functions
Derivatives of Inverse Trigonometric FunctionsArcsineArccosineArctangentArcsecant
Applications
V63.0121.041, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 1, 2010 16 / 32
. . . . . .
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:
V63.0121.041, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 1, 2010 17 / 32
. . . . . .
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:
V63.0121.041, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 1, 2010 17 / 32
. . . . . .
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
V63.0121.041, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 1, 2010 18 / 32
. . . . . .
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
V63.0121.041, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 1, 2010 18 / 32
. . . . . .
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
V63.0121.041, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 1, 2010 18 / 32
. . . . . .
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
ddx
arcsin(x) =1√
1− x2 .
.1 .x
. .y = arcsin x
.√1− x2
V63.0121.041, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 1, 2010 19 / 32
. . . . . .
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
ddx
arcsin(x) =1√
1− x2
.
.1 .x
. .y = arcsin x
.√1− x2
V63.0121.041, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 1, 2010 19 / 32
. . . . . .
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
ddx
arcsin(x) =1√
1− x2
.
.1 .x
. .y = arcsin x
.√1− x2
V63.0121.041, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 1, 2010 19 / 32
. . . . . .
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
ddx
arcsin(x) =1√
1− x2
.
.1 .x
. .y = arcsin x
.√1− x2
V63.0121.041, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 1, 2010 19 / 32
. . . . . .
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
ddx
arcsin(x) =1√
1− x2
.
.1 .x
. .y = arcsin x
.√1− x2
V63.0121.041, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 1, 2010 19 / 32
. . . . . .
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
ddx
arcsin(x) =1√
1− x2
.
.1 .x
. .y = arcsin x
.√1− x2
V63.0121.041, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 1, 2010 19 / 32
. . . . . .
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
ddx
arcsin(x) =1√
1− x2 .
.1 .x
. .y = arcsin x
.√1− x2
V63.0121.041, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 1, 2010 19 / 32
. . . . . .
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
V63.0121.041, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 1, 2010 20 / 32
. . . . . .
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
V63.0121.041, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 1, 2010 21 / 32
. . . . . .
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
V63.0121.041, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 1, 2010 21 / 32
. . . . . .
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
ddx
arccos(x) = − 1√1− x2 .
.1.√1− x2
.x. .y = arccos x
V63.0121.041, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 1, 2010 22 / 32
. . . . . .
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
ddx
arccos(x) = − 1√1− x2 .
.1.√1− x2
.x. .y = arccos x
V63.0121.041, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 1, 2010 22 / 32
. . . . . .
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.
V63.0121.041, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 1, 2010 23 / 32
. . . . . .
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.
V63.0121.041, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 1, 2010 23 / 32
. . . . . .
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
ddx
arctan(x) =1
1+ x2 .
.x
.1. .y = arctan x
.√1+ x2
V63.0121.041, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 1, 2010 24 / 32
. . . . . .
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
ddx
arctan(x) =1
1+ x2
.
.x
.1. .y = arctan x
.√1+ x2
V63.0121.041, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 1, 2010 24 / 32
. . . . . .
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
ddx
arctan(x) =1
1+ x2
.
.x
.1
. .y = arctan x
.√1+ x2
V63.0121.041, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 1, 2010 24 / 32
. . . . . .
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
ddx
arctan(x) =1
1+ x2
.
.x
.1. .y = arctan x
.√1+ x2
V63.0121.041, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 1, 2010 24 / 32
. . . . . .
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
ddx
arctan(x) =1
1+ x2
.
.x
.1. .y = arctan x
.√1+ x2
V63.0121.041, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 1, 2010 24 / 32
. . . . . .
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
ddx
arctan(x) =1
1+ x2
.
.x
.1. .y = arctan x
.√1+ x2
V63.0121.041, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 1, 2010 24 / 32
. . . . . .
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
ddx
arctan(x) =1
1+ x2 .
.x
.1. .y = arctan x
.√1+ x2
V63.0121.041, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 1, 2010 24 / 32
. . . . . .
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
V63.0121.041, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 1, 2010 25 / 32
. . . . . .
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
V63.0121.041, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 1, 2010 26 / 32
. . . . . .
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
V63.0121.041, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 1, 2010 26 / 32
. . . . . .
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
ddx
arcsec(x) =1
x√x2 − 1 .
.x
.1. .y = arcsec x
.√
x2 − 1
V63.0121.041, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 1, 2010 27 / 32
. . . . . .
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
ddx
arcsec(x) =1
x√x2 − 1 .
.x
.1. .y = arcsec x
.√
x2 − 1
V63.0121.041, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 1, 2010 27 / 32
. . . . . .
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
ddx
arcsec(x) =1
x√x2 − 1
.
.x
.1. .y = arcsec x
.√
x2 − 1
V63.0121.041, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 1, 2010 27 / 32
. . . . . .
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
ddx
arcsec(x) =1
x√x2 − 1
.
.x
.1. .y = arcsec x
.√
x2 − 1
V63.0121.041, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 1, 2010 27 / 32
. . . . . .
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
ddx
arcsec(x) =1
x√x2 − 1
.
.x
.1
. .y = arcsec x
.√
x2 − 1
V63.0121.041, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 1, 2010 27 / 32
. . . . . .
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
ddx
arcsec(x) =1
x√x2 − 1
.
.x
.1. .y = arcsec x
.√
x2 − 1
V63.0121.041, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 1, 2010 27 / 32
. . . . . .
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
ddx
arcsec(x) =1
x√x2 − 1
.
.x
.1. .y = arcsec x
.√
x2 − 1
V63.0121.041, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 1, 2010 27 / 32
. . . . . .
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
ddx
arcsec(x) =1
x√x2 − 1 .
.x
.1. .y = arcsec x
.√
x2 − 1
V63.0121.041, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 1, 2010 27 / 32
. . . . . .
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
V63.0121.041, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 1, 2010 28 / 32
. . . . . .
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
V63.0121.041, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 1, 2010 28 / 32
. . . . . .
Outline
Inverse Trigonometric Functions
Derivatives of Inverse Trigonometric FunctionsArcsineArccosineArctangentArcsecant
Applications
V63.0121.041, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 1, 2010 29 / 32
. . . . . .
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.
V63.0121.041, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 1, 2010 30 / 32
. . . . . .
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.
V63.0121.041, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 1, 2010 30 / 32
. . . . . .
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
V63.0121.041, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 1, 2010 31 / 32
. . . . . .
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
V63.0121.041, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 1, 2010 31 / 32
. . . . . .
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
V63.0121.041, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 1, 2010 31 / 32
. . . . . .
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
V63.0121.041, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 1, 2010 31 / 32
. . . . . .
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!)
V63.0121.041, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 1, 2010 32 / 32