Calculus/Derivatives of Hyperbolic Functions

The hyperbolic functions are functions that are related to the trigonometric functions, largely due to the consequences of their definitions. While not necessarily related to the triangular or circular definitions of the trigonometric functions, their identities make them look quite similar.

Nevertheless, these functions are important in calculus because they form a large subset of solutions to some problems in calculus and are important in many areas of applied mathematics.

Introducing the Hyperbolic Functions


Section defines hyperbolic functions according to the parametric definition, similar to trigonometric functions.

That is, rotating a ray from the direction of the positive half of the $$x$$-axis by an angle $$\theta$$ (counterclockwise for $$\theta > 0,$$ and clockwise for $$\theta < 0$$) yields intersection points of this ray with the unit hyperbola: $$A=(x_A,y_A)$$.

It is defined that: $$ $$ $$

where $$a$$ is twice the area between the ray, the hyperbola, and the $$x$$-axis. As a consequence of these definitions, it becomes clear

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Keep in mind that similar to $$\sec (x) $$, $$\csc (x) $$, and $$\cot (x) $$, the definitions of the hyperbolic versions are as follows: $$ $$ $$

These definitions are limited in that applying calculus to this would be difficult without additional tools. Using the exponential definition of hyperbolic functions (not defined in Section ) allow us to more easily find derivatives, the goal of this section.

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Properties of the Hyperbolic Functions
Proof: By equation, $$ \sinh x = \frac{e^x - e^{-x}}{2} $$. If a function is odd, then $$f(-x)=-f(x)$$.
 * $$\begin{align}

\sinh (-x) &= \frac{e^{(-x)} - e^{-(-x)}}{2} \\ &= \frac{e^{-x} - e^{x}}{2} \\ &= -\frac{e^{x} - e^{-x}}{2} = -\sinh (x) \qquad \blacksquare \end{align}$$

Proof: By equation, $$ \cosh x = \frac{e^x + e^{-x}}{2} $$. If a function is even, then $$f(-x)=f(x)$$.
 * $$\begin{align}

\cosh (-x) &= \frac{e^{(-x)} + e^{-(-x)}}{2} \\ &= \frac{e^{-x} + e^{x}}{2} = \frac{e^{x} + e^{-x}}{2} = \cosh (x) \qquad \blacksquare \end{align}$$

Proof: Using equations and, $$ \cosh x = \frac{e^x + e^{-x}}{2} $$,
 * $$\begin{align}

\sinh (2x) &= \frac{e^{(2x)} + e^{-(2x)}}{2} \\ &= \frac{e^{4x}-1}{2e^{2x}} \\ &= \frac{\left(e^{2x}-1\right)\left(e^{2x}+1\right)}{2e^{2x}} \end{align}$$ Notice that $$\sinh (x)\cosh (x) = \frac{e^{4x}-1}{4e^{2x}}$$. Multiplying this by two gives us $$\sinh (2x)$$. Hence,
 * $$\sinh(2x) = 2\sinh(x)\cosh(x) \qquad \blacksquare$$

Proof:

Derivatives of the Hyperbolic Functions
Proof:

Proof: