Derivative of Reciprocal

Theorem

Let $f: \R_{\ne 0} \to \R$ be the reciprocal function defined as:

$\map f x = \dfrac 1 x$

Then its derivative is given by:

$\map {f'} x = -\dfrac 1 {x^2}$

We have:

$\ds \dfrac 1 x$

$=$

$\ds x^{-1}$

$\ds \leadsto \ \ $

$\ds \map {\dfrac \d {\d x} } {\dfrac 1 x}$

$=$

$\ds \paren {-1} x^{-2}$

$\ds $

$=$

$\ds -\dfrac 1 {x^2}$

$\blacksquare$

1989: Ephraim J. Borowski and Jonathan M. Borwein: Dictionary of Mathematics ... (previous) ... (next): Appendix $2$: Table of derivatives and integrals of common functions