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AP Calculus BC
Unit 3: Differentiation: Composite, Implicit, and Inverse Functions
3.4 Differentiating Inverse Trigonometric Functions
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What is the primary purpose of inverse trigonometric functions?
Find corresponding angles
The inverse trigonometric function Arcsin finds the angle whose sine is
x
x
x
.
What is the domain of the arcsin(x) function?
[
−
1
,
1
]
[ - 1, 1]
[
−
1
,
1
]
The range of the arctan(x) function is
( - π / 2, π / 2)
.
arcsin(1 / 2) is equal to
π
/ 6.
The derivative of
a
r
c
s
i
n
(
x
)
arcsin(x)
a
rcs
in
(
x
)
is
1
1
−
x
2
\frac{1}{\sqrt{1 - x^{2}}}
1
−
x
2
1
.
The derivative of
a
r
c
c
o
s
(
x
)
arccos(x)
a
rccos
(
x
)
is
−
1
1
−
x
2
\frac{ - 1}{\sqrt{1 - x^{2}}}
1
−
x
2
−
1
.
The derivative of
a
r
c
t
a
n
(
x
)
arctan(x)
a
rc
t
an
(
x
)
is
1
1
+
x
2
\frac{1}{1 + x^{2}}
1
+
x
2
1
.
Match the inverse trigonometric function with its derivative:
a
r
c
s
i
n
(
x
)
arcsin(x)
a
rcs
in
(
x
)
↔️
1
1
−
x
2
\frac{1}{\sqrt{1 - x^{2}}}
1
−
x
2
1
a
r
c
c
o
s
(
x
)
arccos(x)
a
rccos
(
x
)
↔️
−
1
1
−
x
2
\frac{ - 1}{\sqrt{1 - x^{2}}}
1
−
x
2
−
1
a
r
c
t
a
n
(
x
)
arctan(x)
a
rc
t
an
(
x
)
↔️
1
1
+
x
2
\frac{1}{1 + x^{2}}
1
+
x
2
1
The chain rule formula is
d
d
x
[
f
(
g
(
x
)
)
]
=
\frac{d}{dx}[f(g(x))] =
d
x
d
[
f
(
g
(
x
))]
=
f
′
(
g
(
x
)
)
⋅
g
′
(
x
)
f'(g(x)) \cdot g'(x)
f
′
(
g
(
x
))
⋅
g
′
(
x
)
.
The derivative of
a
r
c
s
i
n
(
3
x
)
arcsin(3x)
a
rcs
in
(
3
x
)
is
3
1
−
9
x
2
\frac{3}{\sqrt{1 - 9x^{2}}}
1
−
9
x
2
3
.
What is the derivative of
a
r
c
t
a
n
(
x
2
)
arctan(x^{2})
a
rc
t
an
(
x
2
)
?
2
x
1
+
x
4
\frac{2x}{1 + x^{4}}
1
+
x
4
2
x
The derivative of
a
r
c
c
o
s
(
x
)
arccos(\sqrt{x})
a
rccos
(
x
)
is
−
1
2
x
1
−
x
\frac{ - 1}{2\sqrt{x}\sqrt{1 - x}}
2
x
1
−
x
−
1
.
A
r
c
s
i
n
(
x
)
Arcsin(x)
A
rcs
in
(
x
)
finds the angle whose sine is
x
x
x
.
Why are the domains and ranges of inverse trigonometric functions restricted?
To ensure unique values
The range of arcsin(x) is
[
−
π
/
2
,
π
/
2
]
[ - π / 2, π / 2]
[
−
π
/2
,
π
/2
]
The range of arccos(x) is
[
0
,
π
]
[0, π]
[
0
,
π
]
What is the domain of arctan(x)?
(
−
∞
,
∞
)
( - ∞, ∞)
(
−
∞
,
∞
)
a
r
c
s
i
n
(
1
/
2
)
=
arcsin(1 / 2) =
a
rcs
in
(
1/2
)
=
π / 6</latex>
a
r
c
c
o
s
(
0
)
=
arccos(0) =
a
rccos
(
0
)
=
π
/
2
π / 2
π
/2
What is the value of
a
r
c
t
a
n
(
1
)
arctan(1)
a
rc
t
an
(
1
)
?
π
/
4
π / 4
π
/4
Order the inverse trigonometric functions by the complexity of their derivatives:
1️⃣
a
r
c
t
a
n
(
x
)
arctan(x)
a
rc
t
an
(
x
)
2️⃣
a
r
c
s
i
n
(
x
)
arcsin(x)
a
rcs
in
(
x
)
3️⃣
a
r
c
c
o
s
(
x
)
arccos(x)
a
rccos
(
x
)
What is the derivative of
a
r
c
s
i
n
(
x
)
arcsin(x)
a
rcs
in
(
x
)
?
1
1
−
x
2
\frac{1}{\sqrt{1 - x^{2}}}
1
−
x
2
1
The derivative of
a
r
c
c
o
s
(
x
)
arccos(x)
a
rccos
(
x
)
is
1
1
−
x
2
\frac{1}{\sqrt{1 - x^{2}}}
1
−
x
2
1
False
The derivative of
a
r
c
t
a
n
(
x
)
arctan(x)
a
rc
t
an
(
x
)
is
1
1
+
x
2
\frac{1}{1 + x^{2}}
1
+
x
2
1
What is the derivative of arcsin(x)</latex> if
y
=
y =
y
=
a
r
c
s
i
n
(
x
)
arcsin(x)
a
rcs
in
(
x
)
?
1
1
−
x
2
\frac{1}{\sqrt{1 - x^{2}}}
1
−
x
2
1
The derivative of
a
r
c
c
o
s
(
x
)
arccos(x)
a
rccos
(
x
)
is negative.
Match the composite inverse trigonometric function with its correct derivative:
a
r
c
s
i
n
(
g
(
x
)
)
arcsin(g(x))
a
rcs
in
(
g
(
x
))
↔️
g
′
(
x
)
1
−
g
(
x
)
2
\frac{g'(x)}{\sqrt{1 - g(x)^{2}}}
1
−
g
(
x
)
2
g
′
(
x
)
a
r
c
c
o
s
(
g
(
x
)
)
arccos(g(x))
a
rccos
(
g
(
x
))
↔️
−
g
′
(
x
)
1
−
g
(
x
)
2
\frac{ - g'(x)}{\sqrt{1 - g(x)^{2}}}
1
−
g
(
x
)
2
−
g
′
(
x
)
a
r
c
t
a
n
(
g
(
x
)
)
arctan(g(x))
a
rc
t
an
(
g
(
x
))
↔️
g
′
(
x
)
1
+
g
(
x
)
2
\frac{g'(x)}{1 + g(x)^{2}}
1
+
g
(
x
)
2
g
′
(
x
)
The derivative of
a
r
c
s
i
n
(
3
x
)
arcsin(3x)
a
rcs
in
(
3
x
)
is
3
1
−
9
x
2
\frac{3}{\sqrt{1 - 9x^{2}}}
1
−
9
x
2
3
The derivative of
a
r
c
t
a
n
(
x
2
)
arctan(x^{2})
a
rc
t
an
(
x
2
)
is
2
x
1
+
x
4
\frac{2x}{1 + x^{4}}
1
+
x
4
2
x
What is the derivative of
a
r
c
c
o
s
(
x
)
arccos(\sqrt{x})
a
rccos
(
x
)
?
−
1
2
x
1
−
x
\frac{ - 1}{2\sqrt{x}\sqrt{1 - x}}
2
x
1
−
x
−
1
Steps to differentiate an implicit function involving inverse trigonometric functions:
1️⃣ Differentiate both sides
2️⃣ Use the chain rule
3️⃣ Solve for
d
y
d
x
\frac{dy}{dx}
d
x
d
y
Given
a
r
c
t
a
n
(
y
)
+
arctan(y) +
a
rc
t
an
(
y
)
+
x
=
x =
x
=
1
1
1
, what is
d
y
d
x
\frac{dy}{dx}
d
x
d
y
?
−
(
1
+
y
2
)
- (1 + y^{2})
−
(
1
+
y
2
)
To differentiate inverse trigonometric functions, the
chain rule
is always necessary.
The derivative of
a
r
c
t
a
n
(
x
2
)
arctan(x^{2})
a
rc
t
an
(
x
2
)
is
2
x
1
+
x
4
\frac{2x}{1 + x^{4}}
1
+
x
4
2
x
The derivative of
a
r
c
c
o
s
(
x
)
arccos(x)
a
rccos
(
x
)
is - 1 / \sqrt{1 - x^{2}}
The derivative of
a
r
c
t
a
n
(
x
)
arctan(x)
a
rc
t
an
(
x
)
is 1 / (1 + x^{2})
Steps to find the derivative of
y
=
y =
y
=
a
r
c
t
a
n
(
x
2
)
arctan(x^{2})
a
rc
t
an
(
x
2
)
1️⃣ Identify the inner function:
g
(
x
)
=
g(x) =
g
(
x
)
=
x
2
x^{2}
x
2
2️⃣ Find its derivative:
g
′
(
x
)
=
g'(x) =
g
′
(
x
)
=
2
x
2x
2
x
3️⃣ Apply the chain rule:
d
y
d
x
=
\frac{dy}{dx} =
d
x
d
y
=
1
1
+
(
x
2
)
2
⋅
2
x
\frac{1}{1 + (x^{2})^{2}} \cdot 2x
1
+
(
x
2
)
2
1
⋅
2
x
4️⃣ Simplify the result:
d
y
d
x
=
\frac{dy}{dx} =
d
x
d
y
=
2
x
1
+
x
4
\frac{2x}{1 + x^{4}}
1
+
x
4
2
x
The derivative of
y
=
y =
y
=
a
r
c
t
a
n
(
x
2
)
arctan(x^{2})
a
rc
t
an
(
x
2
)
is \frac{2x}{1 + x^{4}}</latex>
The inverse trigonometric functions have a restricted domain and
range
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