Inverse Hyperbolic Functions

In mathematics, inverse hyperbolic functions are the inverse functions of the hyperbolic functions. Just as the inverse trigonometric functions (like arcsin, arccos) "undo" the trigonometric functions (sin, cos), the inverse hyperbolic functions "undo" the hyperbolic functions (sinh, cosh, tanh, etc.). These functions are crucial in calculus, differential equations, and various areas of physics and engineering.

Definition and Derivation

The hyperbolic functions are defined using the exponential function, $e^x$. Their inverse functions can also be expressed in terms of logarithms. Let's look at the definitions and how we derive the logarithmic forms.

1. Inverse Hyperbolic Sine (arsinh or sinh^-1)

Let $y = \sinh x$. We want to find $x$ in terms of $y$.

We know that $\sinh x = \frac{e^x - e^{-x}}{2}$.

So, $y = \frac{e^x - e^{-x}}{2}$.

Multiply by 2: $2y = e^x - e^{-x}$.

Multiply by $e^x$ to get rid of the negative exponent: $2ye^x = (e^x)^2 - 1$.

Rearrange this into a quadratic equation in terms of $e^x$: $(e^x)^2 - 2ye^x - 1 = 0$.

Let $u = e^x$. The equation becomes $u^2 - 2yu - 1 = 0$.

Using the quadratic formula ($u = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$), where $a=1$, $b=-2y$, $c=-1$:

$u = \frac{-(-2y) \pm \sqrt{(-2y)^2 - 4(1)(-1)}}{2(1)}$

$u = \frac{2y \pm \sqrt{4y^2 + 4}}{2}$

$u = \frac{2y \pm \sqrt{4(y^2 + 1)}}{2}$

$u = \frac{2y \pm 2\sqrt{y^2 + 1}}{2}$

$u = y \pm \sqrt{y^2 + 1}$.

Since $u = e^x$, and $e^x$ must be positive, we need to choose the correct sign. $\sqrt{y^2 + 1}$ is always greater than $|y|$. Thus, $y - \sqrt{y^2 + 1}$ will always be negative. Therefore, we must take the positive sign.

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

Taking the natural logarithm of both sides:

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

So, the inverse hyperbolic sine function is:

$\text{arsinh } y = \ln(y + \sqrt{y^2 + 1})$.

Replacing $y$ with $x$ for the standard function notation:

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

The domain of arsinh $x$ is all real numbers ($-\infty < x < \infty$), and its range is also all real numbers.

2. Inverse Hyperbolic Cosine (arccosh or cosh^-1)

Let $y = \cosh x$. We know that $\cosh x = \frac{e^x + e^{-x}}{2}$.

So, $y = \frac{e^x + e^{-x}}{2}$.

Multiply by 2: $2y = e^x + e^{-x}$.

Multiply by $e^x$: $2ye^x = (e^x)^2 + 1$.

Rearrange into a quadratic in $e^x$: $(e^x)^2 - 2ye^x + 1 = 0$.

Let $u = e^x$. The equation is $u^2 - 2yu + 1 = 0$.

Using the quadratic formula:

$u = \frac{-(-2y) \pm \sqrt{(-2y)^2 - 4(1)(1)}}{2(1)}$

$u = \frac{2y \pm \sqrt{4y^2 - 4}}{2}$

$u = \frac{2y \pm 2\sqrt{y^2 - 1}}{2}$

$u = y \pm \sqrt{y^2 - 1}$.

So, $e^x = y \pm \sqrt{y^2 - 1}$.

Taking the natural logarithm: $x = \ln(y \pm \sqrt{y^2 - 1})$.

Now, we need to consider the domain and range. The range of $\cosh x$ is $y \ge 1$. For $y=1$, $e^x = 1 \pm 0 = 1$, so $x=0$.

For $y > 1$, notice that $\ln(y + \sqrt{y^2 - 1})$ and $\ln(y - \sqrt{y^2 - 1})$ are related.

Let's check: $\ln(y - \sqrt{y^2 - 1}) = \ln\left(\frac{(y - \sqrt{y^2 - 1})(y + \sqrt{y^2 - 1})}{y + \sqrt{y^2 - 1}}\right)$

$= \ln\left(\frac{y^2 - (y^2 - 1)}{y + \sqrt{y^2 - 1}}\right) = \ln\left(\frac{1}{y + \sqrt{y^2 - 1}}\right) = -\ln(y + \sqrt{y^2 - 1})$.

So, the two possible solutions for $x$ are negatives of each other. Since $\cosh x = \cosh (-x)$, the inverse function is not unique unless we restrict the domain of $\cosh x$.

By convention, the principal value of $\text{arccosh } y$ is taken for $x \ge 0$. In this case, $e^x \ge 1$. The expression $y + \sqrt{y^2 - 1}$ is always greater than or equal to 1 (for $y \ge 1$), while $y - \sqrt{y^2 - 1}$ is between 0 and 1. Thus, we choose the positive logarithm.

$\text{arccosh } y = \ln(y + \sqrt{y^2 - 1})$.

Replacing $y$ with $x$:

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

The domain of arccosh $x$ is $x \ge 1$, and its range is $y \ge 0$.

3. Inverse Hyperbolic Tangent (artanh or tanh^-1)

Let $y = \tanh x$. We know that $\tanh x = \frac{e^x - e^{-x}}{e^x + e^{-x}}$.

So, $y = \frac{e^x - e^{-x}}{e^x + e^{-x}}$.

Multiply by $e^x$: $y = \frac{(e^x)^2 - 1}{(e^x)^2 + 1}$.

Let $u = (e^x)^2$. Then $y = \frac{u - 1}{u + 1}$.

Solve for $u$:

$y(u + 1) = u - 1$

$yu + y = u - 1$

$y + 1 = u - yu$

$y + 1 = u(1 - y)$

$u = \frac{1 + y}{1 - y}$.

Substitute back $u = (e^x)^2$: $(e^x)^2 = \frac{1 + y}{1 - y}$.

Since $(e^x)^2 = e^{2x}$, we have $e^{2x} = \frac{1 + y}{1 - y}$.

Take the natural logarithm: $2x = \ln\left(\frac{1 + y}{1 - y}\right)$.

$x = \frac{1}{2} \ln\left(\frac{1 + y}{1 - y}\right)$.

So, the inverse hyperbolic tangent function is:

$\text{artanh } y = \frac{1}{2} \ln\left(\frac{1 + y}{1 - y}\right)$.

Replacing $y$ with $x$:

$\text{artanh } x = \frac{1}{2} \ln\left(\frac{1 + x}{1 - x}\right)$.

The domain of artanh $x$ is $-1 < x < 1$, and its range is all real numbers.

4. Inverse Hyperbolic Cotangent (arccoth or coth^-1)

Let $y = \coth x$. We know that $\coth x = \frac{e^x + e^{-x}}{e^x - e^{-x}}$.

So, $y = \frac{e^x + e^{-x}}{e^x - e^{-x}}$.

Multiply by $e^x$: $y = \frac{(e^x)^2 + 1}{(e^x)^2 - 1}$.

Let $u = (e^x)^2$. Then $y = \frac{u + 1}{u - 1}$.

Solve for $u$:

$y(u - 1) = u + 1$

$yu - y = u + 1$

$yu - u = y + 1$

$u(y - 1) = y + 1$

$u = \frac{y + 1}{y - 1}$.

Substitute back $u = (e^x)^2$: $(e^x)^2 = \frac{y + 1}{y - 1}$.

$e^{2x} = \frac{y + 1}{y - 1}$.

Take the natural logarithm: $2x = \ln\left(\frac{y + 1}{y - 1}\right)$.

$x = \frac{1}{2} \ln\left(\frac{y + 1}{y - 1}\right)$.

So, the inverse hyperbolic cotangent function is:

$\text{arccoth } y = \frac{1}{2} \ln\left(\frac{y + 1}{y - 1}\right)$.

Replacing $y$ with $x$:

$\text{arccoth } x = \frac{1}{2} \ln\left(\frac{x + 1}{x - 1}\right)$.

The domain of arccoth $x$ is $|x| > 1$ (i.e., $x > 1$ or $x < -1$), and its range is all real numbers except 0.

5. Inverse Hyperbolic Secant (arsech or sech^-1)

Let $y = \text{sech } x$. We know $\text{sech } x = \frac{1}{\cosh x} = \frac{2}{e^x + e^{-x}}$.

So, $y = \frac{2}{e^x + e^{-x}}$.

$y(e^x + e^{-x}) = 2$.

Multiply by $e^x$: $y((e^x)^2 + 1) = 2e^x$.

Rearrange into a quadratic in $e^x$: $y(e^x)^2 - 2e^x + y = 0$.

Let $u = e^x$. Then $yu^2 - 2u + y = 0$.

Using the quadratic formula:

$u = \frac{-(-2) \pm \sqrt{(-2)^2 - 4(y)(y)}}{2y}$

$u = \frac{2 \pm \sqrt{4 - 4y^2}}{2y}$

$u = \frac{2 \pm 2\sqrt{1 - y^2}}{2y}$

$u = \frac{1 \pm \sqrt{1 - y^2}}{y}$.

So, $e^x = \frac{1 \pm \sqrt{1 - y^2}}{y}$.

Taking the natural logarithm: $x = \ln\left(\frac{1 \pm \sqrt{1 - y^2}}{y}\right)$.

The range of $\text{sech } x$ is $0 < y \le 1$. For $y=1$, $x=0$. For $0 < y < 1$, we have two solutions.

We use the convention that the principal value of $\text{arsech } y$ corresponds to $x \ge 0$. In this case, $e^x \ge 1$. The term $\frac{1 + \sqrt{1 - y^2}}{y}$ will be greater than or equal to 1. The term $\frac{1 - \sqrt{1 - y^2}}{y}$ will be less than 1.

So, we choose the '+' sign:

$\text{arsech } y = \ln\left(\frac{1 + \sqrt{1 - y^2}}{y}\right)$.

Replacing $y$ with $x$:

$\text{arsech } x = \ln\left(\frac{1 + \sqrt{1 - x^2}}{x}\right)$.

The domain of arsech $x$ is $0 < x \le 1$, and its range is $y \ge 0$.

6. Inverse Hyperbolic Cosecant (arcsch or csch^-1)

Let $y = \text{csch } x$. We know $\text{csch } x = \frac{1}{\sinh x} = \frac{2}{e^x - e^{-x}}$.

So, $y = \frac{2}{e^x - e^{-x}}$.

$y(e^x - e^{-x}) = 2$.

Multiply by $e^x$: $y((e^x)^2 - 1) = 2e^x$.

Rearrange into a quadratic in $e^x$: $y(e^x)^2 - 2e^x - y = 0$.

Let $u = e^x$. Then $yu^2 - 2u - y = 0$.

Using the quadratic formula:

$u = \frac{-(-2) \pm \sqrt{(-2)^2 - 4(y)(-y)}}{2y}$

$u = \frac{2 \pm \sqrt{4 + 4y^2}}{2y}$

$u = \frac{2 \pm 2\sqrt{1 + y^2}}{2y}$

$u = \frac{1 \pm \sqrt{1 + y^2}}{y}$.

So, $e^x = \frac{1 \pm \sqrt{1 + y^2}}{y}$.

Since $e^x$ must be positive, and $\sqrt{1+y^2} > |y|$, we must choose the '+' sign if $y>0$ and the '-' sign if $y<0$. This can be compactly written as $e^x = \frac{1 + \sqrt{1 + y^2}}{y}$ if $y>0$ and $e^x = \frac{1 - \sqrt{1 + y^2}}{y}$ if $y<0$. Notice that $\frac{1 - \sqrt{1 + y^2}}{y} = \frac{1 - \sqrt{1 + y^2}}{y} \times \frac{1 + \sqrt{1 + y^2}}{1 + \sqrt{1 + y^2}} = \frac{1 - (1 + y^2)}{y(1 + \sqrt{1 + y^2})} = \frac{-y^2}{y(1 + \sqrt{1 + y^2})} = \frac{-y}{1 + \sqrt{1 + y^2}}$. So, if $y>0$, $e^x = \frac{1 + \sqrt{1 + y^2}}{y}$. If $y<0$, $e^x = \frac{-y}{1 + \sqrt{1 + y^2}} = \frac{|y|}{1 + \sqrt{1 + y^2}}$. This means we can always write $e^x = \frac{1 + \sqrt{1 + y^2}}{|y|}$? No, this is not quite right. Let's re-examine $e^x = \frac{1 \pm \sqrt{1 + y^2}}{y}$. If $y > 0$, then $y$ is positive. $1 + \sqrt{1+y^2}$ is positive. $1 - \sqrt{1+y^2}$ is negative. So we must choose the '+' sign to get a positive $e^x$. $e^x = \frac{1 + \sqrt{1 + y^2}}{y}$. If $y < 0$, then $y$ is negative. Let $y = -z$ where $z > 0$. $e^x = \frac{1 \pm \sqrt{1 + (-z)^2}}{-z} = \frac{1 \pm \sqrt{1 + z^2}}{-z}$. To make $e^x$ positive, we must choose the '-' sign: $e^x = \frac{1 - \sqrt{1 + z^2}}{-z} = \frac{\sqrt{1 + z^2} - 1}{z}$. So, for $y>0$, $x = \ln\left(\frac{1 + \sqrt{1 + y^2}}{y}\right)$. For $y<0$, $x = \ln\left(\frac{\sqrt{1 + y^2} - 1}{|y|}\right)$. This can be simplified. Notice that $\ln\left(\frac{1 + \sqrt{1 + y^2}}{y}\right) = \ln\left(\frac{\sqrt{1 + y^2} - 1}{|y|}\right)$ if $y<0$. Let's check the relationship: $\ln\left(\frac{\sqrt{1 + y^2} - 1}{|y|}\right) = \ln\left(\frac{(\sqrt{1 + y^2} - 1)(\sqrt{1 + y^2} + 1)}{|y|(\sqrt{1 + y^2} + 1)}\right) = \ln\left(\frac{(1 + y^2) - 1}{|y|(\sqrt{1 + y^2} + 1)}\right) = \ln\left(\frac{y^2}{|y|(\sqrt{1 + y^2} + 1)}\right)$. If $y>0$, $|y|=y$, so $\ln\left(\frac{y}{\sqrt{1 + y^2} + 1}\right) = \ln\left(\frac{y(\sqrt{1+y^2}-1)}{(\sqrt{1+y^2}+1)(\sqrt{1+y^2}-1)}\right) = \ln\left(\frac{y(\sqrt{1+y^2}-1)}{(1+y^2)-1}\right) = \ln\left(\frac{y(\sqrt{1+y^2}-1)}{y^2}\right) = \ln\left(\frac{\sqrt{1+y^2}-1}{y}\right)$. This is not matching. Let's use the property that $\text{csch } x = \sinh (\ln u)$ implies $u = \frac{1+\sqrt{1+y^2}}{y}$ or $u = \frac{1-\sqrt{1+y^2}}{y}$. We are looking for $x = \text{arcsch } y$. Consider $e^x = \frac{1 + \sqrt{1+y^2}}{y}$ for $y>0$. Consider $e^x = \frac{1 - \sqrt{1+y^2}}{y}$ for $y<0$. This is negative. This cannot be $e^x$. Let's restart the derivation for arcsch $y$. $y = \frac{2}{e^x - e^{-x}}$ $y(e^x - e^{-x}) = 2$ $y e^x - y e^{-x} = 2$ Multiply by $e^x$: $y (e^x)^2 - y = 2 e^x$ $y (e^x)^2 - 2 e^x - y = 0$ Quadratic formula for $e^x$: $e^x = \frac{2 \pm \sqrt{4 - 4(y)(-y)}}{2y} = \frac{2 \pm \sqrt{4 + 4y^2}}{2y} = \frac{2 \pm 2\sqrt{1 + y^2}}{2y} = \frac{1 \pm \sqrt{1 + y^2}}{y}$. Since $e^x > 0$: If $y > 0$, then $1 + \sqrt{1+y^2} > 0$ and $1 - \sqrt{1+y^2} < 0$. So we must choose $e^x = \frac{1 + \sqrt{1 + y^2}}{y}$. If $y < 0$, then $1 + \sqrt{1+y^2} > 0$ and $1 - \sqrt{1+y^2} < 0$. The denominator $y$ is negative. So $\frac{1 + \sqrt{1+y^2}}{y}$ is negative. We must choose $e^x = \frac{1 - \sqrt{1 + y^2}}{y}$. This is also negative. There must be a mistake in the algebra or understanding. Let's check the identity: $\text{arcsch } y = \text{arsinh } (1/y)$. So, $\text{arcsch } y = \ln\left(\frac{1}{y} + \sqrt{\left(\frac{1}{y}\right)^2 + 1}\right) = \ln\left(\frac{1}{y} + \sqrt{\frac{1}{y^2} + 1}\right) = \ln\left(\frac{1}{y} + \sqrt{\frac{1+y^2}{y^2}}\right)$. If $y>0$, $\sqrt{\frac{1+y^2}{y^2}} = \frac{\sqrt{1+y^2}}{|y|} = \frac{\sqrt{1+y^2}}{y}$. So, $\text{arcsch } y = \ln\left(\frac{1}{y} + \frac{\sqrt{1+y^2}}{y}\right) = \ln\left(\frac{1 + \sqrt{1+y^2}}{y}\right)$. This matches our previous result for $y>0$. If $y<0$, $|y| = -y$. So $\sqrt{\frac{1+y^2}{y^2}} = \frac{\sqrt{1+y^2}}{-y}$. Then, $\text{arcsch } y = \ln\left(\frac{1}{y} + \frac{\sqrt{1+y^2}}{-y}\right) = \ln\left(\frac{1 - \sqrt{1+y^2}}{y}\right)$. This matches our previous result for $y<0$. So the formula is: $\text{arcsch } y = \ln\left(\frac{1 + \sqrt{1+y^2}}{y}\right)$ if $y>0$. $\text{arcsch } y = \ln\left(\frac{1 - \sqrt{1+y^2}}{y}\right)$ if $y<0$. Let's simplify $\ln\left(\frac{1 - \sqrt{1+y^2}}{y}\right)$ for $y<0$. Let $y=-z$ where $z>0$. $\ln\left(\frac{1 - \sqrt{1+(-z)^2}}{-z}\right) = \ln\left(\frac{1 - \sqrt{1+z^2}}{-z}\right) = \ln\left(\frac{\sqrt{1+z^2} - 1}{z}\right)$. This means: $\text{arcsch } y = \ln\left(\frac{1 + \sqrt{1+y^2}}{y}\right)$ for $y>0$. $\text{arcsch } y = \ln\left(\frac{\sqrt{1+y^2} - 1}{|y|}\right)$ for $y<0$. This can be written more compactly using the sign function, $\text{sgn}(y)$. $\text{arcsch } y = \ln\left(\frac{1 + \sqrt{1+y^2}}{\text{sgn}(y) y}\right)$? No. Consider the identity $\text{arcsch } y = \text{arsinh } (1/y)$. The domain of arcsch $y$ is all real numbers except 0. Its range is also all real numbers except 0. Let's write the final form for arcsch $x$: $\text{arcsch } x = \ln\left(\frac{1 + \sqrt{1+x^2}}{x}\right)$ for $x>0$. $\text{arcsch } x = \ln\left(\frac{\sqrt{1+x^2} - 1}{|x|}\right)$ for $x<0$. A common alternative form is $\text{arcsch } x = \text{arsinh}(1/x)$. $\text{arsinh}(1/x) = \ln(1/x + \sqrt{(1/x)^2 + 1}) = \ln(1/x + \sqrt{(1+x^2)/x^2})$. If $x>0$, $\ln(1/x + \sqrt{1+x^2}/x) = \ln((1+\sqrt{1+x^2})/x)$. If $x<0$, $\ln(1/x + \sqrt{1+x^2}/|x|) = \ln(1/x + \sqrt{1+x^2}/(-x)) = \ln((1-\sqrt{1+x^2})/x)$. This matches our derived forms.

Summary of Logarithmic Forms:

Function	Logarithmic Form	Domain	Range
$\text{arsinh } x$	$\ln(x + \sqrt{x^2 + 1})$	$(-\infty, \infty)$	$(-\infty, \infty)$
$\text{arccosh } x$	$\ln(x + \sqrt{x^2 - 1})$	$[1, \infty)$	$[0, \infty)$
$\text{artanh } x$	$\frac{1}{2} \ln\left(\frac{1 + x}{1 - x}\right)$	$(-1, 1)$	$(-\infty, \infty)$
$\text{arccoth } x$	$\frac{1}{2} \ln\left(\frac{x + 1}{x - 1}\right)$	$(-\infty, -1) \cup (1, \infty)$	$(-\infty, 0) \cup (0, \infty)$
$\text{arsech } x$	$\ln\left(\frac{1 + \sqrt{1 - x^2}}{x}\right)$	$(0, 1]$	$[0, \infty)$
$\text{arcsch } x$	$\ln\left(\frac{1 + \sqrt{1+x^2}}{x}\right)$ for $x>0$; $\ln\left(\frac{\sqrt{1+x^2}-1}{|x|}\right)$ for $x<0$.	$(-\infty, 0) \cup (0, \infty)$	$(-\infty, 0) \cup (0, \infty)$

Derivatives of Inverse Hyperbolic Functions

The derivatives of inverse hyperbolic functions are quite simple and often resemble the derivatives of inverse trigonometric functions, but without the alternating negative signs for cos, sec, csc.

1. Derivative of arsinh $x$

If $y = \text{arsinh } x$, then $x = \sinh y$.

Differentiating implicitly with respect to $x$: $1 = \cosh y \frac{dy}{dx}$.

So, $\frac{dy}{dx} = \frac{1}{\cosh y}$.

We know that $\cosh^2 y - \sinh^2 y = 1$.

So, $\cosh^2 y = 1 + \sinh^2 y$.

Since $\cosh y = \sqrt{1 + \sinh^2 y}$ (cosh is always positive),

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

Therefore, $\frac{d}{dx}(\text{arsinh } x) = \frac{1}{\sqrt{1 + x^2}}$.

2. Derivative of arccosh $x$

If $y = \text{arccosh } x$, then $x = \cosh y$.

Differentiating implicitly: $1 = \sinh y \frac{dy}{dx}$.

So, $\frac{dy}{dx} = \frac{1}{\sinh y}$.

We know that $\cosh^2 y - \sinh^2 y = 1$.

So, $\sinh^2 y = \cosh^2 y - 1$.

$\sinh y = \pm \sqrt{\cosh^2 y - 1}$.

Since the range of arccosh $x$ is $y \ge 0$, $\sinh y \ge 0$. Thus, we take the positive root: $\sinh y = \sqrt{\cosh^2 y - 1}$.

$\sinh y = \sqrt{x^2 - 1}$.

Therefore, $\frac{d}{dx}(\text{arccosh } x) = \frac{1}{\sqrt{x^2 - 1}}$. (Note: Domain $x>1$ for the derivative).

3. Derivative of artanh $x$

If $y = \text{artanh } x$, then $x = \tanh y$.

Differentiating implicitly: $1 = \text{sech}^2 y \frac{dy}{dx}$.

So, $\frac{dy}{dx} = \frac{1}{\text{sech}^2 y}$.

We know that $1 - \tanh^2 y = \text{sech}^2 y$.

So, $\text{sech}^2 y = 1 - x^2$.

Therefore, $\frac{d}{dx}(\text{artanh } x) = \frac{1}{1 - x^2}$. (Note: Domain $|x|<1$).

4. Derivative of arccoth $x$

If $y = \text{arccoth } x$, then $x = \coth y$.

Differentiating implicitly: $1 = -\text{csch}^2 y \frac{dy}{dx}$.

So, $\frac{dy}{dx} = -\frac{1}{\text{csch}^2 y}$.

We know that $\coth^2 y - 1 = \text{csch}^2 y$.

So, $\text{csch}^2 y = x^2 - 1$.

Therefore, $\frac{d}{dx}(\text{arccoth } x) = -\frac{1}{x^2 - 1} = \frac{1}{1 - x^2}$. (Note: Domain $|x|>1$).

5. Derivative of arsech $x$

If $y = \text{arsech } x$, then $x = \text{sech } y$.

Differentiating implicitly: $\frac{dx}{dy} = -\text{sech } y \tanh y$.

So, $\frac{dy}{dx} = \frac{1}{-\text{sech } y \tanh y}$.

We know $\tanh^2 y = 1 - \text{sech}^2 y = 1 - x^2$.

So, $\tanh y = \pm \sqrt{1 - x^2}$.

Since the range of arsech $x$ is $y \ge 0$, and $\text{sech } y$ is always positive, $y$ is in $[0, \infty)$. For $y \in [0, \infty)$, $\tanh y$ can be positive or negative. However, the principal value of arsech $x$ is defined for $x \in (0,1]$. For $x \in (0,1]$, $y = \text{arsech } x \in [0, \infty)$. If $y=0$, $x=1$. If $y>0$, $x = \text{sech } y > 0$. The derivative $\frac{d}{dx}(\text{sech } x) = -\text{sech } x \tanh x$. For $x \in (0,1]$, $y = \text{arsech } x$, so $\text{sech } y = x$. We need $\tanh y$. Since $y \ge 0$, $\tanh y \ge 0$. So we take the positive root: $\tanh y = \sqrt{1 - x^2}$.

$\frac{dy}{dx} = \frac{1}{-x \sqrt{1 - x^2}}$.

Therefore, $\frac{d}{dx}(\text{arsech } x) = -\frac{1}{x \sqrt{1 - x^2}}$. (Note: Domain $0

6. Derivative of arcsch $x$

If $y = \text{arcsch } x$, then $x = \text{csch } y$.

Differentiating implicitly: $\frac{dx}{dy} = -\text{csch } y \coth y$.

So, $\frac{dy}{dx} = \frac{1}{-\text{csch } y \coth y}$.

We know $\coth^2 y = 1 + \text{csch}^2 y = 1 + x^2$.

So, $\coth y = \pm \sqrt{1 + x^2}$.

If $y>0$, $x = \text{csch } y > 0$, and $\coth y > 0$. So $\coth y = \sqrt{1+x^2}$. If $y<0$, $x = \text{csch } y < 0$, and $\coth y < 0$. So $\coth y = -\sqrt{1+x^2}$.

We need to consider the sign of $\coth y$. If $x>0$, then $y>0$, so $\coth y = \sqrt{1+x^2}$. $\frac{dy}{dx} = \frac{1}{-x \sqrt{1+x^2}}$. If $x<0$, then $y<0$, so $\coth y = -\sqrt{1+x^2}$. $\frac{dy}{dx} = \frac{1}{-x (-\sqrt{1+x^2})} = \frac{1}{x \sqrt{1+x^2}}$.

This can be combined using the absolute value:

$\frac{d}{dx}(\text{arcsch } x) = -\frac{1}{|x| \sqrt{1 + x^2}}$. (Note: Domain $x \ne 0$).

Integrals of Inverse Hyperbolic Functions

The integrals of inverse hyperbolic functions can be found by reversing the differentiation process or by using the logarithmic forms.

Integral of arsinh $x$: $\int \text{arsinh } x \, dx = x \text{ arsinh } x - \sqrt{x^2 + 1} + C$ This can be derived using integration by parts with $u = \text{arsinh } x$ and $dv = dx$.

Integral of arccosh $x$: $\int \text{arccosh } x \, dx = x \text{ arccosh } x - \sqrt{x^2 - 1} + C$ This can be derived using integration by parts with $u = \text{arccosh } x$ and $dv = dx$.

Integral of artanh $x$: $\int \text{artanh } x \, dx = x \text{ artanh } x + \frac{1}{2} \ln|1 - x^2| + C$ This can be derived using integration by parts with $u = \text{artanh } x$ and $dv = dx$. Alternatively, use the logarithmic form $\text{artanh } x = \frac{1}{2} \ln\left(\frac{1+x}{1-x}\right)$ and integrate.

Integral of arccoth $x$: $\int \text{arccoth } x \, dx = x \text{ arccoth } x - \frac{1}{2} \ln|x^2 - 1| + C$ This can be derived using integration by parts with $u = \text{arccoth } x$ and $dv = dx$.

Integral of arsech $x$: $\int \text{arsech } x \, dx = x \text{ arsech } x - \text{arcsin } x + C$ (for $0 < x \le 1$) This can be derived using integration by parts.

Integral of arcsch $x$: $\int \text{arcsch } x \, dx = x \text{ arcsch } x + \ln|x + \sqrt{x^2 + 1}| + C$ This can be derived using integration by parts. Note that $\ln|x + \sqrt{x^2+1}| = \text{arsinh}(x)$.

Applications of Inverse Hyperbolic Functions

Inverse hyperbolic functions appear in various mathematical and scientific contexts:

• Calculus: Used in integration, especially for integrals involving terms like $\sqrt{x^2 \pm a^2}$ or $\frac{1}{a^2 \pm x^2}$.
• Differential Equations: Solutions to certain second-order linear differential equations often involve inverse hyperbolic functions.
• Physics:
  • Relativity: In special relativity, the velocity addition formula can be simplified using the Gudermannian function, which relates hyperbolic and trigonometric functions.
  • Fluid Dynamics: Used in modeling fluid flow and wave phenomena.
  • Electromagnetism: Appear in solutions to problems involving magnetic fields.
• Engineering: Applications in areas like electrical engineering (e.g., transmission line theory) and mechanical engineering.
• Statistics: The logit transformation, $y = \ln(\frac{p}{1-p})$, which maps probabilities to the real line, is related to artanh $x$. Specifically, $\text{artanh}(x) = \frac{1}{2} \ln(\frac{1+x}{1-x})$, and if $x=p$, this is related to the logit of $p$.

Example Problems

Example 1: Evaluate arsinh(2)

Using the logarithmic form: $\text{arsinh}(2) = \ln(2 + \sqrt{2^2 + 1}) = \ln(2 + \sqrt{4 + 1}) = \ln(2 + \sqrt{5})$.

Example 2: Solve for x: $\cosh(2x) = 3$

Using the definition of arccosh: $2x = \text{arccosh}(3)$ $2x = \ln(3 + \sqrt{3^2 - 1})$ $2x = \ln(3 + \sqrt{9 - 1})$ $2x = \ln(3 + \sqrt{8})$ $2x = \ln(3 + 2\sqrt{2})$ $x = \frac{1}{2} \ln(3 + 2\sqrt{2})$

Example 3: Find the derivative of $f(x) = x \text{artanh}(x^2)$

We need to use the product rule and the chain rule. Let $u = x$ and $v = \text{artanh}(x^2)$. $\frac{du}{dx} = 1$. For $\frac{dv}{dx}$, let $w = x^2$. Then $\text{artanh}(w)$. $\frac{d}{dw}(\text{artanh } w) = \frac{1}{1 - w^2}$. $\frac{dw}{dx} = 2x$. So, $\frac{dv}{dx} = \frac{d}{dw}(\text{artanh } w) \cdot \frac{dw}{dx} = \frac{1}{1 - (x^2)^2} \cdot (2x) = \frac{2x}{1 - x^4}$.

Using the product rule $f'(x) = u \frac{dv}{dx} + v \frac{du}{dx}$: $f'(x) = x \left(\frac{2x}{1 - x^4}\right) + \text{artanh}(x^2) \cdot 1$ $f'(x) = \frac{2x^2}{1 - x^4} + \text{artanh}(x^2)$.

Example 4: Evaluate the integral $\int_0^1 \frac{1}{1-x^2} dx$

This integral is improper because the integrand $\frac{1}{1-x^2}$ has a vertical asymptote at $x=1$. We know that $\frac{d}{dx}(\text{artanh } x) = \frac{1}{1-x^2}$ for $|x|<1$. So, $\int \frac{1}{1-x^2} dx = \text{artanh } x + C$.

Evaluating the definite integral: $\int_0^1 \frac{1}{1-x^2} dx = \lim_{b \to 1^-} \int_0^b \frac{1}{1-x^2} dx$ $= \lim_{b \to 1^-} [\text{artanh } x]_0^b$ $= \lim_{b \to 1^-} (\text{artanh } b - \text{artanh } 0)$ $= \lim_{b \to 1^-} \text{artanh } b - 0$ As $b \to 1^-$, $\text{artanh } b \to \infty$ because the range of artanh is $(-\infty, \infty)$ and it approaches infinity as its argument approaches 1 from below. So, the integral diverges to infinity.

Alternatively, using the logarithmic form: $\int_0^1 \frac{1}{1-x^2} dx = \lim_{b \to 1^-} \left[\frac{1}{2} \ln\left|\frac{1+x}{1-x}\right|\right]_0^b$ $= \lim_{b \to 1^-} \left(\frac{1}{2} \ln\left|\frac{1+b}{1-b}\right| - \frac{1}{2} \ln\left|\frac{1+0}{1-0}\right|\right)$ $= \lim_{b \to 1^-} \left(\frac{1}{2} \ln\left(\frac{1+b}{1-b}\right) - \frac{1}{2} \ln(1)\right)$ $= \lim_{b \to 1^-} \frac{1}{2} \ln\left(\frac{1+b}{1-b}\right)$ As $b \to 1^-$, $1+b \to 2$, and $1-b \to 0^+$. So $\frac{1+b}{1-b} \to \infty$. Therefore, $\frac{1}{2} \ln\left(\frac{1+b}{1-b}\right) \to \infty$. The integral diverges.