CHAPTER 25 Derivatives of Inverse Trig Functions
CHAPTER 25
Derivatives of Inverse Trig Functions
Our goal is simple, and the answers will come quickly. We will derive six new derivative formulas for the six inverse trigonometric functions:
h
i
d
?
1
sin (x)
dx
h
i
d
?
1
cos (x)
dx
h
i
d
?
1
tan (x)
dx
h
i
d
?
1
cot (x)
dx
h
i
d
?
1
sec (x)
dx
h
i
d
?
1
csc (x)
dx
These formulas will flow from the inverse rule from Chapter 24 (page 278):
h
i
d?
1
=
f (x)
0?
1
?
? .
1
dx
f f (x)
(25.1)
25.1 Derivatives of Inverse Sine and Cosine
Applying the inverse rule (25.1) with f (x) = sin(x) yields
h
i
d
? 1
=
sin (x)
?
1
?
?.
1
dx
cos sin (x)
(25.2)
Wcose?sairne?1a(xlm)? oinstththeedreen. omWienajtuosrt.
have to simplify To do this recall
the
!
? 1
=
the angle ? ? ? ?
2
2
sin (x) for which sin(?) = x .
Thus
? 1
sin (x)
It
is
the
angle
(between
??
and
?)
of
a
the
2
2
triangle on cause sin of
the this
uannigtlecierqclueawlshxo.)seTohpepnocsoiste?ssinid?e1(ixs)?xi.s(tBhee-
length of this side
the adjacpent side. By the
length is
? 1
2
x
.
Putting
Pyt?hag?o1rean? =thpeor?em2 cos sin (x) 1 x
into the above Equation (25.2), we get or latest rule:
1
?1
x
sin (x)
|
?
{z
?
?}
1
cos sin (x)
h
i
Rule 20
d
?
1
sin (x)
dx
= p1 ?2
1x
Derivatives of Inverse Sine and Cosine
287
We
reviewed
? 1
sin (x)
In
Section
6.1
and
presented
its
graph
on
page
101.
Figure 25.1 repeats the graph, along with the derivative from Rule 20.
=
? 1
y
f (x) cos (x)
? 2
1
-1
0 =p 1
f (x)
?2
1x
=
? 1
f (x) sin (x)
x 1
??
2
Figure 25.1.
The
function
? 1
sin (x)
and
its
derivative.
The derivativeis
always
positive,
reflecting
the
fact
that
the
tangents
to
? 1
sin
(x)
have
positive
slope.
The
derivative
has
vertical
asymptotes
at
= x
?1,
as
the
tangents
to
? 1
sin (x)
become
increasingly
steep
as
x
approaches
?1.
Now
consider
? 1
cos
(
x).
The
tangents to
its graph (Figure 25.2
below)
have negative slope, and the geometry suggests that its derivative is negative
the derivative of sin?1(x). Indeed this turns out to be exactly the case. This
chapter's Exercise 1 asks you to prove our next rule:
Rule 21
h
i
d
? 1
cos (x)
dx
=
? p1
?2 1x
=
? 1
?
f (x) cos (x)
-1 ?
x 1
1
0
? =p 1
f (x)
?2
1x
Figure
25.2.
The
function
? 1
cos (x)
and
its
derivative.
288
Derivatives of Inverse Trig Functions
25.2 Derivatives of Inverse Tangent and Cotangent
Now
let's
find
the
derivative
of
? 1
tan
(x).
Putting
f
(x)
=
tan(x)
into
the
inverse
rule
(25.1),
we
have
? 1
=
? 1
f (x) tan (x)
and
f
0
(x)
=
2
sec (
x),
and
we
get
h
i
d
?
1
tan (x)
dx
=
?
1
?
?
2
1
sec tan (x)
=
?
?
1
?
1
?? .
2
sec tan (x)
(25.3)
The
expression
sec
?? ?
1
tan (x)
in
the
denominator
is
the
length of the hypotenuse of the triangle to the right.
(See example Ppythagorean
6.3 in Chapter 6, page theorem, the length is
114? .)
By
? 1
t?he =
sec tan (x)
1
+
2
x
.
Inserting this into the above Equation (25.4)
yields
h
i
d dx
? 1
tan (x)
=
?
?
1
?
1
??
2
sec tan (x)
=
?p 1 ?
+
2 2
1x
=
1
+
.
2
1x
?
?
?1
) (x
n
ta
sec ?
x
1
tan (x)
1
We now have:
h
i
Rule 22
d
? 1
=1
tan (x)
+2
dx
1x
We
discussed
? 1
tan (x)
in
Chapter
6,
and
its
graph
is
in
Figure
6.3.
Below
Figure 25.3 repeats the graph, along with the derivative
1 2+
.
x1
?
=
? 1
f (x) tan (x)
2
1
x
0 =1
f (x) + 2
??
1x
2
Figure 25.3.
and
!lim?1
x
1 +2 1x
The = 0,
function
? 1
tan (
reflecting the
x) and its fact that
derivative
1 +
.
2
Note
1x
the tangent lines to
l!im1
x
=
1 +
2
=
1x
? 1
0
y tan (x)
become closer and closer to horizontal as ! ?1. The derivative bumps up
x
to 1 at = , where the tangent to =
? 1
is steepest, with slope 1
x0
y tan (x)
Exercise 3 below asks you to mirror the above arguments to deduce:
Rule 23
h d
? 1
i =
? 1
cot (x)
+2
dx
1x
Derivatives of Inverse Secant and Cosecant
289
25.3 Derivatives of Inverse Secant and Cosecant
We
reviewed
? 1
sec (x)
in
Section
6.3.
For
its
derivative,
put
= f (x) sec(x)
into
the
inverse
rule
(25.1),
with
? 1
=
? 1
f (x) sec (x)
and
f 0(x) = sec(x) tan(x).
We
get
h
i
d
? 1
=
sec (x)
0?
1
?
?
1
dx
f f (x)
=
??
1
?1 ?
?? ?
1
sec f (x) tan f (x)
=
?
? 1
?1 ?
??
? .
1
sec sec (x) tan sec (x)
Because
sec
?? ?
1
sec (x)
=
x,
the
above
becomes
h
i
d
? 1
sec (x)
dx
= ?
?1 ?
? .
1
x tan sec (x)
(p
In
Example
6.5
we
showed
that
?? ?
1
tan sec (x)
=
px2 ? 1 ? 2?
x1
With this, Equation 25.4 above becomes
(25.4)
if x is positive if x is negative
8
h
i
d
?
1
sec (x)
dx
=
>>< >>:
p1 2?
xx 1 p1
? 2?
xx 1
if x is positive if x is negative.
But
if
x
is
negative,
then
? x
is
positive,
and
the
above
consolidates
to
h
i
Rule 24
d
?
1
sec (x)
dx
= p1 | | 2? xx 1
This
graph
of
? 1
sec (x)
and
its
derivative
is
shown
in
Figure
25.3.
y
?
? 2
? 11
=
=
? 1
y f (x) sec (x)
= 0 = p1 y f (x) | | 2 ?
xx 1
Figure 25.4.
The graph of
? 1
and its derivative.
sec (x)
The domain of
both
functions
is
?1 (,
? 1]
[
[1,
1).
Note that the derivative has vertical
asymptotes
at
x = ?1,
where
the
tangent
line
to
=
? 1
y sec (x)
is
vertical.
290
Derivatives of Inverse Trig Functions
This chapter's Exercise 2 asks you to use reasoning similar to the above to deduce our final rule.
Rule 25
h
i
d
?
1
csc (x)
dx
=
? p1 | | 2? xx 1
Each of our new rules has a chain rule generalization. For example, Rule 25 generalizes as
h d
??
?i
1
csc g(x)
dx
=
??
? ??q 1
2?
0
g (x)
=
??
?0 ??qg (x) 2 ? .
g(x) (g(x)) 1
g(x) (g(x)) 1
Here is a summary of this Chapter's new rules, along with their chain rule generalizations.
h
i
d
?
1
= p1
sin (x) dx
?2 1x
h d
? 1
i
?
= p1
cos (x) dx
?2 1x
h
i
d
? 1
=1
tan (x)
+2
dx
1x
h
i
d
?
1
=
? 1
cot (x)
+2
dx
1x
h
i
d
?
1
= p1
sec (x) dx
| | 2? xx 1
h
i
d
? 1
csc (x)
dx
=
? p1 | | 2? xx 1
h d
??
?i
1
sin g(x)
=
q
1
0
g (x)
dx
?
2
h d
??
?i
1
cos g(x)
=
1 (g(x))
? q1
0
g (x)
dx
?
2
h d
??
1
?i =
1 (g(x))
1
0
tan g(x) dx
+
g (x)
2
1 (g(x))
h d
??
1
?i =
? 1
0
cot g(x) dx
+
g (x)
2
1 (g(x))
h d
??
?i
1
sec g(x)
dx
=
??
0
??qg (x) 2 ?
h d
??
1
?i
csc g(x)
dx
=
g(x) (g(x)) 1
?0
??
??qg (x) 2 ?
g(x) (g(x)) 1
Example 25.1
hp d
? 1
i
h?
=d
? 1
?1 i ? 2 =1
?
?? 1
h d
1
2
i
? 1
cos (x)
cos (x)
cos (x)
cos (x)
dx
dx
2
dx
? =1
? 1
?? 1
2
? p1
=
cos (x) 2
?2 1x
?
p
1p
.
? 1
?2
2 cos (x) 1 x
h
i
h
i
? 1
Example 25.2
?
?
d
1
tan (x)
=
1d
tan (x)
? 1
?
=
1
tan (x)
1
tan (x)
=e
.
e
e
tan (x) e
+2
+2
dx
dx
1x 1x
Example 25.3
h d
??
1
?i x=
1
hi d x = 1 x=
tan e dx
+ x2 1 (e )
dx
e
+
e
2x
1e
x
e +
.
2x
1e
................
................
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