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Higher Order Derivatives of Exponential Containing Functions

April 2nd, 2009 |

Author: Merrin

The exponential function enjoys special status in calculus because it is its own derivative. It comes as no surprise that the elementary functions expressed in terms of exponentials have closed forms for their nth derivatives. Some elementary functions which contain exponentials are

$e^{x} , b^{x} , \sin x,\cos x,\sinh x, \cosh x$

The derivative of an exponential with arbitrary base can be written as a natural exponential.

$a^{x} =e^{(\ln a)x} =e^{\nu x}$

The derivatives of such a natural exponential with a given rate can be written as

$\cfrac{d^{n} }{dx^{n} } e^{\nu x} =D^{n} e^{\nu x} =\nu ^{n} e^{\nu x}$

The same goes for complex numbers.

$D^{n} e^{ix} =i^{n} e^{ix}$

The derivatives of sine and consine alternate between each other and change signs with periods of four. This can be understood in the following formula for the nth derivative.

${D^{n} \cos x=\frac{1}{2} D^{n} (e^{ix} +e^{-ix} )=\frac{1}{2} (i^{n} e^{ix} +(-i)^{n} e^{-ix} )}$

${=\frac{1}{2} (e^{ix+in\pi /2} +e^{-ix-in\pi /2} )=\cos (x+n\pi/2 )}$

${D^{n} \cos x=\cos (x+n\pi/2 )}$

A similar formula for sine can be derived.

${D^{n} \sin x=\cfrac{1}{2i} D^{n} (e^{ix} -e^{-ix} )=\cfrac{1}{2i} (i^{n} e^{ix} -(-i)^{n} e^{-ix} )}$

${=\cfrac{1}{2i} (e^{ix+in\pi /2} -e^{-ix-in\pi /2} )=\sin (x+n\pi/2 )}$

${D^{n} \sin x = \sin (x+n\pi/2 )}$

Each derivative of these functions is equivalent to advancing the phase by a quarter of a circle. After four derivatives the function returns to its original value.

The hyperbolic sine and hyperbolic cosine also have equations that represent the derivatives in terms of the phase except it is complex.

${D^{n} \cosh x=\cfrac{1}{2} (e^{x} +(-1)^{n} e^{-x} )=\cfrac{1}{2} (-1)^{n/2} ((-1)^{-n/2} e^{x} +(-1)^{n/2} e^{-x} )}$

${=\cfrac{1}{2} (i)^{n} (e^{x-in\pi /2} +e^{-x+in\pi /2} )=i^{n} \cosh (x-in\pi /2)}$

${D^{n} \cosh x =i^{n} \cosh (x-in\pi /2)}$

The nth derivative of the hyperbolic sine can be written several ways, but we choose a form that closely matches the previous formula.

${D^{n} \sinh x=\cfrac{1}{2} (e^{x} -(-1)^{n} e^{-x} )=\cfrac{1}{2} (-1)^{n/2} ((-1)^{-n/2} e^{x} -(-1)^{n/2} e^{-x} )}$

$=\cfrac{1}{2} (i)^{n} (e^{x-i\pi n/2} -e^{-x+in\pi /2} )=i^{n} \sinh (x-in\pi /2)$

$D^{n} \sinh x = i^{n} \sinh (x-in\pi /2)$

The derivatives of the exponential functions in the section form an aesthetically pleasing set of functions which can be differentiated to nth order.

In fractional calculus, a possible starting point is that the derivative of an exponential function to any order should be an exponential function. In this case, all the results of this section can be considered as fractional derivatives where the order can be extended to any positive value. We will find later that this is true only in a particular limit of fractional differentiation, but it is still a logical starting point.

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