Mathematical Tools for Physics

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