Page 197 - Calculus Demystified

CHAPTER 6
Transcendental Functions
184
√
2 π
Sin −1 − =− .
2 4
√
Notice that even though the sine function takes the value 3/2 at many different
values of the variable x, the function Sine takes this value only at x = π/3.
Similar comments apply to the other two examples.
We also have
√
3 5π
Cos −1 − = ,
2 6
π
−1
Cos 0 = ,
2
√
2 π
Cos −1 = .
2 4
We calculate the derivative of f(t) = Sin −1 t by using the usual trick for inverse
functions. The result is
d −1 1 1
(Sin (x)) = = √ .
dx 1 − sin (Sin −1 x) 1 − x 2
2
The derivative of the function Cos −1 t is calculated much like that of Sin −1 t. We
find that
d −1 1
(Cos (x)) =−√ .
dx 1 − x 2
EXAMPLE 6.36
Calculate the following derivatives:

d d −1 2 d −1 1
x
Sin −1 , Sin (x + x) , Sin .
dx √ dx dx x √
x= 2/2 x=1/3 x=− 3
SOLUTION
We have

d 1 √
Sin −1 √ = √ = 2,
x
√
dx x= 2/2 1 − x 2 x= 2/2

d −1 2 1 15
Sin x + x = · (2x + 1) = √ ,
dx 2 2 65
x=1/3 1 − (x + x)
x=1/3

d −1 1 1 1
Sin (1/x) = · − =−√ .
dx x=− 3 1 − (1/x) 2 x 2 x=− 3 6
√
√