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Area functions

Area functions are inverse functions of hyperbolic functions.

Area hyperbolic sine, arsinh

Hyperbolic sine is a bijective function $sinh:\mathbb{R} \rightarrow \mathbb{R}$  and it has inverse function, $arsinh:\mathbb{R} \rightarrow \mathbb{R}$. Sometimes the inverse function is also written $sinh^{-1}$.

Let’s set $x=sinh(y)$, and solve it for $y$.

$x= sinh(y)$
$2\cdot x=e^{y}-e^{-y}$

Now, by multiplying with $e^{y}$ we get

$2\cdot x \cdot e^{y}=e^{2y}-1$
$(e^{y})^{2}-2\cdot x \cdot e^{y}-1= 0$

We can see this is quadriatic equation, so by treating $x$ as a constant and using the quadratic formula, we obtain

$e^{y}=x \pm \sqrt{x^{2}+1}$

Due to $e^{y}$ being positive for every $y$ and because $\sqrt{x^{2}+1} > x$ for all $x$, we’re only left with positive square root,

$e^{y}=x + \sqrt{y^{2}+1}$

By taking the natural logarithm we get the expression for inverse hyperbolic sine

$arsinhx=\ln(x+\sqrt{x^{2}+1})$.

arsinh

 

As we already know, graphs of the mutually inverse functions are symmetric with respect to the bisector of the first and third quadrant, line $y=x$.

Derivate of $arsinh$ is

$\displaystyle{\frac{d}{dx} arsinhx=\frac{1}{ \sqrt{x^{2}+1}}}$, for all real $x$.

Example. Find the derivate of $y=x^{2}\cdot arsinh(2x)$.

We can see that the expression is a product, and that’s how we’ll derivate it – by  using the product rule of differentiation.

$$y’= (x^{2})’ \cdot arsinh(2x) + x^{2}\cdot (arsinh(2x))’$$

$$y’= 2x \cdot arsinh(2x) + x^{2}\cdot\frac{1}{ \sqrt{(2x)^{2}+1}} \cdot (2x)’$$

$$y’= 2x \cdot arsinh(2x) + x^{2}\cdot\frac{1}{ \sqrt{(2x)^{2}+1}} \cdot 2$$

There is not much we can do to simplify it, so let’s just factor out $2x$

$$y’= 2x \left[ arsinh(2x) + \frac{x}{ \sqrt{ 4(x)^{2} +1}} \right].$$

 

Area hyperbolic cosine, arcosh

Hyperbolic cosine, $cosh:\mathbb{R} \rightarrow [1, +\infty)$ is not an injection, which means it’s also not a bijection. As a result, it doesn’t have an inverse function. However, we can restrict domain of $cosh$ to get bijective function, and find its inverse.

Let’s define $Cosh: [0, +\infty) \rightarrow [1, +\infty)$, as a restriction of $cosh$ to the positive real numbers, $Cosh= cosh{\big|}_{[0, +\infty)}$. It’s bijection and its inversion will be function $arcosh: [1, +\infty) \rightarrow [0, +\infty)$, or $Cosh^{-1}$.

Finally, by following the steps we used to get $arsinh$, we get the expression for inversion of hyperbolic cosine,

$$arcoshx=\ln(x+\sqrt{x^{2}-1}).$$

Graph of $arcosh$ is a symmetric image of graph of $cosh$ with respect to the $y=x$ line.

arcosh

On the image above, we can see hyperbolic cosine in color red, $cosh(x)$, and its restriction $Cosh(x)$, colored green. Blue function, $arcosh(x)$ is an inverse function of $Cosh(x)$.

Derivate of $arcosh$ is

$\displaystyle{\frac{d}{dx} arcoshx=\frac{1}{ \sqrt{x^{2}-1}}}$, for all real $x>1$

Area hyperbolic tangent, artanh

Due to hyperbolic tangent,  $tanh: \mathbb{R} \rightarrow (-1,1)$, being a bijective function, we don’t have to restrict its domain to get inversion like we did with $cosh$. Inverse function of hyperbolic tangent is called area hyperbolic tangent, $artanh: (-1,1) \rightarrow \mathbb{R}$ or $tanh^{-1}$.

By following the same process of getting an inverse function, we get the expression for area hyperbolic tangent $$artanh(x)=\displaystyle{\frac{1}{2} \ln\left(\frac{1+x}{1-x}\right)}$$.

artanh

Derivate of $artanh$ is

$\displaystyle{\frac{d}{dx} artanhx=\frac{1}{ 1-x^{2}}}$, for all real $|x|<1$.

Example. Find the derivate of $\displaystyle{y=artanh\left(\frac{1}{x^{2}}\right)}, |x|<1$.

Using the derivative of inverse tangent hyperbolic function and the chain rule, we have

$y’= \left[\displaystyle{artanh\left(\frac{1}{x^{2}}\right)}\right]’, |x|<1$

$\displaystyle{y’=\frac{1}{1-\left(\frac{1}{x^{2}}\right)^{2}} \cdot \left(\frac{1}{x^{2}}\right)}’$

$\displaystyle{y’=\frac{1}{1-\frac{1}{x^{4}}} \cdot (-2x^{-3})}$

$\displaystyle{y’= – \frac{2x^{4}}{(x^{4}-1)x^{3}}}$

Now just simplify the fraction,

$\displaystyle{y’=-\frac{2x}{x^{4}-1}}$

$\displaystyle{y’=\frac{2x}{1-x^{4}}}$

Area hyperbolic cotangent, arcoth

The bijective function, hyperbolic cotangent $coth: \mathbb{R} \backslash \{0\} \rightarrow (-\infty,-1) \cup (1,+\infty)$, has an inverse function for every $x$ from its domain. Its called area hyperbolic cotangent, $arcoth: (-\infty,-1) \cup (1,+\infty) \rightarrow \mathbb{R} \backslash \{0\}$ or $coth^{-1}$. Its expression is

$$arcoth(x)=\displaystyle{\frac{1}{2} \ln\left(\frac{1+x}{1-x}\right)}, x \in (-\infty, -1) \cup (1, +\infty)$$.

arcoth

Derivate of $arcoth$ is

$\displaystyle{\frac{d}{dx} arcothx=\frac{1}{ 1-x^{2}}}$, for all real $|x|>1$.

 

Area hyperbolic secant, $arsech$

Since hyperbolic secant, $sech: \mathbb{R}  \rightarrow (0,1) $, isn’t a bijective function, we have to restrict its domain like we did with $cosh$.

Let’s define $Sech: [0, +\infty) \rightarrow (0,1)$  as a restriction of $sech$ to the positive real numbers. Therefore, $Sech= sech{\big|}_{[0, +\infty)}$. It’s a bijection, and its inversion will be $arsech$.

The expressiong of $arsech$ is

$$arsech:  \displaystyle{\ln\left(\frac{1} {x} + \sqrt{\frac{1 }{ x^{2}} + 1}\right)}$$

 

arsech

Graphic of $arsech$ is a symmetric image of graph of $sech$ with respect to the $y=x$ line.

Derivate of $arsech$ is

$\displaystyle{\frac{d}{dx} arsechx=\frac{-1}{ x\sqrt{1-x^{2}}}}$, for all real $x \in (0,1)$

 

Area hyperbolic cosecant, $arcsch$

Hyperbolic cosecant, $csch: (-\infty,-1) \cup (1,+\infty) \rightarrow \mathbb{R} \backslash \{0\}$, is a bijective function. As a result, its inverse exists for every $x$ from its domain. Area hyperbolic cosecant, $arcsch: \mathbb{R} \backslash \{0\} \rightarrow (-\infty,-1) \cup (1,+\infty)$, satisfies

$\displaystyle{\frac{d}{dx} arsechx=\frac{-1}{ x\sqrt{1+x^{2}}}}$, for all real $x \in (0,1)$

 

 

arcsch

$$arcsch: \displaystyle{\ln\left(\frac{1} {x} + \sqrt{\frac{1 }{ x^{2}} + 1}\right)}$$

Derivate of $arcsch$ is

$\displaystyle{\frac{d}{dx} arcschx=\frac{-1}{|x| \sqrt{1+x^{2}}}}$, for all real $x$ except $0$.