Calculus 2nd Edition Rogawski

400 CHAPTER 7 EXPONENTIAL FUNCTIONS
1
y
y
y
1
1
−2
−1
2
x
−2
−1
2
x
−4
−2
y = sech x
y
2 4
x
y = tanh x
y = coth x
FIGURE 5 The hyperbolic tangent and cotangent.
2
EXAMPLE 1 Verifying the Basic Identity Verify Eq. (1).
−2
2
x
Solution It follows from the definitions that
cosh x + sinh x = e x ,
cosh x − sinh x = e −x
−2
y = csch x
FIGURE 4 The hyperbolic secant and
cosecant.
We obtain Eq. (1) by multiplying these two equations together:
cosh 2 x − sinh 2 x = (cosh x + sinh x)(cosh x − sinh x) = e x · e −x = 1
Derivatives of Hyperbolic Functions
The formulas for the derivatives of the hyperbolic functions are similar to those for the
corresponding trigonometric functions, differing at most by a sign. For hyperbolic sine
and cosine we have
d
d
sinh x = cosh x,
dx
cosh x = sinh x
dx
It is straightforward to check this directly. For example,
d
dx sinh x = d ( e x − e −x ) ( e x − e −x ) ′
=
= ex + e −x
dx 2
2
2
= cosh x
REMINDER
tanh x = sinh x
cosh x = ex − e −x
e x + e −x ,
sech x = 1
cosh x = 2
e x + e −x
coth x = cosh x
sinh x = ex + e −x
e x − e −x ,
csch x = 1
sinh x = 2
e x − e −x
These formulas are similar to the formulas
dx d
d
sin x = cos x, dx
cos x = − sin x. The
derivatives of the hyperbolic tangent, cotangent, secant, and cosecant functions are computed
in a similar fashion. We find that their derivatives also differ from their trigonometric
counterparts by a sign at most.
Derivatives of Hyperbolic and Trigonometric Functions
d
dx tanh x = sech2 x,
d
dx coth x = − csch2 x,
d
d
sech x = − sech x tanh x,
dx
d
d
csch x = − csch x coth x,
dx
d
dx tan x = sec2 x
d
dx cot x = − csc2 x
sec x = sec x tan x
dx
csc x = − csc x cot x
dx