1—Basic Stuff 4
The solutions to this areex=y±
√
y^2 + 1, and because
√
y^2 + 1is always greater than|y|, you must in this
case take the positive sign to get a positiveex. Take the logarithm ofexand
sinh
sinh−^1
x= sinh−^1 y= ln
(
y+
√
y^2 + 1
)
(−∞< y <+∞)
Asxgoes through the values−∞to+∞, the values thatsinhxtakes on go over the range−∞to+∞. This
implies that the domain ofsinh−^1 yis−∞< y <+∞. The graph of an inverse function is the mirror image
of the original function in the 45 ◦liney=x, so if you have sketched the graphs of the original functions, the
corresponding inverse functions are just the reflections in this diagonal line.
The other inverse functions are found similarly; see problem 3
cosh−^1 y= ln
(
y±
√
y^2 − 1
)
, y≥ 1
tanh−^1 y=
1
2
ln
1 +y
1 −y
, |y|< 1 (4)
coth−^1 y=
1
2
ln
y+ 1
y− 1
, |y|> 1
Thecosh−^1 function is commonly written with only the+sign before the square root. What does the other sign
do? Draw a graph and find out. Also, what happens if you add the two versions of thecosh−^1?
The calculus of these functions parallels that of the circular functions.
d
dx
sinhx=
d
dx
ex−e−x
2
=
ex+e−x
2
= coshx