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Derivatives of the Inverse Hyperbolic Functions

April 8th, 2009 |

Author: Merrin

The derivatives of the inverse hyperbolic functions can be derived similarly to the derivatives of the inverse trigonometric functions. Alternatively the inverse hyperbolic functions have function definitions which can just be differentiated. These definitions can be found by deriving the inverses.

$\begin{array}{l} {\text{arcsinh}\, x=\ln (x+\sqrt{x^{2} +1} )} \\\\ {\text{arccosh}\, x=\ln (x+\sqrt{x^{2} -1} )} \\\\ {\text{arctanh}\, x=\cfrac{1}{2} \ln \cfrac{1+x}{1-x} } \\\\ {\text{arccoth}\, x=\cfrac{1}{2} \ln \cfrac{x+1}{x-1} } \\\\ {\text{arcsech}\, x=\ln \left(\cfrac{1}{x} +\sqrt{\cfrac{1}{x^{2} } -1} \right)} \\\\ {\text{arccsch}\, x=\ln \left(\cfrac{1}{x} +\sqrt{1+\cfrac{1}{x^{2} } } \right)\, \, } \end{array}$

The first four functions are easily differentiated in one line. The last two functions have different forms depending on whether x is positive or negative. The roots give an absolute value sign. We can use the properties of their graphs to see the domains of these functions and their ranges. The process is similar to what we did for trigonometric functions.

Example 1. Find the derivative of

$\text{arcsinh}\, x$

Solution 1. We differentiate the functional form of

$\text{arcsinh}\, x$

$\cfrac{d}{dx} \text{arcsinh}\, x=\cfrac{d}{dx}\ln(x+\sqrt{1+x^2})=\cfrac{1}{x+\sqrt{x^{2} +1} } \left(1+\cfrac{x}{\sqrt{x^{2} +1} } \right)$

$\cfrac{d}{dx} \text{arcsinh}\, x=\cfrac{1}{\sqrt{x^{2} +1} }$

Example 2. Find the derivative of

$\text{arccosh}\, x$

Solution 2. We differentiate the functional form of

$\text{arccosh} x$

$\cfrac{d}{dx} \text{arccosh}\, x=\cfrac{d}{dx}\ln(x+\sqrt{x^2-1})=\cfrac{1}{x+\sqrt{x^{2} -1} } \left(1+\cfrac{x}{\sqrt{x^{2} -1} } \right)$

$\cfrac{d}{dx} \text{arccosh}\, x=\cfrac{1}{\sqrt{x^{2} -1} }$

Example 3. Find the derivative of

$\text{arctanh}\, x$

Solution 3. We differentiate the functional form of

$\text{arctanh}\, x$

$\cfrac{d}{dx} \text{arctanh}\, x=\cfrac{d}{dx}\left(\cfrac{1}{2} \ln \cfrac{1+x}{1-x}\right)=\cfrac{1}{2} \left(\cfrac{1}{1+x} +\cfrac{1}{1-x} \right)=\cfrac{1}{1-x^{2} }$

Example 4. Find the derivative of

$\text{arccoth}\, x$

Solution 4. The derivative of the hyperbolic arccotangent is the same as for the hyperbolic arctangent, however their domains are different.

$\cfrac{d}{dx} \text{arccoth}\, x=\cfrac{d}{dx}\left(\cfrac{1}{2} \ln \cfrac{1+x}{x-1}\right)=\cfrac{1}{2} \left(\cfrac{1}{1+x} -\cfrac{1}{x-1} \right)$

$\cfrac{d}{dx} \text{arccoth}\, x=-\cfrac{1}{x^{2} -1} =\cfrac{1}{1-x^{2} }$

Example 5. Find the derivative of

$\text{arcsech}\, x$

Solution 5. With implicit differentiation we can zero in on the derivative fairly quickly. If you try to differentiate the functional form, the algebra is longer. We use the slope of the graph of the function to fix the sign of the derivative. Notice that real values of the hyperbolic arcsecant only occur for positive x.

$\begin{array}{l} {y=\text{arcsech}\, x} \\ {x=\text{sech}\, y} \\ {1=-\text{sech}\, y\text{tanh}\, y\cfrac{dy}{dx} } \\\\ {1=\pm \text{sech}\, y\sqrt{1-\text{sech}\,^{2} y} \cfrac{dy}{dx} } \\ {\cfrac{dy}{dx} =-\cfrac{1}{x\sqrt{1-x^{2} } } } \end{array}$

Example 6. Find the derivative of

$\text{arccsch}\, x$

Solution 6. Also the algebra for finding the derivative is simpler ifwe use implicit differentiation. If you try to differentiate the functionalform you have to consider two cases when x > 0 and when x < 0 because there is an absolute value implied by the square root. Using the standard methodof implicit differentiation we fix the sign of the derivative from the graph.

$\begin{array}{l}{y=\text{arccsch}\, x}\\ {x=\text{csch}\, y} \\{1=-\text{csch}\, y \,\text{coth}\, y\cfrac{dy}{dx} }\\{1=\pm \text{csch}\, y\sqrt{1+\text{csch}\,^{2} y} \cfrac{dy}{dx} } \\{\cfrac{dy}{dx} =-\cfrac{1}{\left|x\right|\sqrt{1+x^{2} } } }\\\end{array}$