1.4B and 3.10: Derivatives of Inverse Trig Functions
Problem 1 Explain what each of the following means:
(a) sin
1
(x)
Solution:
1
(b) (sin(x))
Solution:
(c) sin x
This denotes the inverse function to sin(x), sometimes denoted by arcsin(x).
This means sin(x) raised to the
1 power, i.e.
1
.
sin(x)
1
Solution:
✓ ◆
1
This means sin
.
x
Problem 2 Without using a calculator, determine if the statement
cos
1
cos(7⇡/6) = 7⇡/6
is true or false.
Solution:
This statement is false: the correct statement is cos
1
cos(7⇡/6) = 5⇡/6. (Why?)
Spoiler Alert: the cosine function is not invertible since its graph fails the horizontal line test.
To produce the inverse cosine we must first restrict the domain of cosine, to the interval [0, ⇡], to produce
an invertible function:
1
So, the range of cos
1
is [0, ⇡]. Since 7⇡/6 is not in this range, 7⇡/6 is never a possible output of cos
Problem 3 True or False: sin
Solution:
1
1
.
(0) = ⇡.
This statement is false: the correct statement is sin
1
(0) = 0. (Why?)
Spoiler Alert: the sine function is not invertible since its graph fails the horizontal line test.
To produce the inverse sine we first restrict the domain of sine, to the interval [ ⇡/2, ⇡/2], to produce an
invertible function:
2
So, the range of sin
1
is [ ⇡/2, ⇡/2]. Since ⇡ is not in this range, ⇡, is never a possible output of sin
1
.
Problem 4 Simplify each of the following expressions.
(a) cos
1
sin(⇡/2)
Solution: By the unit circle, sin(⇡/2) = 1, and so we are looking for cos
is [0, ⇡], and so, by properties of inverse functions, cos 1 (1) = 0.
(b) tan sin
1
1
(1). The range of cos
1
(x/4)
Solution:
triangle:
Let ✓ = sin
1
(x/4), then sin(✓) = x/4. We can then draw the corresponding right
Calling the adjacent side y, by the Pythagorean Theorem we obtain
42 = x2 + y 2 =) y =
Remark: Since ✓ is in the range of sin
Therefore, y > 0.
1
, it follows that
3
p
16
x2
⇡/2 < ✓ < ⇡/2. Therefore, cos(✓) = y/4 > 0.
Then
tan sin
1
(4/x) = tan(✓),
x
= ,
y
x
=p
.
16 x2
Note: tan ✓ has the same sign as x, since y > 0.
Problem 5 A table of values for f and f 0 is shown below. Suppose that f is a one-to-one function and f
is its inverse.
x
1
3
4
(a) Evaluate f
1
Solution:
(b) Evaluate
f 0 (x)
4
5
3
f (x)
3
4
6
(f (x)) at x = 3.
f
1
(f (3)) = f
1
(4) = 3
d
f (f (x)) at x = 3.
dx
Solution:
d
f (f (x)) = f 0 (f (x)) · f 0 (x) =)
dx

d
f (f (x))
dx
x=3
= f 0 (f (3)) · f 0 (3)
= f 0 (4) · 5 = 3 · 5 = 15
(c) Evaluate
d
ln((f (x)) at x = 3.
dx
Solution:
d
f 0 (x)
ln((f (x)) =
dx
f (x)
=)
(d) Evaluate f
Solution:
(e) Evaluate
1
(x) at x = 3.
f
d
f
dx
1
1
(3) = 1 () 3 = f (1)
(x) at x = 3.
4

d
ln(f (x))
dx
=
x=3
f 0 (3)
5
=
f (3)
4
1
Solution:
d
f
dx
1
(x) =
1
f 0 (f
=)
=
(f) Evaluate lim
x!4
Solution:
f (x)
x
lim
x!4

1 (x))
d
f
dx
1
(x)
=
x=3
1
f 0 (f
1 (3))
1
1
=
f 0 (1)
4
f (4)
4
f (x)
x
f (4)
= f 0 (4) = 3
4
Problem 6 Find the derivative of f 1 at the following points on the graph of f
Verify your answer by evaluating the derivative of f 1 at the given point.
1
without solving for f
1
.
Delete.We shall only ask this in part (a).
2
1
(a) f (x) = x + 1 (for x 0) at the point (5, 2) on the graph of f
derivative of f 1 at the given point.
Solution:
(f
Verifying: f
1
1 0
) (5) =
(x) =
(b) f (x) = x2
2x
Solution:
(f
p
. Verify your answer by evaluating the
1
. Since f 0 (x) = 2x, f 0 (2) = 4. Thus, (f

d
1
d
1 =)
f 1 (x) = p
=)
f 1
dx
dx
2 x 1
f 0 (2)
x
3 (for x  1) at the point (12, 3) on the graph of f
1 0
) (12) =
1
f 0(
3)
. Since f 0 (x) = 2x
2, f 0 ( 3) =
6
1 0
1
.
4
1
= p
2 5
) (5) =
x=5
1
1
=
1
4
.
2=
8. Thus, (f
1 0
) (12) =
1
.
8
Notice that for this problem, finding the inverse of f (x) involves complicated algebra. In this case, it’s
much easier to find the derivative of f 1 without solving for f 1
Problem 7 Find the slope of the tangent line to the curve y = f
2
line to the curve y = f (x) at (7, 4) is .
3
1
(x) at (4, 7) if the slope of the tangent
2
Note that the statement “the slope of the tangent line to the curve y = f (x) at (7, 4) is ”
3
2
specifically means that f 0 (7) = . The slope of the tangent line to the curve y = f 1 (x) at (4, 7) is (f 1 )0 (4),
3
1
1
3
and so we use the formula for the derivative of the inverse function to compute: (f 1 )0 (4) = 0
= 2 = .
f (7)
2
3
Solution:
Problem 8 Find the derivatives of the following functions:
5
(a) f (x) = sec
1
p
( x).
1
f 0 (x) = p p
x x
Solution:
(b) g(x) = ln(sin
tan
Solution:
1
1
2
1
p
2x x
=
1
(x)).
g 0 (x) =
Solution:
(c) h(x) =
1
1
· x
1 2
1
sin
1
(x)
·p
1
1
x2
.
1
.
(x2 + 4)
h0 (x) =
tan
1
(x2 + 4)
2
·
1
· (2x).
1 + (x2 + 4)2
Problem 9 A boat sails directly toward a 100-meter skyscraper that stands on the edge of a harbor. The
angular size ✓ of the building is the angle formed by lines from the top and bottom of the building to the
observer on the boat (see figure below).
(a) Express the angle ✓ as the function of x, the distance of the boat from the building.
Solution:
tan ✓ =
100
=) ✓ = tan
x
1
✓
100
x
◆
d✓
(b) The boat is sailing directly toward the skyscraper at 3 m/s. Find
when the boat is x = 300m from
dt
the building.
6
Solution:
✓
◆
d✓
d
100
=
tan 1
dt
dt
x
✓
◆
d
100 dx
=
tan 1
dx
x
dt
1
100 dx
=
2 · x2 dt
1 + 100
x
1
100
=
·
· ( 3)
2
100 2
x
1+ x
300
= 2
x + 1002

d✓
dt
300
3002 + 1002
3
=
900 + 100
3
=
1000
= 0.003 rad/sec
=
x=300
Alternatively, this problem could be done without inverse functions.
100
x✓
◆
d
d 100
(tan ✓) =
dt
dt
x
d✓
100
dx
sec2 ✓
=
2
dt
x dt
tan ✓ =
We have to evaluate sec2 ✓ at the moment when x = 300m
Using the triangle above, we have:
p
p
p
3002 + 1002
32 + 12
10
sec ✓ =
= 100
=
, we obtain:
300
300
3
7
p
10
3
!2 

d✓
dt
d✓
dt
=
x=300
=
x=300
8
100
( 3)
3002
1
9
3
·
=
rad/sec
300 10
1000