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Derivatives of the Hyperbolic Functions

April 2nd, 2009 |

Author: Merrin

In this section, we will calculate the derivatives of the hyperbolic functions. The hyperbolic functions are defined analogously to the trigonometric functions.

$\sinh x = \cfrac{1}{2}\left(e^x-e^{-x}\right) \qquad \cosh x = \cfrac{1}{2}\left(e^x+e^{-x}\right)$

$\tanh x = \cfrac{\sinh x}{\cosh x} \qquad \coth x = \cfrac{\cosh x}{\sinh x}$

$\text{sech x}\,\, = \cfrac{1}{\cosh x} \qquad \text{csch}\,\, x = \cfrac{1}{\sinh x}$

We worked out “all” the hyperbolic identities analogous to the trigonometric identities using Osborne’s rule. Check back if you are unfamiliar with this. We will use several of these hyperbolic identities to simplify the calculations of our derivatives of the hyperbolic functions. Analagous to the derivatives of the trigonometric functions the derivatives of the rest of the hyperbolic functions can be found from the derivatives of the hyperbolic cosine and hyperbolic sine. We know how to differentiate exponential functions so finding these derivatives is not a problem.

$\cfrac{d}{dx} \sinh x=\cfrac{1}{2} \cfrac{d}{dx} e^{x} -\cfrac{1}{2} \cfrac{d}{dx} e^{-x} =\cfrac{1}{2} (e^{x} +e^{-x} )=\cosh x$

$\cfrac{d}{dx} \cosh x=\cfrac{1}{2} \cfrac{d}{dx} e^{x} +\cfrac{1}{2} \cfrac{d}{dx} e^{-x} =\cfrac{1}{2} (e^{x} -e^{-x} )=\sinh x$

With these two derivatives we can start our work finding the other four derivatives of hyperbolic functions. We can use the quotient rule for the hyperbolic tangent and the hyperbolic cotangent. We can use the reciprocal rule for the hyperbolic secant and the hyperbolic cosecant. The derivative of the hyperbolic tangent is

$\cfrac{d}{dx} \tanh x = \cfrac{d}{dx} \cfrac{\sinh x}{\cosh x} = \cfrac{(\sinh x)'(\cosh x)-(\sinh x)(\cosh x)'}{\cosh ^{2} x}$