MATH 242 – HYPERBOLIC FUNCTIONS
Disclaimer: This assignment is different from the others. This assignment presupposes no knowledge about
hyperbolic functions. We didn’t discuss them in class and you don’t necessarily need to read section 7.7.
You will be learning about hyperbolic functions as you do the assignment.
(1) The hyperbolic sine function is defined to be
ex − e−x
,
2
and the hyperbolic cosine function is defined to be
sinh x =
ex + e−x
.
2
What are the domains of sinh x and cosh x?
Compute the values
(a) sinh(0)
(b) cosh(0)
(c) sinh(ln 2)
(d) sinh(ln x) (Simplify it.)
Show that cosh2 x − sinh2 x = 1.
Compute the limits
(a) lim sinh x
cosh x =
(i)
(ii)
(iii)
(iv)
x→∞
(b)
lim sinh x
x→−∞
(c) lim cosh x
x→∞
(d)
lim cosh x
x→−∞
(v) What is the range of sinh x? (Hint: Think about the previous part.)
d
[sinh x] = cosh x.
(vi) Show that dx
d
(vii) Show that dx
[cosh x] = sinh x.
(2) The hyperbolic tangent, cotangent, secant, and cosecant are defined to be
sinh x
cosh x
1
1
, coth x =
, sech x =
, csch x =
.
cosh x
sinh x
cosh x
sinh x
Using parts (iii), (vi), (vii) of the previous question (and perhaps the Quotient Rule), show that
tanh x =
d
[tanh x] = sech2 x,
dx
d
[coth x] = − csch2 x,
dx
(3)
(i)
(ii)
(iii)
(iv)
d
[sech x] = − sech x tanh x,
dx
d
[csch x] = − csch x coth x.
dx
Explain why cosh x >
p0 for all x.
Show that cosh x = 1 + sinh2 x. (Hint: Use something from question (1).)
Explain why f (x) = sinh x is an increasing function. (Hint: What is its derivative?)
Since f (x) = sinh x is increasing, it is one-to-one and therefore has an inverse function
f −1 (x) = sinh−1 x.
What is the domain and range of f −1 (x) = sinh−1 x?
(v) Use part (ii) to show that
p
cosh(sinh−1 x) = 1 + x2 .
1
2
MATH 242 – HYPERBOLIC FUNCTIONS
(vi) Use the formula for the derivative [f −1 ]0 (x) of an inverse function (Theorem 1, page 370) to
show that
d
1
[sinh−1 x] = √
.
dx
1 + x2
(4) One can attempt to find a formula for sinh−1 x by solving
ex − e−x
2
for x. However, this proves to be difficult algebraically. We will follow a different route to get a
formula.
√
(i) Carefully compute the derivative of ln(x + 1 + x2 ) and show that
p
d
1
[ln(x + 1 + x2 )] = √
.
dx
1 + x2
(You’ll need to do some careful
simplification to get this.)
√
(ii) Since sinh−1 x and ln(x+ 1 + x2 ) have the same derivative, they differ by a constant (Corollary
2, page 200). That is,
p
sinh−1 x = ln(x + 1 + x2 ) + C
y = sinh x =
for some constant C. Plug in x = 0 to both sides to see that C = 0. (What is sinh−1 (0)?
You essentially figured it out somewhere in (1).) Conclude with the explicit formula for inverse
hyperbolic sine:
p
sinh−1 x = ln(x + 1 + x2 ).