⋔αth ∫∆rm

3.11

Hyperbolic Functions

Definition of Hyperbolic Functions

$$\begin{gathered} \sin h x = \frac{e^x-e^{-x}}{2} \\ \cos h x = \frac{e^x+e^{-x}}{2} \\ \tan h x = \frac{\sin h x}{\cos h x} \\ csch x = \frac{1}{\sin h x} \\ sech x = \frac{1}{\cos h x} \\ coth x = \frac{\cos h x}{\sin h x} \end{gathered}$$

Certain even and odd combinations of the exponential functions e^x and e^-x arise so frequently in mathematics and its applications that they deserve to be given special names. In many ways they are analogous to the trigonomtric functions, and they have the same relationship to the hyperbola that the trigonometric functions have to the circle. For this reason they are collectively called hyperbolic functions and individually called hyperbolic sin, hyperbolic cosine, and so on.

Inverse Hyperbolic Functions

$$\begin{gathered} y = \sin h^{-1}x \iff \sin h y = x \\ y = \cos h^{-1}x \iff \cos h y = x and y \ge 0 \\ y = \tan h^{-1}x \iff \tan h y = x \end{gathered}$$

You can see from the figures above that sinh and tanh are one-to-one functions and so they have inverse functions denoted by sinh^-1 and tanh^-1. Figure 2 shows that cosh is not one-to-one, but when restricted to the domain [0, ∞) it becomes one-to-one. The inverse hyperbolic cosine function is defined as the inverse of this restricted function.

Since the hyperbolic functions are defined in terms of exponential functions, it's not surprising to learn that the inverse hyperbolic functions can be expressed in terms of logarithms. In particular, we have:

$$\begin{gathered} \sin h^{-1}x = \ln (x + \sqrt{x^2+1}) x\in \mathrm{ℝ} \\ \cos h^{-1}x = \ln (x + \sqrt{x^2-1}) x\ge 1 \\ \tan h^{-1}x = \frac{1}{2}\ln \left(\frac{1+x}{1-x}\right) -1<x<1 \end{gathered}$$

Derivatives of Inverse Hyperbolic Functions

$$\begin{gathered} \frac{d}{dx}\left(\sin h^{-1}x\right) = \frac{1}{\sqrt{1+x^2}} \\ \frac{d}{dx}\left(\cos h^{-1}x\right) = \frac{1}{\sqrt{x^2-1}} \\ \frac{d}{dx}\left(\tan h^{-1}x\right) = \frac{1}{1-x^2} \\ \frac{d}{dx}\left(csc{h}^{-1}x\right) = -\frac{1}{\left|x\right|\sqrt{x^2+1}} \\ \frac{d}{dx}\left(sec{h}^{-1}x\right) = -\frac{1}{x\sqrt{1-x^2}} \\ \frac{d}{dx}\left(cot{h}^{-1}x\right) = \frac{1}{1-x^2} \end{gathered}$$

The inverse hyperbolic functions are all differentiable because the hyperbolic functions are differentiable. The formulas above can be proved either by the method for inverse functions or by differentiating Formulas 3,4, and 5 (formulas from the previous boxes).

Links

• Home
• 3.1: Derivatives of Polynomials and Exponential Functions
• 3.2: The Product and Quotient Rules
• 3.3: Derivatives of Trigonometric Functions
• 3.4: The Chain Rule
• 3.5: Implicit Differentiation
• 3.6: Steps in Logarithmic Differentiation
• 3.7: Rates of Change in the Natural and Social Sciences
• 3.8: Exponential Growth and Decay
• 3.9: Related Rates
• 3.10: Linear Approximations and Differentials
• 3.11: Hyperbolic Functions