Calculus 2nd Edition Rogawski

392 CHAPTER 7 EXPONENTIAL FUNCTIONS
THEOREM 1 Derivatives of Arcsine and Arccosine
d
1
dx sin−1 x = √ ,
1 − x 2
d
1
dx cos−1 x = −√ 1 − x 2
2
REMINDER If g(x) is the inverse of
f (x), then
g ′ (x) =
1
f ′ (g(x))
3
Proof Apply Eq. (3) in the margin to f (x) = sin x and g(x) = sin −1 x:
d
1
dx sin−1 x =
f ′ (sin −1 x) = 1
cos(sin −1 x) = 1
√
1 − x 2
In the last step, we use Eq. (1) from Example 2. The derivative of cos −1 x is similar (see
Exercise 49 or the next example).
EXAMPLE 3 Complementary Angles The derivatives of sin −1 x and cos −1 x are equal
up to a minus sign. Explain this by proving that
1
y
x
sin −1 x + cos −1 x = π 2
Solution In Figure 5, we have θ = sin −1 x and ψ = cos −1 x. These angles are comple-
mentary, so θ + ψ = π/2 as claimed. Therefore,
q
FIGURE 5 The angles θ = sin −1 x and
ψ = cos −1 x are complementary and thus
sum to π/2.
d
dx cos−1 x = d ( π
x)
dx 2 − sin−1 = − d
dx sin−1 x
π
2
− π 2
y = tan −1 x
y
π
π
2
y
x
x
EXAMPLE 4 Calculate f ′( 1
2
)
, where f (x) = arcsin(x 2 ).
Solution Recall that arcsin x is another notation for sin −1 x. By the Chain Rule,
d
dx arcsin(x2 ) = d
dx sin−1 (x 2 ) =
( )
12
( ) 1 2
f ′ = √
2
1 −
( )
= √
1
4
12
15
16
1
√
1 − x 4
= 4 √
15
d
dx x2 =
2x
√
1 − x 4
We now address the remaining trigonometric functions. The function f(θ) = tan θ
is one-to-one on ( − π 2 , π 2
)
, and f(θ) = cot θ is one-to-one on (0, π) [see Figure 10 in
Section 1.4]. We define their inverses by restricting them to these domains:
θ = tan −1 x is the unique angle in
(
− π 2 , π )
such that tan θ = x
2
θ = cot −1 x is the unique angle in (0, π) such that cot θ = x
FIGURE 6
y = cot −1 x
The range of both tan θ and cot θ is the set of all real numbers R. Therefore, θ = tan −1 x
and θ = cot −1 x have domain R (Figure 6).