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4.6 Inverse Trigonometric Functions
Remember that only one-to-one functions have inverses. So, in order to find the inverse functions for sine,
cosine, and tangent, we must restrict their domains to intervals where they are one-to-one.
To find the inverse sine function, we restrict the domain of sine to [−π/2, π/2].
We define the inverse sine function, sin−1 x by sin−1 x = y ↔ sin y = x.
sin−1 x, or arcsin x, has domain
and range
.
sin−1 x is the ANGLE in the interval [−π/2, π/2] whose sine is x.
Examples:
sin−1
√
3
2
arcsin(− 21 )
What is the domain of f (x) = arcsin(5x − 2)?
1
sin−1 2
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In order to have an inverse for cosine, we restrict the domain of cosine to the interval [0, π].
The inverse cosine function cos−1 is defined by cos−1 x = y ↔ cos y = x.
cos−1 x, or arccos x, has domain
and range
.
cos−1 x is the ANGLE in the interval [0, π] whose cosine is x.
Examples:
arccos(0)
cos−1 (
√
2
2 )
lim cos
x→∞
What is the domain of f (x) = arccos(ln x)?
2
−1
x−9
2x + 8
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In order to have an inverse for tangent, we restrict the domain of tangent to the interval (−π/2, π/2).
The inverse tangent function tan−1 is defined by tan−1 x = y ↔ tan y = x.
tan−1 x, or arctan x, has domain
lim arctan x =
x→∞
and range
.
lim arctan x =
x→−∞
tan−1 x is the ANGLE in the interval (−π/2, π/2) whose tangent is x.
Examples:
tan−1
√1
3
x
lim arctan
+
3
−
x
x→3
arctan(−1)
What is the domain of f (x) = arctan(x2 − 9x)?
3
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Examples: Evaluate the following expressions.
tan(sin−1 45 )
sin(cos−1 (− 32 ))
sin(tan−1 x)
cot(cos−1 x)
Derivatives of Inverse Trig Functions
1
d
sin−1 x = √
dx
1 − x2
1
d
cos−1 x = − √
dx
1 − x2
1
d
tan−1 x =
dx
1 + x2
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Show that
1
d
.
sin−1 x = √
dx
1 − x2
Find the derivatives of the following functions.
f (x) = (arcsin(3x2 + ln x))2
y = arctan(5x2 + 9x) + x arccos(5x)
Find the equation of the tangent line to the graph of f (x) = arctan(−2x) at the point where x =
5
√
3
2 .
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4.8 Indeterminate Forms and L’Hospital’s Rule
We have seen limits in the past that take the form 00 , ∞
∞ , and ∞ − ∞. When we encountered these, we had
to do something else...algebra, simplification, factoring...to be able to find the limit. These types of limits
are examples of indeterminate forms.
0
∞
f (x)
= or
, then we can use L’Hospital’s Rule to find the limit.
x→a g(x)
0
∞
If lim
L’Hospital’s Rule: Suppose f and g are differentiable functions. If lim
x→a
f (x)
0
∞
= or
, then
g(x)
0
∞
f (x)
f ′ (x)
= lim ′
x→a g(x)
x→a g (x)
lim
Notes: The limit could also be of the form
−∞ −∞
−∞ , ∞ ,
or
∞
−∞ .
f (x)
x→a g(x)
=
0
∞,
this is NOT indeterminate: The limit is 0.
f (x)
x→a g(x)
=
∞
0 ,
this is NOT indeterminate: The limit will be ∞ or −∞.
If lim
If lim
ex − 1
x→0 sin 3x
(1) lim
(ln x)3
x→∞
x2
(2) lim
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Indeterminate Products: If lim f (x)g(x) = 0 · ∞, this limit is indeterminate. Why?
x→a
1 2
·x
x→∞ x
3
·x
x→∞ x2
lim
5
· x2
x→∞ x2
lim
lim
To find the limit, the goal is to write the indeterminate product in the form
Rule.
(1) lim+ csc x ln(1 + sin 7x)
x→0
(2) (#40, 4.8) lim xex
x→−∞
7
0
0
or
∞
∞
and use L’Hospital’s
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Indeterminate Difference: If lim [f (x) − g(x)] = ∞ − ∞, this limit is indeterminate. To find the limit, the
x→a
goal is once again to convert this difference into a quotient that we can use L’Hospital’s Rule on if necessary.
lim
x→0+
2x + 1 1
−
sin x
x
Indeterminate Powers: If lim [f (x)]g(x) is of the form 00 , ∞0 , or 1∞ , these are indeterminate. These cases
x→a
are treated by first taking the natural logarithm, which will make the limit of the form 0 · ∞. Then, proceed
as we did with indeterminate products. However, we must remember to “undo” the natural logarithm to
find our final answer. (Note that 0∞ is NOT an indeterminate form. A limit of this form will be 0.)
(1) lim
x→∞
4
1+ 2
x
x2
8
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(2) lim xtan x
x→0+
(3) (#62, 4.8) lim (ex + x)1/x
x→∞
Summary: There are 7 basic indeterminate forms:
0 ±∞
,
, ∞ − ∞, 0 · ∞, 00 , ∞0 , 1∞ .
0 ±∞
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