Page 204 - Calculus Demystified

Transcendental Functions
CHAPTER 6
y 191

_ 1
y = Sec x

1 x

Fig. 6.24

y

_ /2 /2 x

Fig. 6.25

It is possible to show that

d 1
Csc −1 x =− √ , |x| > 1.
dx |x|· x − 1
2
√ √
You Try It: What is Sec −1 (−2/ 3)? What is Csc −1 (− 2)?