# Reciprocal derivative

The reciprocal difference quotient is

$\frac{\Delta&space;x}{\Delta&space;f(x)}=\frac{h}{f(x+h)-f(x)}$

The reciprocal derivative of f(x), symbolized by a reversed prime, is the limit of the reciprocal difference quotient as h approaches zero:

$\grave{}f(x)=\lim_{h\rightarrow&space;0}\frac{\Delta&space;x}{\Delta&space;f(x)}=&space;\lim_{h\rightarrow&space;0}\frac{h}{f(x+h)-f(x)}$

An alternate limit form of the reciprocal derivative definition of f(a) is

$\grave{}f(a)=\lim_{x\rightarrow&space;a}=&space;\frac{x-a}{f(x)-f(a)}$

The reciprocal derivative of a linear function, f(x) = ax + b, is

$\grave{}f(x)=\lim_{h\rightarrow&space;0}\frac{h}{f(x+h)-f(x)}=&space;\lim_{h\rightarrow&space;0}\frac{h}{a(x+h)+b-ax-b}=&space;\lim_{h\rightarrow&space;0}\frac{h}{ah}=\frac{1}{a}$

The reciprocal derivative of a power function, f(x) = xn, is

$\grave{}f(a)=\lim_{x\rightarrow&space;a}\frac{x-a}{f(x)-f(a)}=&space;\lim_{x\rightarrow&space;a}\frac{x-a}{x^{n}-a^{n}}=&space;\frac{1}{na^{n-1}}=\frac{1}{n}a^{1-n}$

In general the reciprocal derivative is the reciprocal of the classical derivative:

$\grave{}f(x)=\lim_{h\rightarrow&space;0}\frac{\Delta&space;x}{\Delta&space;f(x)}=&space;\left&space;(&space;\lim_{h\rightarrow&space;0}&space;\frac{\Delta&space;f(x)}{\Delta&space;x}&space;\right&space;)^{-1}&space;=\frac{1}{f'(x)}=\left&space;(&space;f^{-1}(x)&space;\right&space;)'$

by the inverse function theorem.

The reciprocal derivative of a constant times a function is the reciprocal constant times the reciprocal derivative of the function:

$\grave{}\left&space;(&space;cf(x)&space;\right&space;)=&space;\lim_{h\rightarrow&space;0}\frac{h}{cf(x+h)-cf(x)}=&space;\frac{1}{c}\lim_{h\rightarrow&space;0}\frac{h}{f(x+h)-f(x)}=&space;\frac{\grave{}f(x)}{c}$

The reciprocal derivative of the sum of two functions is the harmonic sum of the two reciprocal derivatives:

$\grave{}\left&space;(&space;f(x)+g(x)&space;\right&space;)=&space;\lim_{h\rightarrow&space;0}\frac{h}{f(x+h)+g(x+h)-\left&space;(&space;f(x)+g(x)&space;\right&space;)}$

$=\lim_{h\rightarrow&space;0}\frac{h}{f(x+h)-f(x)+g(x+h)-g(x)}=&space;\grave{}f(x)&space;\oplus\,&space;\grave{}g(x)$

in which the circled plus represents harmonic addition. One may easily prove the following for differentiable functions f and g:

$\grave{}f&space;+&space;\grave{}g=\grave{}(f&space;\oplus&space;g)^{-1}$

$\grave{}(f&space;\oplus&space;g)=&space;(\grave{}f&space;+&space;\grave{}g)^{-1}$

$\grave{}(f&space;+&space;g)=&space;\grave{}f&space;\oplus&space;\,&space;\grave{}g$

$(\,&space;\grave{}f)^{-1}&space;\oplus&space;\,&space;(\,&space;\grave{}g)^{-1}&space;=&space;(\,&space;\grave{}(f+g))^{-1}$

$\grave{}(fg)=\frac{\grave{}f}{g}&space;\oplus&space;\,&space;\frac{\grave{}g}{f}$

$\grave{}(f/g)&space;=\left&space;(\,&space;\frac{\grave{}f}{g}&space;\ominus&space;\frac{\grave{}g}{f}&space;\right&space;)v^{2}$

If the functions f and g and inverses of one another, then by the inverse-function theorem:

$f\,&space;\acute{}&space;\:&space;g\acute{}\,&space;=\,&space;\grave{}f&space;\;&space;\grave{}g=1$

Why would someone use the reciprocal derivative instead of the derivative? The reason is that they might rather work with a function than with its inverse. In the neighborhood of a non-zero derivative according to the inverse-function theorem, the reciprocal of a derivative equals the derivative of the inverse function (assuming non-zero derivatives). So one can get at the inverse function through the reciprocal derivative.

This applies to instantaneous rates of change such as speed, dx(t)/dt. These rates are appropriate if time is the independent variable, but if distance is the independent variable, then the inverse rate, dt(x)/dx, would be appropriate. But we are accustomed to working with speed rather than its inverse, pace. In that case we can use the time speed, (dx(t)/dt)-1 = dt(x)/dx, which is the reciprocal derivative of x(t) but the derivative of t(x).

Reference: “Properties of Reciprocal Derivatives” M. M. Pahirya and R. A. Katsala, Ukrainian Mathematical Journal, Vol. 62, No. 5, 2010, pages 816-823.