Reciprocal derivative

The reciprocal difference quotient is

\frac{\Delta x}{\Delta 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 0}\frac{\Delta x}{\Delta f(x)}= \lim_{h\rightarrow 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 a}= \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 0}\frac{h}{f(x+h)-f(x)}= \lim_{h\rightarrow 0}\frac{h}{a(x+h)+b-ax-b}= \lim_{h\rightarrow 0}\frac{h}{ah}=\frac{1}{a}

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

\grave{}f(a)=\lim_{x\rightarrow a}\frac{x-a}{f(x)-f(a)}= \lim_{x\rightarrow a}\frac{x-a}{x^{n}-a^{n}}= \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 0}\frac{\Delta x}{\Delta f(x)}= \left ( \lim_{h\rightarrow 0} \frac{\Delta f(x)}{\Delta x} \right )^{-1} =\frac{1}{f'(x)}=\left ( f^{-1}(x) \right )'

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 ( cf(x) \right )= \lim_{h\rightarrow 0}\frac{h}{cf(x+h)-cf(x)}= \frac{1}{c}\lim_{h\rightarrow 0}\frac{h}{f(x+h)-f(x)}= \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 ( f(x)+g(x) \right )= \lim_{h\rightarrow 0}\frac{h}{f(x+h)+g(x+h)-\left ( f(x)+g(x) \right )}

=\lim_{h\rightarrow 0}\frac{h}{f(x+h)-f(x)+g(x+h)-g(x)}= \grave{}f(x) \oplus\, \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 + \grave{}g=\grave{}(f \oplus g)^{-1}

\grave{}(f \oplus g)= (\grave{}f + \grave{}g)^{-1}

\grave{}(f + g)= \grave{}f \oplus \, \grave{}g

(\, \grave{}f)^{-1} \oplus \, (\, \grave{}g)^{-1} = (\, \grave{}(f+g))^{-1}

\grave{}(fg)=\frac{\grave{}f}{g} \oplus \, \frac{\grave{}g}{f}

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

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

f\, \acute{} \: g\acute{}\, =\, \grave{}f \; \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.