Calculus Of One Real Variable – By Pheng Kim Ving Chapter 7: The Exponential And Logarithmic Functions – Section 7.7: The Inverse Hyperbolic Functions 7.7 The Inverse Hyperbolic Functions

 1. Definitions

# The Inverse Hyperbolic Sine Function

The graph of the hyperbolic sine function y = sinh x is sketched in Fig. 1.1. Clearly sinh is one-to-one, and so has an
inverse, denoted sinh–1. The inverse hyperbolic sine function sinh–1 is defined as follows: The graph of y = sinh–1 x is the mirror image of that of y = sinh x in the line y = x. It's shown in Fig. 1.1. We have
dom(sinh–1) = R and range(sinh–1) = R. Fig. 1.1   Graph of y = sinh–1 x.

# The Inverse Hyperbolic Cosine Function  Fig. 1.2   Graph of y = cosh–1 x.

# The Inverse Hyperbolic Tangent Function

The graph of the hyperbolic tangent function y = tanh x is sketched in Fig. 1.3. Clearly tanh is one-to-one, and so has an
inverse, denoted tanh–1. The inverse hyperbolic tangent function tanh–1 is defined as follows:  Fig. 1.3   Graph of y = tanh–1 x.

# The Inverse Hyperbolic Cotangent Function

The graph of the hyperbolic cotangent function y = coth x is sketched in Fig. 1.4. Clearly coth is one-to-one, and thus has
an inverse, denoted coth–1. The inverse hyperbolic cotangent function coth–1 is defined as follows:  Fig. 1.4   Graph of y = coth–1 x.

# The Inverse Hyperbolic Secant Function  Fig. 1.5   Graph of y = sech–1 x.

# The Inverse Hyperbolic Cosecant Function

The graph of the hyperbolic cosecant function y = csch x is sketched in Fig. 1.6. Clearly csch is one-to-one, and so has
an inverse, denoted csch–1. The inverse hyperbolic cosecant function csch–1 is defined as follows:  Fig. 1.6   Graph of y = csch–1 x.

# Example 1.1

Prove the identity: ### Note

Recall that the inverse of the natural exponential function is the natural logarithm function. Since the hyperbolic functions
are defined in terms of the natural exponential function, it's not surprising that their inverses can be expressed in terms
of the natural logarithm function. Also see Problem & Solution 1 and Problem & Solution 2.

### Solution

Let y = sinh–1 x. Then x = sinh y = (eyey)/2. So eyey – 2x = 0. Multiplying both sides by ey yields e2y – 1 – 2xey = 0,
or e2y – 2xey – 1 = 0, which is a quadratic equation in ey. Its roots are: EOS

 2. Differentiation We prove formula [2.1] as follows. Let y = sinh–1 x. Then x = sinh y. Differentiating this equation implicitly with respect to x
we get: The remaining differentiation formulas are proved in a similar way.

# Example 2.1

Differentiate sinh–1 tan x.

### Solution ## Problems & Solutions

1. In Example 1.1 we proved the identity: Also see Problem & Solution 2.

Solution

Let y = cosh–1 x. Then x = cosh y = (ey + ey)/2. So ey + ey – 2x = 0. Multiplying both sides by ey yields e2y + 1 – 2xey =
0, or e2y – 2xey + 1 = 0, which is a quadratic equation in ey. Its roots are:  2. In Example 1.1 we proved 1 identity and in Problem & Solution 1 you were asked to prove another identity. Now again
you're asked to prove the following 2 identities: Solution

a.  Let y = tanh–1 x. So x = tanh y and |x| < 1. We have: xe2yx = e2y + 1,

e2y(x – 1) = x + 1,  3. Differentiate the following functions.
a.  sinh–1 (x/a), a > 0.
b.  cosh–1 (x/a), a > 0.

Solution  4. Differentiate the following functions.
a.  y = sech–1 (x2).
b.  f(t) = csch–1 tan t.

Solution  5. Prove that: Solution

Let y = csch–1 x. Then x = csch y. Let z = sinh–1 (1/x), so that 1/x = sinh z, or:  