Page 198 - Calculus Demystified

CHAPTER 6
Transcendental Functions
2
You Try It: Calculate (d/dx)Cos −1 [x + x]. Also calculate (d/dx)Sin −1 × 185
3
[ln x − x ].
EXAMPLE 6.37
Calculate each of the following derivatives:

d d −1 d −1 √
x
Cos −1 , Cos (ln x) , Cos ( x) .
dx dx √ dx
x=1/2 x= e x=1/2
SOLUTION
We have

d 1 2
x
Cos −1 =− √ =−√ ,
dx x=1/2 1 − x 2 x=1/2 3

d −1 1 1 −2
Cos (ln x) =− · =−√ ,
dx x= e 1 − (ln x) 2 x x= e 3e
√
√

d −1 √ 1 1
Cos ( x) =− √ · x −1/2 =−1.
dx 2 2
x=1/2 1 − ( x)
x=1/2
You Try It: Calculate (d/dx) ln[Cos −1 x] and (d/dx) exp[Sin −1 x].
6.6.3 THE INVERSE TANGENT FUNCTION
Define the function Tan x to be the restriction of tan x to the interval (−π/2,π/2).
Observe that the tangent function is undefined at the endpoints of this interval.
Since
d 2
Tan x = sec x
dx
we see that Tan x is increasing, hence it is one-to-one (Fig. 6.19). Also Tan takes
arbitrarily large positive values when x is near to, but less than, π/2.And Tan takes
negative values that are arbitrarily large in absolute value when x is near to, but
greater than, −π/2.ThereforeTan takes all real values. SinceTan : (−π/2,π/2) →
(−∞, ∞) is one-to-one and onto, the inverse function Tan −1 : (−∞, ∞) →
(−π/2,π/2) exists. The graph of this inverse function is shown in Fig. 6.20. It
is obtained by the usual procedure of reflecting in the line y = x.
EXAMPLE 6.38

Calculate
√ √
−1 −1 −1
Tan 1, Tan 1/ 3, Tan (− 3).