## Hyperbolic Functions

### Hyperbolic Sine, Hyperbolic Cosine, and Hyperbolic Tangent

In many applications, exponential functions appear in combinations in the form of $e^x+e^{-x}\quad\text{and}\quad e^x-e^{-x}.$ It is convenient to introduce some new functions $\bbox[#F2F2F2,5px,border:2px solid black]{\cosh x=\frac{e^x+e^{-x}}{2},\quad\sinh x=\frac{e^x-e^{-x}}{2} }$ These functions are called hyperbolic cosine and hyperbolic sine. Although these are combined exponential functions, in some ways they resemble the trigonometric functions.

• An easy way to remember the above formulas is to notice that the hyperbolic cosine, like the cosine function, is an even function and the hyperbolic sine, like the sine function, is an odd function \begin{align} \cosh(-x)&=\frac{e^{-x}+e^{-(-x)}}{2}=\cosh x\\ \sinh(-x)&=\frac{e^{-x}-e^{-(-x)}}{2}=-\frac{e^{x}-e^{-x}}{2}=-\sinh x \end{align}

• Because both $$e^x$$ and $$e^{-x}$$ are continuous and differentiable everywhere, so are the functions $$\cosh x$$ and $$\sinh x$$.

• The graphs of $$y=\cosh x$$, $$y=\sinh x$$, as well as the graphs of $$y=\frac{1}{2}e^x$$ and $$y=\frac{1}{2}e^{-x}$$ are illustrated in Figure 1.

• Notice that $\cosh 0=1\quad \sinh 0=0$

The definition of the hyperbolic tangent resembels the definition of the tangent function $\bbox[#F2F2F2,5px,border:2px solid black]{ \tanh x=\frac{\sinh x}{\cosh x}=\frac{e^x-e^{-x}}{e^x+e^{-x}}}$ This function is also continuous and differentiable everywhere. The graph of $$y=\tanh x$$ is shown in Figure 2.

• As we can see from Figure 2, the hyperbolic tangent is an odd function, because \begin{align} \tanh (-x)&=\frac{\sinh (-x)}{\cosh(-x)}\\ &=\frac{-\sinh x}{\cosh x}\\ &=-\tanh x. \end{align}

• Because when $$x$$ is positive large, $$e^{-x}$$ is extremely small $e^x\pm e^{-x}\approx e^x$ and we have $\tanh x=\frac{e^x-e^{-x}}{e^x+e^{-x}}\approx \frac{e^x}{e^x}=1.$ Therefore, as we can see in Figure 2, we have $\lim_{x\to +\infty}\tanh x=1.$ Using the same reasoning we can show $\lim_{x\to -\infty }\tanh x=-1.$

### Other Hyperbolic Functions

We also have the hyperbolic cotangent, hyperbolic secant, and hyperbolic cosecant functions: \begin{align} \coth x&=\frac{\cosh x}{\sinh x}=\frac{e^x+e^{-x}}{e^x-e^{-x}}\\[6pt] \text{sech} x&=\frac{1}{\cosh x}=\frac{2}{e^x+e^{-x}}\\[6pt] \text{csch} x&=\frac{1}{\sinh x}=\frac{2}{e^x-e^{-x}} \end{align}

• Note that because $$\sinh 0=0$$, the hyperbolic cotangent and hyperbolic cosecant functions are not defined for $$x=0$$.
• The graph of these functions is shown in Figure 3.

### Hyperbolic Identities

The properties of the hyperbolic functions closely resemble the corresponding properties of the trigonometric functions.

Using the definitions, we can show $\bbox[#F2F2F2,5px,border:2px solid black]{\cosh^2 x-\sinh^2x=1}$

\begin{align} \cosh x-\sinh x&=e^{-x}\\ \cosh x+\sinh x&=e^{x} \end{align}

\begin{align} 1-\tanh^{2}x&=\frac{1}{\cosh^{2}x}\\[6pt] \cosh^{2}x-1&=\frac{1}{\sinh^{2}x}, \end{align}

\bbox[#F2F2F2,5px,border:2px solid black]{\begin{align} \sinh(x\pm y)=\sinh x\cosh y\pm\cosh x\sinh y,\\[6pt]\cosh(x\pm y)=\cosh x\cosh y\pm\sinh x\sinh y \end{align}}

Example 1

Show that $$\cosh^2 x-\sinh^2x=1$$.

Solution

Recall that $$(A\pm B)^2 = A^2\pm 2 AB+B^2$$. So \begin{align} \cosh^2 x-\sinh^2 x=&\frac{(e^x+e^{-x})^2}{4}-\frac{(e^x-e^{-x})^2}{4}\\[6pt] =&\frac{e^{2x}+2 \overbrace{e^x e^{-x}}^{=1}+e^{-2x}}{4}\\[6pt] &\quad-\frac{e^{2x}-2 \overbrace{e^x e^{-x}}^{=1}+e^{-2x}}{4}\\ =&\frac{1}{4}(e^{2x}+2+e^{-2x}-e^{2x}+2-e^{-2x})\\ =&\frac{4}{4}=1. \end{align}

Example 2

Show that $$1-\tanh^2 x=\text{sech}^2 x$$.

Solution
\begin{align} 1-\tanh^2 x=&1-\frac{\sinh^2 x}{\cosh^2 x}\\[6pt] =&\frac{\cosh^2 x}{\cosh^2 x}-\frac{\sinh^2 x}{\cosh^2 x}\\[6pt] =&\frac{\cosh^2 x-\sinh^2 x}{\cosh^2 x}\\[6pt] =&\frac{1}{\cosh^2 x}=\text{sech}^2 x \end{align}
Example 3

Prove $$\sinh(x+ y)=\sinh x\cosh y+\cosh x\sinh y$$

Solution

Let’s simplify the right hand side \begin{align} \sinh x\cosh y+\cosh x\sinh y=&\frac{e^x-e^{-x}}{2}\frac{e^y+e^{-y}}{2}\\[6pt] &\quad +\frac{e^x+e^{-x}}{2}\frac{e^y-e^{-y}}{2}\\[6pt] =&\frac{1}{4}\left[e^{x+y}+\cancel{e^{x-y}}-\bcancel{e^{-x+y}}-e^{-x-y}\right]\\[6pt] &\quad+\frac{1}{4}\left[e^{x+y}-\cancel{e^{x-y}}+\bcancel{e^{-x+y}}-e^{-x-y}\right]\\[6pt] =&\frac{1}{2}\left[e^{x+y}-e^{-(x+y)}\right]\\[6pt] =&\sinh(x+y) \end{align}

### Why the Prefix “Hyper”?

#### Click to see where the prefix hyper comes from

The point $$P(\cos\theta,\sin\theta)$$ lies on the unit circle $$x^2 + y^2=1$$. In this case, $$\theta$$ is the radian measure of $$\angle QOP$$.

• Consider the shaded sector in Figure 4. Its area is $$A=\frac{1}{2}\underbrace{r^2}_{=1}\theta$$. That is, $$\theta=2A$$

If $$t$$ is a real number, the point $$P(\cosh t,\sinh t)$$ lies on the right branch of the hyperbola $$x^2 – y^2=1$$ because $$\cosh^2\theta-\sinh^2\theta=1$$ and $$\cosh x\geq 1$$. In this case, $$t$$ does not represent the measure of an angle. However, $$t$$ is twice the area of the shaded hyperbolic sector (Figure 5).

## Derivatives of Hyperbolic Functions

### Derivatives of cosh x, sinh x, tanh x

We can show that \bbox[#F2F2F2,5px,border:2px solid black]{ \begin{align}\frac{d}{dx}\sinh x&=\cosh x\\[6pt] \frac{d}{dx}\cosh x&=\sinh x\\[6pt] \frac{d}{dx}\tanh x&=\text{sech}^2x=1-\tanh^2x \end{align}}

Example 4

Show $\frac{d}{dx}\sinh x=\cosh x.$

Solution
\begin{align} \frac{d}{dx}\sinh x=&\frac{d}{dx}\frac{e^x-e^{-x}}{2}\\ =&\frac{1}{2}\frac{d}{dx}(e^x-e^{-x})\\ =&\frac{1}{2}\left(\frac{d e^x}{dx}-\frac{d e^{-x}}{dx}\right)\\ =&\frac{1}{2}(e^x-(-e^{-x}))\\ =&\cosh x \end{align}
Example 5

Show $\frac{d}{dx}\tanh x=\text{sech}^2x$

Solution
\begin{align} \frac{d}{dx}\tanh x=&\frac{d}{dx}\frac{\overbrace{\sinh x}^u}{\underbrace{\cosh x}_v}\\ =&\frac{\overbrace{\cosh x}^{u’}\ \overbrace{\cosh x}^v-\overbrace{\sinh x}^{v’}\ \overbrace{\sinh x}^u}{\cosh^2 x}\\ =&\frac{1}{\cosh^2 x} &{\small (\cosh^2x-\sinh^2 x=1)}\\[6pt] =&\text{sech}^2x \end{align}

### Derivatives of Other Hyperbolic Functions

\bbox[#F2F2F2,5px,border:2px solid black]{ \begin{align} \frac{d}{dx}\coth x =&-\text{csch}^2 x\\[6pt] \frac{d}{dx}\text{sech } x =&-\text{sech }x\tanh x\\[6pt] \frac{d}{dx}\text{csch } x =&-\text{csch }x \coth x \end{align} }

## Inverse Hyperbolic Functions

Let’s look at the graphs of $y=\sinh x$, $y=\cosh x$, and $y=\tanh x$ (Figure 6). It is clear from this figure that $\sinh$ and $\tanh$ are one-to-one functions. Therefore, they have inverse functions.

Although $\cosh$ is not one-to-one on its domain, we can define its inverse function if we restrict the domain of cosh to $[0,+\infty)$. The inverses of $\sinh$, $\cosh$, and $\tanh$ are denoted by $\text{arcsinh}$, $\text{arccosh}$, and $\text{arctanh}$ (or $\sinh^{-1}$, $\cosh^{-1}$, and $\tanh^{-1}$), respectively:

\bbox[#F2F2F2,5px,border:2px solid black]{\begin{aligned}y=\sinh x & \Leftrightarrow x=\text{arcsinh }y & & (x,y\in\mathbb{R})\\ y=\cosh x & \Leftrightarrow x=\text{arccosh }y & & (x\geq0,y\geq1)\\ y=\tanh x & \Leftrightarrow x=\text{arctanh }y & & (x\in\mathbb{R},-1<y<1) \end{aligned} }

The remaining inverse hyperbolic functions are defined similarly. The graphs of $y=\text{arcsinh }x$, $y=\text{arccosh }x$, and $y=\text{arctanh }x$ are illustrated in the Figure 7.

We can find explicit formulas for the inverse hyperbolic functions.

Example 6

Show that $\sinh^{-1}x=\ln\left(x+\sqrt{x^{2}+1}\right)$.

Solution

Let $$y=\sinh x=\frac{e^{x}-e^{-x}}{2}.$$ To find the inverse function, we have to solve this equation for $x$. Multiplying both sides of the equation by $2e^{x}$, we obtain $$2ye^{x}=\left(e^{x}\right)^{2}-1$$ or $$\left(e^{x}\right)^{2}-2ye^{x}-1=0$$ This is a quadratic equation in terms of $e^{x}$. Therefore, by the quadratic formula, $$e^{x}=\frac{2y\pm\sqrt{4y^{2}+4}}{2}=y\pm\sqrt{y^{2}+1}$$ We must select the positive root, because $e^{x}>0$. Thus, $$e^{x}=y+\sqrt{y^{2}+1}$$ Taking the natural logarithm of each side produces $$x=\ln\left(y+\sqrt{y^{2}+1}\right)$$ Therefore the inverse of $y=\sinh x$ is $$\sinh^{-1}(y)=\ln\left(y+\sqrt{y^{2}+1}\right)$$ Here $y$ shows the input of $\sinh^{-1}$, and if we wish we can replace it with $x$ and write $$\sinh^{-1}(x)=\ln\left(x+\sqrt{x^{2}+1}\right).$$

In a similar fashion, we can derive explicit formulas for the other inverse hyperbolic functions. \bbox[#F2F2F2,5px,border:2px solid black]{\begin{aligned} & \text{ arcsinh }x=\ln\left(x+\sqrt{x^{2}+1}\right) & & (x\in\mathbb{R})\\ & \text{ arccosh }x=\ln\left(x+\sqrt{x^{2}+1}\right) & & (x\geq1)\\ & \text{ arctanh }x=\frac{1}{2}\ln\frac{1+x}{1-x} & & (-1<x<1) \end{aligned} } and \bbox[#F2F2F2,5px,border:2px solid black]{\begin{aligned} & \text{arccoth }x=\frac{1}{2}\ln\frac{x+1}{x-1} & & (|x|>1)\\ & \text{arcsech }x=\ln\left(\frac{1+\sqrt{1-x^{2}}}{x}\right) & & (0<x<1)\\ & \text{arccsch }x=\begin{cases} \ln\left(\frac{1+\sqrt{1+x^{2}}}{x}\right) & (x>0)\\ \\ -\ln\left(\frac{1+\sqrt{1+x^{2}}}{-x}\right) & (x<0) \end{cases} \end{aligned} }

## Derivatives of the Inverse Hyperbolic Functions

We can show that \bbox[#F2F2F2,5px,border:2px solid black]{\begin{aligned} & \frac{d}{dx}\text{arcsinh }x=\frac{d}{dx}\sinh^{-1}x=\frac{1}{\sqrt{x^{2}+1}} & & (x\in\mathbb{R})\\ & \frac{d}{dx}\text{arccosh }x=\frac{d}{dx}\cosh^{-1}x=\frac{1}{\sqrt{x^{2}-1}} & & (x>1)\\ & \frac{d}{dx}\text{arctanh }x=\frac{d}{dx}\tanh^{-1}x=\frac{1}{1-x^{2}} & & (-1<x<1) \end{aligned} } \bbox[#F2F2F2,5px,border:2px solid black]{\begin{aligned} & \frac{d}{dx}\text{arccoth }x=\frac{d}{dx}\coth^{-1}x=\frac{1}{1-x^{2}}, & & (|x|>1)\\ & \frac{d}{dx}\text{arcsech }x=\frac{d}{dx}\text{sech}^{-1}x=\frac{-1}{x\sqrt{1-x^{2}}}, & & (0<x<1)\\ & \frac{d}{dx}\text{arccsch }x=\frac{d}{dx}\text{csch}^{-1}x=\begin{cases} \dfrac{-1}{x\sqrt{1+x^{2}}} & (x>0)\\ \dfrac{1}{x\sqrt{1+x^{2}}} & (x<0) \end{cases} \end{aligned} }

Example 7
Show that $$\boxed{\frac{d}{dx}\text{arcsinh }x=\frac{d}{dx}\sinh^{-1}x=\frac{1}{\sqrt{x^{2}+1}}\quad( x\in \mathbb{R})}$$

Solution

Let $y=\sinh x$. Then $x=\text{arcsinh }y$, and \begin{aligned} \frac{d}{dy}\text{arcsinh }y & =\frac{1}{\frac{d}{dx}\sinh x}\\ & =\frac{1}{\cosh x}\\ & =\frac{1}{\sqrt{\sinh^{2}x+1}} && \left(\cosh^{2}x-\sinh^{2}x=1\text{ and } \cosh x>0\right)\\ & =\frac{1}{\sqrt{y^{2}+1}}&& \left(y=\sinh x\right)\end{aligned}

With a change in notation, we can express the above result as $$\frac{d}{dx}\text{arcsinh }x=\frac{1}{\sqrt{x^{2}+1}}$$

Example 8

Show that $$\boxed{\frac{d}{dx}\text{arccosh }x{=\frac{d}{dx}\cosh^{-1}x}{=\frac{1}{\sqrt{x^{2}-1}}}{\quad(x>1)}}$$

Solution

Let $y=\cosh x$ and $x>0$. Then $x=\text{arccosh }y$ and \begin{aligned} \frac{d}{dy}(\text{arccosh }y) & =\frac{1}{\frac{d}{dx}\cosh x}\\ & =\frac{1}{\sinh x}&&\left(\frac{d}{dx}\cosh x=\sinh x\right)\\ & =\frac{1}{\sqrt{\cosh^{2}x-1}}&&\left(\cosh^{2}x-\sinh^{2}x=1, \sinh x>0\text{ when }x>1\right)\\ & =\frac{1}{\sqrt{y^{2}-1}}&&\left(\cosh x=y\right)\ \end{aligned} $$\therefore\frac{d}{dy}(\text{arccosh }y)=\frac{1}{\sqrt{y^{2}-1}}$$ Here $y$ shows the input of $\text{arccosh}$, and if we wish we can replace it by any letter including $x$. Therefore $$\frac{d}{dx}\text{arccosh }x=\frac{1}{x^{2}-1}$$

Example 9

Show that $$\boxed{\frac{d}{dx}\text{arctanh }x=\frac{d}{dx}\tanh^{-1}x=\frac{1}{1-x^{2}}\quad(-1<x<1)}$$

Solution

Let $y=\tanh x$. Then $x=\text{arctanh }y$, and \begin{aligned} \frac{d}{dy}(\text{arctanh }y) & =\frac{1}{\frac{d}{dx}\tanh x}\\ & =\frac{1}{\text{sech}^{2}x}\\ & =\frac{1}{1-\tanh^{2}x}&&\left(1-\tanh^{2}x=\text{sech}^{2}x\right)\\ & =\frac{1}{1-y^{2}}&&\left(\tanh x=y\right)\end{aligned}

With a change in notation $$\frac{d}{dx}(\text{arctanh }x)=\frac{1}{1-x^{2}}\quad(-1<x<1)$$