3.6 Hyperbolic Functions

Arkansas Tech University MATH 2914: Calculus I Dr. Marcel B. Finan

3.6 Hyperbolic Functions The hyperbolic functions possess many parallels with the trigonometric functions, both in naming and in their properties and inter-relationships. The function ex and its reciprocal e−x can be combined to create the following new functions:

Hyperbolic Cosine ex + e−x cosh (x) = . 2 Hyperbolic Sine ex − e−x sinh (x) = . 2 Hyperbolic Tangent ex − e−x sinh (x) tanh (x) = = . ex + e−x cosh (x) Hyperbolic Cosecant 1 csch(x) = sinh (x) Hyperbolic Secant 1 sech(x) = cosh (x) Hyperbolic Cotangent 1 cosh (x) coth (x) = = . tanh x sinh (x)

Example 3.6.1 Find cosh (0) and sinh (0).

Solution. Substituting 0 for x in the formulas for cosh x and sinh x to obtain

e0 + e−0 1 + 1 cosh (0) = = = 1 2 2 and e0 − e−0 1 − 1 sinh (0) = = = 0 2 2

1 Hyperbolic Functions Identities Hyperbolic functions have similar identities than that of trigonometric functions. Substituting −x for x in the formula of cosh (x) to obtain e−x + ex cosh (−x) = = cosh (x). 2 1 This says that cosh (x) is an even function. Since sech(x) = cosh (x) , the function sech(x) is also even. Similarly, substituting −x for x in the formula of sinh (x) to obtain e−x − ex sinh (−x) = = − sinh (x). 2 That is, sinh (x) is an odd function. One can easily show, that tanh (x), csch(x), and coth (x) are odd functions. Next, we derive an identity for the hyperbolic functions similar to the Pythagorean identity for the trigonometric functions. ex + e−x 2 ex − e−x 2 cosh2 (x) − sinh2 (x) = − 2 2 e2x + 2 + e−2x e2x − 2 + e−2x = − = 1. 4 4 Thus, (cosh (x), sinh (x)) is a point on the hyperbola x2 − y2 = 1, thus, the name hyperbolic. (Recall that the functions sin x and cos x are called circular functions since (cos x, sin x) is a point on the unit circle!) Simple calculations show the following identities 1 − tanh2 (x) =sech2(x) coth2 (x) − 1 =csch2(x) sinh (x + y) = sinh (x) cosh (y) + cosh (x) sinh (y) sinh (x − y) = sinh (x) cosh (y) − cosh (x) sinh (y) cosh (x + y) = cosh (x) cosh (y) + sinh (x) sinh (y) cosh (x − y) = cosh (x) cosh (y) − sinh (x) sinh (y).

Derivatives of Hyperbolic Functions Since (ex)0 = ex and (e−x)0 = −e−x, we have d d ex + e−x ex − e−x (cosh (x)) = = = sinh (x). dx dx 2 2

2 dy du Now, if y = cosh (u(x)) then by the chain rule, we have dx = sinh (u) · dx . Similarly,

d d ex − e−x ex + e−x (sinh (x)) = = = cosh (x) dx dx 2 2 and d du (sinh (u(x))) = cosh (u(x)) . dx dx To ﬁnd the derivative of tanh (x), we use the quotient rule

d d sinh (x) cosh2 (x) − sinh2 (x) 1 (tanh (x)) = = = . dx dx cosh (x) cosh2 (x) cosh2 (x)

Also, d du (tanh (u(x))) = csch2(u(x)) · . dx dx We leave it to the reader to show the following derivatives d (sech(x)) = sech(x) tanh (x) dx d du (sech(u(x))) = sech(u(x)) tanh (u(x)) · dx dx d (csch(x)) = −csch(x) coth (x) dx d du (csch(u(x))) = −csch(u(x)) coth (u(x)) · dx dx d (coth (x) = −csch2(x). dx d du (coth (u(x)) = −csch2(u(x)) · . dx dx Graphs of Hyperbolic Functions Since cosh (x) is even, the graph is symmetric about the y−axis so we restrict ourselves to x ≥ 0 and then we complete the graph by symmetry. Since 0 ex−e−x x cosh (0) = 1, the y−intercept is (0, 1). Since (cosh (x)) = 2 and e > e−x for x > 0, the derivative is positive and so the function is increasing for 00 ex+e−x x > 0. Since (cosh (x)) = 2 > 0, the graph is concave up. The graph of cosh (x) is shown in Figure 3.6.1(a). Likewise, the function sinh (x) is odd so its graph is symmetric about the origin so we restrict ourselves to x ≥ 0 and then we complete the graph by

3 symmetry. Since sinh (0) = 0, the y−intercept is (0, 0). Since (sinh (x))0 = ex+e−x 2 > 0, the derivative is positive and so the function is increasing for x > 00 ex−e−x x −x 00 0. Since (sinh (x)) = 2 , and e > e for x > 0, we have (sinh (x)) > 0 so that the graph is concave up. The graph of sinh (x) is shown in Figure 3.6.1(b).

Figure 3.6.1

The function tanh (x) is odd so its graph is symmetric about the origin so we restrict ourselves to x ≥ 0 and then we complete the graph by symmetry. Since tanh (0) = 0, the y−intercept is (0, 0). Since (tanh (x))0 = sech2(x) > 0, the derivative is positive and so the function is increasing. Since (tanh (x))00 = −2sech2(x) tanh (x), and tanh (x) > 0 for x > 0, we have (tanh (x))00 < 0 so that the graph is concave down. From the fact that

ex − e−x 1 − e−2x lim = lim = 1 x→∞ ex + e−x x→∞ 1 + e−2x we conclude that y = 1 is a horizontal asymptote. The graph of tanh (x) is shown in the ﬁgure below.