Page 200 - Calculus Demystified

CHAPTER 6
Transcendental Functions
As with the first two trigonometric functions, we note that the tangent func- 187
√ √
tion takes each of the values 1, 1/ 3, − 3 at many different points of its
domain. But Tan x takes each of these values at just one point of its domain.
The derivative of our new function may be calculated in the usual way. The
result is
d −1 1
Tan t = .
dt 1 + t 2
Next we calculate some derivatives:
EXAMPLE 6.39
Calculate the following derivatives:

d d −1 d −1
Tan −1 , Tan (x ) , Tan (e ) .
x
3
x
dx dx √ dx
x=1 x= 2 x=0
SOLUTION
We have

d 1 1
Tan −1 = = ,
x
dx x=1 1 + x 2 x=1 2

d −1 1 2
Tan (x ) = · 3x 2 = ,
3
3 2
√
√
dx x= 2 1 + (x ) x= 2 3

d −1 1 1
Tan (e ) = · e x = .
x
x 2
dx 1 + (e ) 2
x=0 x=0
3
You Try It: Calculate (d/dx)Tan −1 [ln x + x ] and (d/dx) ln[Tan −1 x].
6.6.4 INTEGRALS IN WHICH INVERSE
TRIGONOMETRIC FUNCTIONS ARISE
Our differentiation formulas for inverse trigonometric functions can be written in
reverse, as antidifferentiation formulas. We have

du −1
√ = Sin u + C;
1 − u 2

du −1
√ =−Cos u + C;
1 − u 2

du −1
du = Tan u + C.
1 + u 2
The important lesson here is that, while the integrands involve only polynomials
and roots, the antiderivatives involve inverse trigonometric functions.